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✓ 9 results · all verified · 4 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 5 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Tensor Coherence and Algebraic Descent

1 · Prerequisites

2 · Summary

The page fixes the tensor conventions shared by the Hopf and Hecke branches: $k$ is a field, $\otimes$ means $\otimes_{k}$ over $k$ -vector spaces, tensor powers are left-associated with the empty tensor $k$ , and parentheses in iterated tensor powers may be dropped only after the coherence lemma below has been proved. It then supplies the coherence, duality and descent facts that the construction pages of these branches consume.

Coherence is proved concretely rather than invoked from a general monoidal theorem: the associator, the symmetry and the unit isomorphisms satisfy naturality, the pentagon, the unit triangle and both symmetry hexagons, each verified on elementary tensors and extended to all linear maps by the spanning property of the tensor product. Finite tensor duality identifies $V^{*} \otimes W^{*}$ with $(V \otimes W)^{*}$ through the product dual basis and exhibits the basis-independent coevaluation element $\sum_{i} v_{i} \otimes v_{i}^{*}$ together with both zigzag identities; no surjectivity is asserted in infinite dimension, and the companion page's counterexample shows that the finite-dimensional hypothesis is necessary.

The Choice assumptions are recorded explicitly. The injection lemma and the kernel computation for a tensor product of quotient maps assume the Axiom of Choice through Every linear subspace $U$ of a vector space $V$ has a complement: a linear subspace $W$ with $V = U \oplus W$ , and coefficient separation for a finite independent family inherits that assumption in infinite ambient dimension while remaining choice-free in finite dimension, where the independent list is extended to a basis.

The descent half constructs the free associative $R$ -algebra $R ⟨ S ⟩$ on the words in $S$ with concatenation as product, proves its universal property and the two-sided-ideal description of a generated relation ideal, and identifies the quotient as the presented algebra. Base change along a commutative ring homomorphism is proved by explicit mutually inverse generator maps, with the image ideal of a relation ideal carrying the corresponding quotient presentation and no flatness hypothesis; free bases transport along any commutative specialization. Polynomial and Laurent rings are built from finitely supported monomials, their universal properties by substitution, and their fraction fields over domains from numerator-denominator pairs. Finite matrix and module preliminaries supply the right-inverse determinant argument, invariance of finite matrix rank under field extension, finite composition series, the splitting of a submodule of a finite direct sum of simple modules without arbitrary Choice, and the vanishing trace of a nilpotent endomorphism. The regular-module detection principle closes the page: evaluation at $1$ and at $1^{\otimes n}$ detects equality of algebra elements, multilinear identities are checked on pure tensors, and a quotient identity requires the descent that the recorded warning makes explicit.

Prerequisite pages: tensor-products-of-modules, modules-and-module-homomorphisms, ideals-and-quotient-rings, dual-spaces-bilinear-forms-and-inertia, linear-independence-bases-and-dimension, linear-maps-rank-nullity-and-quotient-spaces, chain-conditions-and-semisimple-modules, relations-functions-and-quotients. The companion tensor-coherence-and-algebraic-descent-examples develops the calculations and failures needed to test these constructions.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

Scalars, tensor powers, the empty tensor, opposite algebras and finite sums

Definition

Fix the following conventions, used on this page and by the later Hopf and Hecke pages.

$k$ is a field (Field) and every vector space is a $k$ -vector space (Vector space over a field); $\otimes$ means $\otimes_{k}$ .
$V \otimes W$ is the tensor product of The tensor product $M \otimes_{R} N$ from the additive group underlying the free $Z$ -module on $M \times N$ , elementary tensors, and finite tensor sums, with the $k$ -vector-space structure of Over a commutative ring, $M \otimes_{R} N$ is an $R$ -module with $r (m \otimes n) = (r m) \otimes n = m \otimes (r n)$ and unit and universal property as in Universal property of the tensor product for balanced maps into abelian groups. Every element of $V \otimes W$ is a finite sum of elementary tensors $v \otimes w$ .
Tensor powers are left-associated: $V^{\otimes 0} : = k$ , $V^{\otimes 1} : = V$ and $V^{\otimes n} : = (V^{\otimes (n - 1)}) \otimes V$ for $n \geq 2$ . The empty tensor $k$ is identified with the tensor unit through the isomorphisms of The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ .
For an algebra $A$ over a commutative ring (Algebras over a commutative ring, central structure maps, and algebra homomorphisms) the opposite algebra $A^{o p}$ has the same underlying set and structure map and product $a \cdot b : = b a$ (The opposite ring $R^{o p}$ ).
Finite sums $\sum_{i \in I} x_{i}$ follow A finite sum in a commutative monoid indexed by an arbitrary finite set; an empty sum is $0$ .

Parentheses in iterated tensor products may be dropped only after Associator naturality, pentagon, unit triangle and symmetry hexagons on elementary tensors ↗ has been proved.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

Associator naturality, pentagon, unit triangle and symmetry hexagons on elementary tensors

Statement

Let $k$ be a field and let $L, M, N, X$ be $k$ -vector spaces. Write $α_{A, B, C} : (A \otimes B) \otimes C \to A \otimes (B \otimes C)$ for the associators of Symmetry and associativity isomorphisms for tensor products over a commutative ring, $σ_{A, B}$ for its symmetries, and $λ, ρ$ for the unit isomorphisms of The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ .

Naturality. $α$ and $σ$ are natural in all variables: for linear maps the usual squares commute; the unit isomorphisms are natural as well.
Pentagon. $({i d}_{L} \otimes α_{M, N, X}) \circ α_{L, M \otimes N, X} \circ (α_{L, M, N} \otimes {i d}_{X}) = α_{L, M, N \otimes X} \circ α_{L \otimes M, N, X}$ as maps $((L \otimes M) \otimes N) \otimes X \to L \otimes (M \otimes (N \otimes X))$ .
Unit triangle. $({i d}_{M} \otimes λ_{N}) \circ α_{M, k, N} = ρ_{M} \otimes {i d}_{N}$ as maps $(M \otimes k) \otimes N \to M \otimes N$ .
First symmetry hexagon. $α_{L, N, M} \circ (σ_{N, L} \otimes {i d}_{M}) \circ α_{N, L, M}^{- 1} \circ σ_{L \otimes M, N} = ({i d}_{L} \otimes σ_{M, N}) \circ α_{L, M, N}$ as maps $((L \otimes M) \otimes N) \to L \otimes (N \otimes M)$ .
Second symmetry hexagon. $α_{N, L, M}^{- 1} \circ σ_{L \otimes M, N} \circ α_{L, M, N}^{- 1} = (σ_{L, N} \otimes {i d}_{M}) \circ α_{L, N, M}^{- 1} \circ ({i d}_{L} \otimes σ_{M, N})$ as maps $L \otimes (M \otimes N) \to (N \otimes L) \otimes M$ .

All five identities are equalities of $k$ -linear maps between iterated tensor products.

Facts & Assumptions

Given: A field $k$ , vector spaces $L, M, N, X$ and linear maps between vector spaces as named in the steps.

[F1]

The conventions: $V \otimes W$ is the tensor product over $k$ with unit and universal property, every element is a finite sum of elementary tensors, and tensor powers are left-associated with $k$ as the empty tensor (Scalars, tensor powers, the empty tensor, opposite algebras and finite sums).

[F2]

The associator and symmetry are isomorphisms acting on elementary tensors by $α_{L, M, N} ((l \otimes m) \otimes n) = l \otimes (m \otimes n)$ and $σ_{M, N} (m \otimes n) = n \otimes m$ , with $σ_{N, M} σ_{M, N} = i d$ (Symmetry and associativity isomorphisms for tensor products over a commutative ring).

[F3]

The unit isomorphisms act by $λ_{N} (r \otimes n) = r n$ and $ρ_{M} (m \otimes r) = m r$ (The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ ).

[F4]

Functoriality: $(f \otimes g) (m \otimes n) = f (m) \otimes g (n)$ defines a linear map, ${i d}_{M} \otimes {i d}_{N} = {i d}_{M \otimes N}$ , and $(f^{'} \circ f) \otimes (g^{'} \circ g) = (f^{'} \otimes g^{'}) \circ (f \otimes g)$ (Module homomorphisms induce tensor-product homomorphisms functorially).

[F5]

Every element of a tensor product is a finite sum of elementary tensors, and the defining relations give $(c m) \otimes n = m \otimes (c n) = c (m \otimes n)$ for $c \in k$ (The tensor product $M \otimes_{R} N$ from the additive group underlying the free $Z$ -module on $M \times N$ , elementary tensors, and finite tensor sums, Scalars, tensor powers, the empty tensor, opposite algebras and finite sums).

Proof

technique · direct

1.1givenF2F4F5algebra

Naturality of $α$ and $σ$ : for linear maps $f : L \to L^{'}$ , $g : M \to M^{'}$ , $h : N \to N^{'}$ and an elementary tensor $(l \otimes m) \otimes n$ one has $α_{L^{'}, M^{'}, N^{'}} (((f \otimes g) \otimes h) ((l \otimes m) \otimes n)) = α_{L^{'}, M^{'}, N^{'}} ((f (l) \otimes g (m)) \otimes h (n)) = f (l) \otimes (g (m) \otimes h (n))$ and $(f \otimes (g \otimes h)) (α_{L, M, N} ((l \otimes m) \otimes n)) = (f \otimes (g \otimes h)) (l \otimes (m \otimes n)) = f (l) \otimes (g (m) \otimes h (n))$ , by [F2] and [F4]; likewise $σ_{M^{'}, N^{'}} ((g \otimes h) (m \otimes n)) = h (n) \otimes g (m) = ((h \otimes g) \circ σ_{M, N}) (m \otimes n)$ . Both sides of each square are $k$ -linear and the elementary tensors span by [F5], so the squares commute on their whole domains.

1.2givenF3F4F5algebra

Naturality of the unit isomorphisms: for linear $f : N \to N^{'}$ and $r \otimes n \in k \otimes N$ one has $λ_{N^{'}} (({i d}_{k} \otimes f) (r \otimes n)) = λ_{N^{'}} (r \otimes f (n)) = r f (n) = f (r n) = f (λ_{N} (r \otimes n))$ by [F3] and [F4], and for linear $g : M \to M^{'}$ likewise $ρ_{M^{'}} ((g \otimes {i d}_{k}) (m \otimes r)) = g (m) r = g (m r) = g (ρ_{M} (m \otimes r))$ ; the elementary tensors span, so both naturality squares commute.

1.3givenF2F5algebra

Pentagon: on an elementary tensor $((l \otimes m) \otimes n) \otimes x$ of $((L \otimes M) \otimes N) \otimes X$ the left composite sends it by [F2] to $(α_{L, M, N} \otimes {i d}_{X}) (((l \otimes m) \otimes n) \otimes x) = (l \otimes (m \otimes n)) \otimes x$ , then to $l \otimes ((m \otimes n) \otimes x)$ , then to $l \otimes (m \otimes (n \otimes x))$ ; the right composite sends it to $(l \otimes m) \otimes (n \otimes x)$ and then to $l \otimes (m \otimes (n \otimes x))$ . Both sides are $k$ -linear maps whose domain is spanned by such elementary tensors [F5], so the two composites agree everywhere.

1.4givenF2F3F5algebra

Unit triangle: for an elementary tensor $(m \otimes c) \otimes n$ of $(M \otimes k) \otimes N$ the left side gives $({i d}_{M} \otimes λ_{N}) (m \otimes (c \otimes n)) = m \otimes (c n)$ by [F2] and [F3], while the right side gives $(ρ_{M} \otimes {i d}_{N}) ((m \otimes c) \otimes n) = (m c) \otimes n$ ; these are equal because $(m c) \otimes n = m \otimes (c n)$ by the balancing relations of [F5]. Both sides are linear on the span of the elementary tensors, so the identity holds.

1.5givenF2F5algebra

First symmetry hexagon: on an elementary tensor $(l \otimes m) \otimes n$ of $(L \otimes M) \otimes N$ the left composite gives successively $n \otimes (l \otimes m)$ (symmetry $σ_{L \otimes M, N}$ ), $(n \otimes l) \otimes m$ (inverse associator), $(l \otimes n) \otimes m$ (symmetry in the first factor), $l \otimes (n \otimes m)$ (associator), while the right composite gives $l \otimes (m \otimes n)$ and then $l \otimes (n \otimes m)$ by the symmetry in the second factor; the two agree on the spanning elementary tensors, hence everywhere.

1.6givenF2F5algebra

Second symmetry hexagon: on an elementary tensor $l \otimes (m \otimes n)$ of $L \otimes (M \otimes N)$ the left composite gives $(l \otimes m) \otimes n$ , then $n \otimes (l \otimes m)$ , then $(n \otimes l) \otimes m$ , while the right composite gives $l \otimes (n \otimes m)$ , then $(l \otimes n) \otimes m$ , then $(n \otimes l) \otimes m$ ; agreement on the spanning elementary tensors gives the identity everywhere.

2.1step 1.1step 1.2step 1.3step 1.4step 1.5step 1.6F1F5∎

Steps 1.1–1.6 verify all five identities on elementary tensors, and each identity is between $k$ -linear maps whose domains are the iterated tensor products of [F1] spanned by elementary tensors [F5]; a linear map is determined by its values on a spanning set, so each identity holds on its whole domain, and no general monoidal coherence theorem was invoked.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

Tensoring injections and the kernel of a tensor product of quotient maps over a field

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $k$ be a field, let $f : V \to V^{'}$ be an injective linear map and let $W$ be a $k$ -vector space. Then $f \otimes {i d}_{W} : V \otimes W \to V^{'} \otimes W$ is injective. Moreover, if $U \subseteq V$ and $Z \subseteq W$ are subspaces and $p : V \to V / U$ , $q : W \to W / Z$ are the quotient maps, then, viewing $U \otimes W$ and $V \otimes Z$ as subspaces of $V \otimes W$ through the injections of the first part and the inclusions $U \subseteq V$ , $Z \subseteq W$ , $\ker (p \otimes q) = U \otimes W + V \otimes Z .$

Facts & Assumptions

Given: A field $k$ , an injective linear map $f : V \to V^{'}$ , a $k$ -vector space $W$ , subspaces $U \subseteq V$ , $Z \subseteq W$ , and the quotient maps $p : V \to V / U$ , $q : W \to W / Z$ .

[A1]

The Axiom of Choice holds, so by Every linear subspace $U$ of a vector space $V$ has a complement: a linear subspace $W$ with $V = U \oplus W$ every linear subspace of a vector space has a complement: for the injections $f$ , and the inclusions $U \subseteq V$ , $Z \subseteq W$ , there are subspaces with $V^{'} = f (V) \oplus C$ , $V = U \oplus V_{1}$ and $W = Z \oplus W_{1}$ (Internal direct sum $V = ⨁_{i < n} U_{i}$ : the sum is everything and each summand meets the sum of the others only in $0_{V}$ ).

[F1]

The conventions: $V \otimes W$ is the tensor product with its universal property, every element is a finite sum of elementary tensors, the defining relations give $(v + v^{'}) \otimes w = v \otimes w + v^{'} \otimes w$ , $v \otimes (w + w^{'}) = v \otimes w + v \otimes w^{'}$ and $0 \otimes w = 0 = v \otimes 0$ , and $k v \otimes w = v \otimes k w$ (Scalars, tensor powers, the empty tensor, opposite algebras and finite sums).

[F2]

Functoriality: $(f \otimes g) (v \otimes w) = f (v) \otimes g (w)$ defines a linear map, $i d \otimes i d = i d$ , and $(f^{'} \circ f) \otimes (g^{'} \circ g) = (f^{'} \otimes g^{'}) \circ (f \otimes g)$ (Module homomorphisms induce tensor-product homomorphisms functorially).

[F3]

Quotient modules consist of cosets, the quotient map is linear with kernel the submodule quotiented by, and kernels and images of linear maps are subspaces (Quotient module $M / N$ with scalar multiplication on additive cosets, Kernel and image of a linear map, Linear subspace of a vector space).

Proof

technique · direct

1.1givenA1F2algebra

Injectivity of $f \otimes {i d}_{W}$ : by [A1] write $V^{'} = f (V) \oplus C$ and define $g : V^{'} \to V$ by $g (f (v) + c) : = v$ ; this is well defined and linear because every element of $V^{'}$ has a unique decomposition, and $g \circ f = {i d}_{V}$ . By [F2], $(g \otimes {i d}_{W}) \circ (f \otimes {i d}_{W}) = (g \circ f) \otimes {i d}_{W} = {i d}_{V} \otimes {i d}_{W} = {i d}_{V \otimes W}$ , so $f \otimes {i d}_{W}$ has a left inverse and is injective.

1.2givenF1F3algebra

Containment: for $u \in U$ and $w \in W$ one has $(p \otimes q) (u \otimes w) = p (u) \otimes q (w) = 0 \otimes q (w) = 0$ by [F1], because $p$ kills $U$ ; likewise $p \otimes q$ kills every $v \otimes z \in V \otimes Z$ because $q$ kills $Z$ . Hence $im (U \otimes W)$ and $im (V \otimes Z)$ lie in the kernel $\ker (p \otimes q)$ , which is a subspace by [F3], so their sum $U \otimes W + V \otimes Z$ lies in the kernel as well.

1.3givenA1F2F3algebra

Complements and an isomorphism: by [A1] write $V = U \oplus V_{1}$ and $W = Z \oplus W_{1}$ . The restrictions $p_{1} : = p ∣_{V_{1}}$ and $q_{1} : = q ∣_{W_{1}}$ are isomorphisms onto $V / U$ and $W / Z$ : a coset $v + U$ with $v = u + v_{1}$ equals $v_{1} + U$ , so $p_{1}$ is surjective, and if $p (v_{1}) = 0$ then $v_{1} \in U \cap V_{1} = 0$ by the direct-sum condition, so $p_{1}$ is injective (and likewise for $q_{1}$ ); by [F2] the map $p_{1} \otimes q_{1} : V_{1} \otimes W_{1} \to (V / U) \otimes (W / Z)$ is then an isomorphism with inverse $s \otimes r$ , where $s = p_{1}^{- 1}$ and $r = q_{1}^{- 1}$ . Every element of $V$ is uniquely $u + x$ with $u \in U$ , $x \in V_{1}$ , so the component maps $v \mapsto u$ and $v \mapsto x$ are well defined and linear, and likewise for $W = Z \oplus W_{1}$ ; consequently each of the three inclusion-induced maps $ι_{U} \otimes {i d}_{W} : U \otimes W \to V \otimes W$ , ${i d}_{V} \otimes ι_{Z} : V \otimes Z \to V \otimes W$ and $ι_{V_{1}} \otimes ι_{W_{1}} : V_{1} \otimes W_{1} \to V \otimes W$ has a left inverse induced by the corresponding component projection (for the first, $(π_{U} \otimes {i d}_{W}) \circ (ι_{U} \otimes {i d}_{W}) = (π_{U} \circ ι_{U}) \otimes {i d}_{W} = {i d}_{U \otimes W}$ by [F2], and similarly for the others), hence is injective and exhibits its domain inside $V \otimes W$ through the injections of the statement; under these identifications the composite $V_{1} \otimes W_{1} \to V \otimes W \to (V / U) \otimes (W / Z)$ is exactly $p_{1} \otimes q_{1}$ , since both send $x \otimes y$ to $p_{1} (x) \otimes q_{1} (y)$ .

2.1step 1.1step 1.2step 1.3F1algebra

Kernel: let $t \in \ker (p \otimes q)$ and write $t = \sum_{i} v_{i} \otimes w_{i}$ as a finite sum of elementary tensors by [F1]. Decompose $v_{i} = u_{i} + x_{i}$ with $u_{i} \in U$ , $x_{i} \in V_{1}$ and $w_{i} = z_{i} + y_{i}$ with $z_{i} \in Z$ , $y_{i} \in W_{1}$ by the direct sums of step 1.3 and expand bilinearly by [F1]: $t = \sum_{i} u_{i} \otimes w_{i} + \sum_{i} x_{i} \otimes z_{i} + \sum_{i} x_{i} \otimes y_{i}$ , where the first summand lies in $im (U \otimes W)$ , the second in $im (V_{1} \otimes Z) \subseteq im (V \otimes Z)$ , and the third is the image of $\sum_{i} x_{i} \otimes y_{i} \in V_{1} \otimes W_{1}$ . Applying $p \otimes q$ kills the first two summands by step 1.2, so $0 = (p \otimes q) (t) = (p_{1} \otimes q_{1}) (\sum_{i} x_{i} \otimes y_{i})$ by the identification of step 1.3; injectivity of $p_{1} \otimes q_{1}$ gives $\sum_{i} x_{i} \otimes y_{i} = 0$ in $V_{1} \otimes W_{1}$ , hence the third summand is $0$ in $V \otimes W$ , and $t \in U \otimes W + V \otimes Z$ .

3.1step 1.1step 1.2step 2.1∎

Steps 1.2 and 2.1 give the two inclusions, so $\ker (p \otimes q) = U \otimes W + V \otimes Z$ for the subspaces exhibited through the injections of step 1.1, and step 1.1 itself is the first assertion of the statement.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-10-08Open item page →

Coefficient separation for an independent family of vectors, with explicit Choice assumptions

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $v_{1}, \dots, v_{n} \in V$ be linearly independent vectors in a $k$ -vector space $V$ and let $w_{1}, \dots, w_{n} \in W$ satisfy $\sum_{i = 1}^{n} v_{i} \otimes w_{i} = 0$ in $V \otimes W$ . Then $w_{i} = 0$ for every $i$ ; equivalently, the linear map $W^{n} \to V \otimes W$ , $(w_{i}) \mapsto \sum_{i} v_{i} \otimes w_{i}$ , is injective.

If $V$ is finite-dimensional the same conclusion is proved without the Axiom of Choice.

Facts & Assumptions

Given: A field $k$ , a $k$ -vector space $V$ with linearly independent vectors $v_{1}, \dots, v_{n}$ , a $k$ -vector space $W$ , and elements $w_{1}, \dots, w_{n} \in W$ with $\sum_{i} v_{i} \otimes w_{i} = 0$ in $V \otimes W$ .

[A1]

The Axiom of Choice holds (The Axiom of Choice).

[F1]

The tensor product conventions: $V \otimes W$ has the universal property, every element is a finite sum of elementary tensors, and the empty tensor $k$ is identified with the tensor unit through the unit isomorphisms, so $k \otimes W \to W$ , $c \otimes w \mapsto c w$ , is an isomorphism (Scalars, tensor powers, the empty tensor, opposite algebras and finite sums).

[F2]

Under [A1], for every injective linear map $f : V \to V^{'}$ and every $k$ -vector space $W$ the map $f \otimes {i d}_{W}$ is injective (Tensoring injections and the kernel of a tensor product of quotient maps over a field).

[F3]

The span of a set is the smallest linear subspace containing it (Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ ); in particular $span (v_{1}, \dots, v_{n})$ is a linear subspace containing each $v_{i}$ , and since it is a subspace it contains every finite linear combination of the $v_{i}$ .

[F4]

The span of a finite list consists of its linear combinations. A finite list is an ordered basis exactly when every vector has a unique expansion in it; addition and scaling of the unique coordinate lists show that its coordinate functionals are linear (A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i < n} λ_{i} v_{i}$ for exactly one $λ : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis).

[F5]

In a finite-dimensional space every linearly independent subset is contained in a basis, with no choice principle used (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ ).

[F6]

Functoriality supplies an additive map $f \otimes g$ with $(f \otimes g) (v \otimes w) = f (v) \otimes g (w)$ and respects composition (Module homomorphisms induce tensor-product homomorphisms functorially). For linear $f, g$ over $k$ , this map is $k$ -linear: the scalar action of [F1] gives $(f \otimes g) (c (v \otimes w)) = f (c v) \otimes g (w) = c (f (v) \otimes g (w))$ , and additivity extends this identity to every finite tensor sum.

Proof

technique · direct

1.1givenA1F1F2F3F4F6algebra

Arbitrary dimension, under [A1]. Put $V_{0} : = span (v_{1}, \dots, v_{n})$ , a linear subspace of $V$ containing every $v_{i}$ by [F3]; the inclusion $ι : V_{0} ↪ V$ is injective and linear, so $ι \otimes {i d}_{W} : V_{0} \otimes W \to V \otimes W$ is injective by [F2]. Since $(ι \otimes {i d}_{W}) (\sum_{i} v_{i} \otimes w_{i}) = \sum_{i} v_{i} \otimes w_{i} = 0$ by [F6], injectivity gives $\sum_{i} v_{i} \otimes w_{i} = 0$ in $V_{0} \otimes W$ . Every element of $V_{0}$ is a linear combination of the finite list $(v_{i})$ by the span clause of [F4], so the list $(v_{1}, \dots, v_{n})$ spans $V_{0}$ , and it is linearly independent by hypothesis, so it is an ordered basis of $V_{0}$ with linear coordinate functionals $φ_{i} : V_{0} \to k$ satisfying $φ_{i} (v_{j}) = δ_{i j}$ by [F4]. For each $i$ , functoriality [F6] gives the linear map $φ_{i} \otimes {i d}_{W} : V_{0} \otimes W \to k \otimes W$ , and applying it to the relation yields $0 = \sum_{j} (φ_{i} \otimes {i d}_{W}) (v_{j} \otimes w_{j}) = \sum_{j} φ_{i} (v_{j}) \otimes w_{j} = 1 \otimes w_{i}$ ; the unit isomorphism of [F1] sends $1 \otimes w_{i}$ to $w_{i}$ , so $w_{i} = 0$ .

1.2givenF1F4F5F6algebra

Finite dimension, without Choice. If $V$ is finite-dimensional, the vectors $v_{1}, \dots, v_{n}$ are distinct and form a linearly independent finite subset of $V$ , so by [F5] this subset is contained in a basis $B$ of $V$ ; listing the finite set $B$ with $v_{1}, \dots, v_{n}$ in the first $n$ positions gives an ordered basis $(b_{1}, \dots, b_{m})$ with $b_{j} = v_{j}$ for $j \leq n$ whose coordinate functionals $ψ_{1}, \dots, ψ_{m}$ are linear by [F4] and satisfy $ψ_{j} (v_{k}) = δ_{j k}$ for $j, k \leq n$ . Applying $ψ_{i} \otimes {i d}_{W}$ to the relation in $V \otimes W$ by [F6] gives $0 = 1 \otimes w_{i}$ , hence $w_{i} = 0$ by the unit isomorphism of [F1]; no complement or Zorn extension is used, only the finite-dimensional extension of [F5].

2.1step 1.1step 1.2givenF1∎

Both cases give $w_{i} = 0$ for all $i$ : step 1.1 under the Axiom of Choice in arbitrary dimension and step 1.2 in finite dimension without it; the map $W^{n} \to V \otimes W$ with $(w_{i}) \mapsto \sum_{i} v_{i} \otimes w_{i}$ is linear by bilinearity of elementary tensors [F1] and has trivial kernel, so subtracting two tuples with the same image proves injectivity, which is the equivalent form of the claim.

Remarks

The proof uses AC through the tensor-injection lemma in arbitrary ambient dimension and avoids it when $V$ is finite-dimensional. It establishes sufficiency of these assumptions, not necessity of AC or a sharp boundary between choice principles.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

Finite tensor duality and basis-independent coevaluation

Statement

Let $V, W$ be finite-dimensional $k$ -vector spaces.

The bilinear map $V^{*} \times W^{*} \to (V \otimes W)^{*}$ , $(f, g) \mapsto [v \otimes w \mapsto f (v) g (w)]$ , induces an isomorphism $V^{*} \otimes W^{*} \to (V \otimes W)^{*}$ . For ordered bases $(v_{1}, \dots, v_{m})$ of $V$ , $(w_{1}, \dots, w_{n})$ of $W$ with dual bases $(v_{i}^{*})$ , $(w_{j}^{*})$ , the image of $v_{i}^{*} \otimes w_{j}^{*}$ is the dual basis vector of $v_{i} \otimes w_{j}$ in the product basis of $V \otimes W$ .
For every ordered basis $(v_{1}, \dots, v_{n})$ of $V$ the element $\sum_{i = 1}^{n} v_{i}^{*} \otimes v_{i} \in V^{*} \otimes V$ is independent of the basis and corresponds to ${i d}_{V}$ under the isomorphism $V^{*} \otimes V \to Hom (V, V)$ , $f \otimes v \mapsto [x \mapsto f (x) v]$ (For finite-dimensional $V$ , the canonical map $V^{*} \otimes_{F} W \to {Hom}_{F} (V, W)$ is an isomorphism). Its image under $σ_{V^{*}, V}$ is the corresponding element $\sum_{i} v_{i} \otimes v_{i}^{*}$ of $V \otimes V^{*}$ .
The evaluation $e v : V^{*} \otimes V \to k$ , $f \otimes v \mapsto f (v)$ , and the coevaluation $c o e v : k \to V \otimes V^{*}$ , $1 \mapsto \sum_{i} v_{i} \otimes v_{i}^{*}$ , are independent of the basis and satisfy $({i d}_{V} \otimes e v) \circ α_{V, V^{*}, V} \circ (c o e v \otimes {i d}_{V}) = {i d}_{V}$ and $(e v \otimes {i d}_{V^{*}}) \circ α_{V^{*}, V, V^{*}}^{- 1} \circ ({i d}_{V^{*}} \otimes c o e v) = {i d}_{V^{*}}$ , the unit isomorphisms $k \otimes V ≅ V$ , $V \otimes k ≅ V$ being understood.

Facts & Assumptions

Given: A field $k$ , finite-dimensional $k$ -vector spaces $V, W$ , ordered bases $(v_{1}, \dots, v_{m})$ of $V$ and $(w_{1}, \dots, w_{n})$ of $W$ with their dual bases, and ordered bases of $V$ written $(v_{1}, \dots, v_{n})$ .

[F1]

The tensor product conventions: $M \otimes N$ is the tensor product over $k$ with its universal property, every element is a finite sum of elementary tensors, the unit isomorphisms $λ_{N} : k \otimes N \to N$ and $ρ_{M} : M \otimes k \to M$ are given by $r \otimes n \mapsto r n$ and $m \otimes r \mapsto m r$ , and tensor powers are left-associated with $V^{\otimes 0} = k$ (Scalars, tensor powers, the empty tensor, opposite algebras and finite sums).

[F2]

The algebraic dual $V^{*}$ is the space of linear functionals on $V$ (Linear functionals and the algebraic dual $V^{*} = L (V, F)$ ).

[F3]

The dual family $(b^{*})_{b \in B}$ of a basis is defined by $b^{*} (c) = δ_{b c}$ and linear extension (The dual family $(b^{*})_{b \in B}$ associated to a Hamel basis $B$ , defined by $b^{*} (c) = δ_{b c}$ ).

[F4]

If $B = (b_{1}, \dots, b_{n})$ is a basis of a finite-dimensional space, its dual family $B^{*}$ is a basis, so it is linearly independent and spans the dual space (The dual family of a finite basis is a basis of the dual space, with the same dimension).

[F5]

The elementary tensors of two bases form a basis of the tensor product (The elementary tensors of two bases form the product basis of the tensor product).

[F6]

Coordinates with respect to an ordered basis are unique, so $x = \sum_{i} λ_{i} b_{i}$ with $λ$ the coordinate list of $x$ ; for the dual basis of a basis this gives $f = \sum_{i} f (b_{i}) b_{i}^{*}$ and $x = \sum_{i} b_{i}^{*} (x) b_{i}$ (A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i < n} λ_{i} v_{i}$ for exactly one $λ : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis).

[F7]

For finite-dimensional $V$ the map $Φ : V^{*} \otimes W \to Hom (V, W)$ , $ϕ \otimes w \mapsto [y \mapsto ϕ (y) w]$ , is an isomorphism with inverse $Ψ (T) = \sum_{i} v_{i}^{*} \otimes T (v_{i})$ (For finite-dimensional $V$ , the canonical map $V^{*} \otimes_{F} W \to {Hom}_{F} (V, W)$ is an isomorphism).

[F8]

The symmetry and associativity isomorphisms act on elementary tensors by $σ_{M, N} (m \otimes n) = n \otimes m$ and $α_{L, M, N} ((l \otimes m) \otimes n) = l \otimes (m \otimes n)$ (Symmetry and associativity isomorphisms for tensor products over a commutative ring).

[F9]

The unit isomorphisms $λ_{N} : k \otimes N \to N$ and $ρ_{M} : M \otimes k \to M$ are isomorphisms with the stated formulas (The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ ).

Proof

technique · direct

1.1givenF1F2F3F4F5algebra

Claim 1. For fixed $(f, g) \in V^{*} \times W^{*}$ the pairing $V \times W \to k$ , $(v, w) \mapsto f (v) g (w)$ , is $k$ -bilinear, so it defines a functional $θ (f, g) \in (V \otimes W)^{*}$ with $θ (f, g) (v \otimes w) = f (v) g (w)$ by the universal property in [F1], the star denoting the space of linear functionals [F2]; the assignment $(f, g) \mapsto θ (f, g)$ is itself $k$ -bilinear, so it induces a $k$ -linear map $θ : V^{*} \otimes W^{*} \to (V \otimes W)^{*}$ by [F1]. On the one hand $(v_{i}^{*} \otimes w_{j}^{*})$ is a basis of $V^{*} \otimes W^{*}$ by [F4] and [F5], on the other hand the dual vectors $(v_{i} \otimes w_{j})^{*}$ in $(V \otimes W)^{*}$ of the product basis $(v_{i} \otimes w_{j})$ (a basis by [F5]) form a basis of $(V \otimes W)^{*}$ by [F4]; and $θ (v_{i}^{*} \otimes w_{j}^{*}) (v_{k} \otimes w_{l}) = v_{i}^{*} (v_{k}) w_{j}^{*} (w_{l}) = δ_{i k} δ_{j l} = (v_{i} \otimes w_{j})^{*} (v_{k} \otimes w_{l})$ by [F3], so $θ (v_{i}^{*} \otimes w_{j}^{*}) = (v_{i} \otimes w_{j})^{*}$ because linear functionals agreeing on the spanning product basis agree. If $z = \sum_{i, j} c_{i j} v_{i}^{*} \otimes w_{j}^{*}$ satisfies $θ (z) = 0$ , then $\sum_{i, j} c_{i j} (v_{i} \otimes w_{j})^{*} = 0$ and linear independence of the dual product basis [F4] gives all $c_{i j} = 0$ , so $θ$ is injective; and every $L \in (V \otimes W)^{*}$ is $L = \sum_{i, j} d_{i j} (v_{i} \otimes w_{j})^{*} = θ (\sum_{i, j} d_{i j} v_{i}^{*} \otimes w_{j}^{*})$ by spanning [F4], so $θ$ is surjective. Hence $θ$ is an isomorphism with the stated values on the dual product basis.

1.2givenF6F7F8algebra

Claim 2. By [F7] the map $Φ : V^{*} \otimes V \to Hom (V, V)$ is an isomorphism, so the preimage of ${i d}_{V}$ is unique and any two bases $(v_{1}, \dots, v_{n})$ give the same element as soon as both give ${i d}_{V}$ . For $y \in V$ the coordinate expansion of [F6] gives $y = \sum_{i} v_{i}^{*} (y) v_{i}$ , hence $Φ (\sum_{i} v_{i}^{*} \otimes v_{i}) (y) = \sum_{i} v_{i}^{*} (y) v_{i} = y$ on a spanning set of $V$ , so $Φ (\sum_{i} v_{i}^{*} \otimes v_{i}) = {i d}_{V}$ and the element $\sum_{i} v_{i}^{*} \otimes v_{i}$ is the unique preimage of ${i d}_{V}$ , independent of the basis. Its image under $σ_{V^{*}, V}$ is $\sum_{i} v_{i} \otimes v_{i}^{*}$ by the elementary-tensor formula of [F8].

2.1step 1.2F1F6F8F9algebra

Claim 3. The pairing $V^{*} \times V \to k$ , $(f, v) \mapsto f (v)$ , is $k$ -bilinear, so it induces $e v : V^{*} \otimes V \to k$ , $f \otimes v \mapsto f (v)$ , by [F1]; evaluation is basis-free. The coevaluation $c o e v : k \to V \otimes V^{*}$ is defined by $1 \mapsto \sum_{i} v_{i} \otimes v_{i}^{*}$ , which by step 1.2 is independent of the basis and equals $σ_{V^{*}, V}$ applied to the unique preimage of ${i d}_{V}$ ; it is $k$ -linear because $k$ is spanned by $1$ . For the first zigzag, use the conventions of [F1] and the formulas of [F8] and [F9]: for $x \in V$ , $(c o e v \otimes {i d}_{V}) (1 \otimes x) = \sum_{i} (v_{i} \otimes v_{i}^{*}) \otimes x$ , the associator sends this to $\sum_{i} v_{i} \otimes (v_{i}^{*} \otimes x)$ , and $({i d}_{V} \otimes e v)$ sends it to $\sum_{i} v_{i} \otimes v_{i}^{*} (x) = x \otimes 1$ by [F6], which the unit isomorphism $ρ_{V}$ identifies with $x$ ; both composites are linear in $x$ , so they agree everywhere. For the second zigzag, $f \otimes 1$ maps under $({i d}_{V^{*}} \otimes c o e v)$ to $\sum_{i} f \otimes (v_{i} \otimes v_{i}^{*})$ , the inverse associator sends this to $\sum_{i} (f \otimes v_{i}) \otimes v_{i}^{*}$ , and $(e v \otimes {i d}_{V^{*}})$ sends it to $\sum_{i} f (v_{i}) (1 \otimes v_{i}^{*}) = 1 \otimes f$ by the dual expansion of [F6], which the unit isomorphism $λ_{V^{*}}$ identifies with $f$ ; again both composites are linear, so equality on the spanning elements $f \otimes 1$ proves the identity.

3.1step 1.1step 1.2step 2.1∎

Steps 1.1, 1.2 and 2.1 prove claims 1, 2 and 3 respectively, with the dual product basis values, the basis independence of the preimage of ${i d}_{V}$ , and both zigzag identities established.

Remarks

Infinite dimension is deliberately outside the claim. No surjectivity of $V^{*} \otimes W^{*} \to (V \otimes W)^{*}$ is claimed in infinite dimension, and the companion page's An infinite-dimensional tensor-dual functional outside the image ↗ exhibits a functional outside the image when $V$ has an infinite basis, so the finite-dimensional hypothesis of part 1 cannot simply be dropped.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-10-08Open item page →

The free associative R-algebra on a set and descent of relations

Statement

Let $R$ be a commutative ring and $S$ a set. There is a unital associative $R$ -algebra $R ⟨ S ⟩$ , the free associative $R$ -algebra on $S$ , whose underlying $R$ -module is free on the set of finite words in $S$ (including the empty word), with product given on words by concatenation and extended $R$ -bilinearly.

Universal property. For every $R$ -algebra $A$ (Algebras over a commutative ring, central structure maps, and algebra homomorphisms) and every map $S \to A$ there is a unique unital $R$ -algebra homomorphism $R ⟨ S ⟩ \to A$ extending it. When $R = k$ is a field and $S$ indexes a basis of a vector space $V$ , $R ⟨ S ⟩$ reproduces the published tensor algebra $T (V)$ (Tensor algebra of a vector space, Universal property of the tensor algebra).
Two-sided ideal description. For $E \subseteq R ⟨ S ⟩$ the two-sided ideal $(E)$ (The ideal generated by a subset and principal ideals) is the set of finite sums $\sum_{s} a_{s} e_{s} b_{s}$ with $a_{s}, b_{s} \in R ⟨ S ⟩$ and $e_{s} \in E$ .
Quotient universal property. If $φ : R ⟨ S ⟩ \to A$ is an $R$ -algebra homomorphism killing $E$ , then $φ$ factors uniquely through the quotient $R ⟨ S ⟩ / (E)$ (A ring homomorphism whose kernel contains a two-sided ideal factors uniquely through the quotient ring); the quotient is the presented $R$ -algebra with generators $S$ and relations $E$ .

Facts & Assumptions

Given: A commutative ring $R$ and a set $S$ .

[F1]

A natural number is a von Neumann natural $n = {0, \dots, n - 1}$ , so a function $n \to S$ is a finite list of elements of $S$ ; function sets, restrictions and functions extended by one value are available (The natural numbers $N$ (von Neumann), The set $B^{A}$ of all functions $A \to B$ ).

[F2]

Addition of natural numbers satisfies $m + 0 = m$ and $m + σ (n) = σ (m + n)$ , is associative and commutative, is cancellative, has $0$ as two-sided identity, and is compatible with the order; $N$ is trichotomously linearly ordered and $i < n$ means $i \in n$ (Addition is associative, Addition is commutative, Left identity for addition, Addition is cancellative, On $N$ the order is membership: $m < n ⟺ m \in n$ , $\leq$ is a linear order on $N$ , Order is compatible with addition).

[F3]

The free $R$ -module on a set $X$ is $R^{(X)} = ⨁_{x \in X} R$ , with standard basis $e_{x}$ and unique finitely supported coefficient families; a set map from a basis extends uniquely to an $R$ -linear map (The free module on a set and its standard basis, Universal property of the free module on a set).

[F4]

Balanced pairings induce unique group homomorphisms out of the tensor product with the prescribed values on elementary tensors (Universal property of the tensor product for balanced maps into abelian groups).

[F5]

An $R$ -algebra is a unital ring with a central unital structure map, and algebra homomorphisms are unital ring homomorphisms over it (Algebras over a commutative ring, central structure maps, and algebra homomorphisms, Commutative ring, Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ).

[F6]

Two-sided ideals are additive subgroups closed under both one-sided multiplications; $(E)$ is the intersection of all two-sided ideals containing $E$ (Left, right and two-sided ideals, The ideal generated by a subset and principal ideals).

[F7]

A ring homomorphism whose kernel contains a two-sided ideal $I$ factors uniquely through $R / I$ (A ring homomorphism whose kernel contains a two-sided ideal factors uniquely through the quotient ring, The quotient ring $R / I$ with $(r + I) (s + I) = r s + I$ ).

[F8]

The tensor algebra $T (V) = ⨁_{n \geq 0} V^{\otimes n}$ of a $k$ -vector space is a unital associative $k$ -algebra, and every linear map $V \to A$ into a unital associative $k$ -algebra extends uniquely to a unital $k$ -algebra homomorphism $T (V) \to A$ ; the degree-one inclusion is $j : V \to T (V)$ (Tensor algebra of a vector space, Universal property of the tensor algebra).

[F9]

Finite sums satisfy reindexing and Fubini (Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule).

Proof

technique · direct

1.1givenF1F2algebraconstruct

Words and concatenation. Call a word in $S$ a function $w : n \to S$ with $n \in N$ its length, and write $W : = ⋃_{n \in N} S^{n}$ for the set of words; the empty word is the unique function $0 \to S$ (there is exactly one, by [F1]). If $w : m \to S$ and $v : n \to S$ , define the concatenation $w v : m + n \to S$ by recursion on $n$ : $w v : = w$ when $n = 0$ ; and when $n = σ (n^{'})$ , write $v^{'}$ for the restriction of $v$ to $n^{'}$ , $a : = v (n^{'})$ for the last letter, and define $w v$ as the function $m + σ (n^{'}) = σ (m + n^{'})$ extending $w v^{'}$ by the value $a$ at the new index $m + n^{'}$ (the case split uses that every index $i < n$ satisfies $i < n - 1$ or $i = n - 1$ by trichotomy [F2], and uniqueness of the decomposition $i = m + j$ for $m \leq i$ follows from commutativity and cancellation [F2]). Unwinding the two recursive clauses, an index of $w v$ lying in the first $m$ positions carries the corresponding letter of $w$ and an index $i = m + j$ with $j < n$ carries $v (j)$ ; associativity $(w v) u = w (v u)$ is then proved by induction on the length of $u$ : for $u$ of length $0$ both sides are $w v$ , and for $u$ with last letter $a$ and predecessor $u^{'}$ one has $w (v u) = w ((v u^{'}) a) = (w (v u^{'})) a = ((w v) u^{'}) a = (w v) u$ by the induction hypothesis and the defining clause, the domains being equal by associativity of addition [F2]; the empty word is neutral since $w \emptyset = w$ by definition and $\emptyset v = v$ by the same induction.

2.1step 1.1F3F4F5F9algebra

The algebra $R ⟨ S ⟩$ . Let $F : = R^{(W)}$ be the free $R$ -module on the set $W$ of words, with basis $(u_{w})_{w \in W}$ ([F3]), and let $μ : F \otimes_{R} F \to F$ be the group homomorphism obtained from the $R$ -bilinear (hence balanced) pairing $(\sum_{w} a_{w} u_{w}, \sum_{v} b_{v} u_{v}) \mapsto \sum_{w, v} a_{w} b_{v} u_{w v}$ by [F4] and [F9]. Define $f g : = μ (f \otimes g)$ ; the displayed formula makes this multiplication $R$ -bilinear and gives $u_{w} u_{v} = u_{w v}$ . Its associativity follows from associativity of concatenation of step 1.1 after expanding finite sums by bilinearity, and $u_{\emptyset}$ is a two-sided identity because $\emptyset$ is neutral for concatenation; hence $F$ with this multiplication is a unital associative ring, and the map $R \to F$ , $r \mapsto r u_{\emptyset}$ , is a central unital structure map because scalars commute with the basis, so $R ⟨ S ⟩ : = F$ is a unital associative $R$ -algebra [F5] with product given on words by concatenation.

3.1step 2.1givenF5F6algebra

Ideal description. Let $R ⟨ S ⟩$ be the $R$ -algebra of step 2.1 and let $E \subseteq R ⟨ S ⟩$ ; let $J$ be the set of finite sums $\sum_{s} a_{s} e_{s} b_{s}$ with $a_{s}, b_{s} \in R ⟨ S ⟩$ , $e_{s} \in E$ . The empty sum shows $0 \in J$ ; a sum of two such finite sums is again one, and $- a e b = (- a) e b$ , so $J$ is an additive subgroup; for $c \in R ⟨ S ⟩$ one has $c (a e b) = (c a) e b$ and $(a e b) c = a e (b c)$ , so $J$ is a two-sided ideal [F6]. It contains $E$ , since $e = 1 e 1$ , and every two-sided ideal $I \supseteq E$ contains every $a e b$ with $e \in E$ by the left and right ideal properties, hence contains all finite sums in $J$ ; therefore $J$ is a two-sided ideal containing $E$ and contained in every such ideal, so $J = (E)$ by the description of $(E)$ as the intersection [F6].

3.2step 2.1F3F5algebra

Universal property. Let $A$ be an $R$ -algebra and $φ_{0} : S \to A$ a map. Define ${\hat{φ}}_{0} (u_{w}) : = φ_{0} (w (0)) φ_{0} (w (1)) \dots φ_{0} (w (n - 1)) \in A$ for a word $w : n \to S$ of length $n$ , the empty product being $1_{A}$ , and extend $R$ -linearly to ${\hat{φ}}_{0} : R ⟨ S ⟩ \to A$ by [F3]. Then ${\hat{φ}}_{0} (u_{w v}) = {\hat{φ}}_{0} (u_{w}) {\hat{φ}}_{0} (u_{v})$ by the defining concatenation rule and induction on the length, and ${\hat{φ}}_{0}$ is unital and $R$ -linear, so it is a unital $R$ -algebra homomorphism [F5] extending $φ_{0}$ on the words of length one. If $ψ$ is any unital $R$ -algebra homomorphism extending $φ_{0}$ , then $ψ (u_{w}) = φ_{0} (w (0)) \dots φ_{0} (w (n - 1)) = {\hat{φ}}_{0} (u_{w})$ for every word, by multiplicativity and induction on the length, so $ψ = {\hat{φ}}_{0}$ on a spanning set and hence everywhere: the extension is unique.

4.1step 2.1step 3.1F5F7algebra

Quotient universal property. Let $I : = (E)$ and let $φ : R ⟨ S ⟩ \to A$ be an $R$ -algebra homomorphism killing $E$ , so $E \subseteq \ker φ$ and hence $I = (E) \subseteq \ker φ$ by step 3.1 since the kernel of a ring homomorphism is a two-sided ideal. By [F7] there is a unique ring homomorphism $\overset{ˉ}{φ} : R ⟨ S ⟩ / I \to A$ with $\overset{ˉ}{φ} (x + I) = φ (x)$ ; it is unital and respects the structure maps, hence is an $R$ -algebra homomorphism, so it is the unique $R$ -algebra factorization, and $R ⟨ S ⟩ / (E)$ is the presented $R$ -algebra on generators $S$ and relations $E$ .

4.2step 3.2F3F8algebra

Tensor algebra. Now let $R = k$ be a field and let $S$ index a basis of a $k$ -vector space $V$ , so $e : S \to V$ , $s \mapsto e_{s}$ is a bijection onto a basis. The map $S \to T (V)$ , $s \mapsto j (e_{s})$ , extends by step 3.2 to a unital $k$ -algebra homomorphism $Θ : k ⟨ S ⟩ \to T (V)$ , and the linear map $V \to k ⟨ S ⟩$ with $e_{s} \mapsto u_{s}$ (existing by [F3]) extends by [F8] to a unital $k$ -algebra homomorphism $Ξ : T (V) \to k ⟨ S ⟩$ . The composite $Ξ Θ$ fixes every one-letter word $u_{s}$ , hence is the identity by the uniqueness in step 3.2; and $Θ Ξ$ restricts to the identity on $V$ , so by the uniqueness in [F8] it is the identity on $T (V)$ . Thus $Θ$ is an isomorphism of unital $k$ -algebras and $k ⟨ S ⟩$ reproduces $T (V)$ .

5.1step 1.1step 2.1step 3.1step 3.2step 4.1step 4.2∎

Collecting: step 1.1 constructs the words and their associative concatenation, step 2.1 the $R$ -algebra $R ⟨ S ⟩$ free on the words on which the product is concatenation, step 3.2 its universal property, step 4.2 the tensor-algebra identification for $R = k$ a field, step 3.1 the two-sided ideal description and step 4.1 the quotient universal property, so all three parts of the claim hold.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-10-08Open item page →

Presentation base change and transport of explicit bases to commutative specializations

Statement

Let $φ : R \to S$ be a homomorphism of commutative rings, let $X$ be a set and let $R ⟨ X ⟩$ be the free $R$ -algebra on $X$ (The free associative R-algebra on a set and descent of relations).

There is a unique $S$ -algebra isomorphism $S \otimes_{R} R ⟨ X ⟩ \to S ⟨ X ⟩$ with $s \otimes x_{i_{1}} \dots x_{i_{n}} \mapsto s x_{i_{1}} \dots x_{i_{n}}$ on words; its inverse sends a word to $1 \otimes$ that word.
If $I \subseteq R ⟨ X ⟩$ is a two-sided ideal, then there is a unique $S$ -algebra isomorphism $S \otimes_{R} (R ⟨ X ⟩ / I) ≅ (S \otimes_{R} R ⟨ X ⟩) / im (S \otimes_{R} I)$ , sending $s \otimes (y + I)$ to $(s \otimes y) + im (S \otimes_{R} I)$ , the image ideal being generated by the images of the relations. No flatness or freeness of $S$ over $R$ is assumed.
If an $R$ -algebra $A$ is free as an $R$ -module with basis $(a_{i})$ , then $S \otimes_{R} A$ is a free $S$ -module with basis $(1 \otimes a_{i})$ . Consequently an explicitly constructed basis isomorphism of a presentation may be tensored with $S$ to give a basis in every commutative specialization.

Facts & Assumptions

Given: A homomorphism $φ : R \to S$ of commutative rings, a set $X$ , and a two-sided ideal $I \subseteq R ⟨ X ⟩$ .

[F1]

$R ⟨ X ⟩$ is the free associative $R$ -algebra on $X$ : it is free as an $R$ -module on the words in $X$ , the product is concatenation, and every map $X \to A$ into a unital $R$ -algebra $A$ extends uniquely to an $R$ -algebra homomorphism $R ⟨ X ⟩ \to A$ (The free associative R-algebra on a set and descent of relations).

[F2]

Extension of scalars: $S$ is an $R$ -module through $φ$ and also an $S$ -module, so $S \otimes_{R} R ⟨ X ⟩$ is an $S$ -module with $s^{'} (s \otimes y) = s^{'} s \otimes y$ , and balanced pairings induce linear maps out of tensor products (Restriction of scalars and extension of scalars $S \otimes_{R} M$ along a ring homomorphism $R \to S$ , Universal property of the tensor product for balanced maps into abelian groups).

[F3]

For $R$ -algebras the tensor product is an $R$ -algebra with $(s \otimes y) (s^{'} \otimes y^{'}) = s s^{'} \otimes y y^{'}$ and unit $1 \otimes 1$ (The tensor product of $R$ -algebras has multiplication $(a \otimes b) (a^{'} \otimes b^{'}) = a a^{'} \otimes b b^{'}$ , Algebras over a commutative ring, central structure maps, and algebra homomorphisms).

[F4]

Tensoring is right exact: the tensor of an exact sequence $A \to B \to C \to 0$ is exact, so the kernel of the tensored surjection is the image of the tensored map on $A$ (Tensoring is right exact).

[F5]

A ring homomorphism killing a two-sided ideal factors uniquely through the quotient ring (A ring homomorphism whose kernel contains a two-sided ideal factors uniquely through the quotient ring, The quotient ring $R / I$ with $(r + I) (s + I) = r s + I$ ).

[F6]

A free module on a set has a standard basis with unique finite expansions, and a set map from a basis extends uniquely to a linear map (The free module on a set and its standard basis, Universal property of the free module on a set).

Proof

technique · direct

1.1givenF1F2F3algebra

The map $Θ$ . Let $ρ : R ⟨ X ⟩ \to S ⟨ X ⟩$ be the $R$ -algebra homomorphism extending the generator map $X \to S ⟨ X ⟩$ (which need not be injective when $S$ is the zero ring), existing by [F1]. The pairing $β : S \times R ⟨ X ⟩ \to S ⟨ X ⟩$ , $β (s, y) : = s ρ (y)$ , is additive in each variable and satisfies $β (s φ (r), y) = s φ (r) ρ (y) = s ρ (r y) = β (s, r y)$ because $ρ$ is $R$ -linear [F3, F1], so by [F2] it induces an $R$ -linear map $Θ : S \otimes_{R} R ⟨ X ⟩ \to S ⟨ X ⟩$ with $Θ (s \otimes y) = s ρ (y)$ ; it is $S$ -linear because $Θ (s^{'} (s \otimes y)) = s^{'} s ρ (y) = s^{'} Θ (s \otimes y)$ . The map $S \to S \otimes_{R} R ⟨ X ⟩$ , $s \mapsto s \otimes 1$ , is unital and central by [F3], so the tensor product is an $S$ -algebra with $(s \otimes y) (s^{'} \otimes y^{'}) = s s^{'} \otimes y y^{'}$ , and $Θ ((s \otimes y) (s^{'} \otimes y^{'})) = s s^{'} ρ (y y^{'}) = s s^{'} ρ (y) ρ (y^{'}) = Θ (s \otimes y) Θ (s^{'} \otimes y^{'})$ and $Θ (1 \otimes 1) = 1$ ; so $Θ$ is a unital $S$ -algebra homomorphism, and $Θ (s \otimes w) = s w$ on a word $w$ .

1.2givenF2F3F4F5algebra

Part 2. The quotient map $π : R ⟨ X ⟩ \to R ⟨ X ⟩ / I$ is a surjective $R$ -algebra homomorphism with kernel $I$ , so the sequence $I \to R ⟨ X ⟩ \to R ⟨ X ⟩ / I \to 0$ is exact, and tensoring with $S$ over $R$ gives an exact sequence whose middle map is ${i d}_{S} \otimes π$ with kernel $J : = im (S \otimes_{R} I)$ by [F4]. The set $J$ is a two-sided ideal of the $S$ -algebra $S \otimes_{R} R ⟨ X ⟩$ : it consists of finite sums $\sum_{i} s_{i} \otimes e_{i}$ with $e_{i} \in I$ and is an additive subgroup, left multiplication by an elementary tensor gives $(s \otimes y) (s_{i} \otimes e_{i}) = s s_{i} \otimes y e_{i}$ with $y e_{i} \in I$ , right multiplication gives $(s_{i} \otimes e_{i}) (s \otimes y) = s_{i} s \otimes e_{i} y$ with $e_{i} y \in I$ , and additivity extends both closures to all of $S \otimes_{R} R ⟨ X ⟩$ . Moreover $J$ is generated as an ideal by the images $1 \otimes e$ of the relations, since $s_{i} \otimes e_{i} = (s_{i} \otimes 1) (1 \otimes e_{i})$ . By [F5] the surjective $S$ -algebra homomorphism ${i d}_{S} \otimes π$ , which kills $J$ , factors through a surjective $S$ -algebra homomorphism $\overline{Θ} : (S \otimes_{R} R ⟨ X ⟩) / J \to S \otimes_{R} (R ⟨ X ⟩ / I)$ with kernel $J / J = 0$ , hence an isomorphism. Its inverse is induced on the quotient by the pairing $(s, y + I) \mapsto s \otimes y + J$ , which is well defined because $y^{'} \in y + I$ gives $s \otimes y^{'} - s \otimes y = s \otimes (y^{'} - y) \in J$ and is bilinear by [F2]; the two maps are inverse because they are inverse on the spanning elements $s \otimes (y + I)$ and $s \otimes y + J$ . The inverse is the unique $S$ -algebra map with these prescribed values, since the tensors $s \otimes (y + I)$ span its source.

1.3givenF2F3F6algebra

Part 3. Let $A$ be free with basis $(a_{i})_{i \in I}$ , so every $y \in A$ has a unique expansion $y = \sum_{i} c_{i} a_{i}$ by [F6]. The pairing $γ : S \times A \to S^{(I)}$ , $γ (s, \sum_{i} c_{i} a_{i}) : = \sum_{i} (s φ (c_{i})) e_{i}$ , is additive in each variable and satisfies $γ (s φ (r), y) = γ (s, r y)$ because the coordinates of $r y$ are $r c_{i}$ , and $φ (r c_{i}) = φ (r) φ (c_{i})$ ; by [F2] it induces an $S$ -linear map $Γ : S \otimes_{R} A \to S^{(I)}$ with $Γ (s \otimes y) = \sum_{i} s φ (c_{i}) e_{i}$ , in particular $Γ (1 \otimes a_{i}) = e_{i}$ , and the $S$ -linear map $Δ : S^{(I)} \to S \otimes_{R} A$ with $Δ (e_{i}) = 1 \otimes a_{i}$ , existing by [F6], satisfies $Γ Δ = i d$ on the standard basis and $Δ Γ (1 \otimes a_{i}) = 1 \otimes a_{i}$ ; since elementary tensors and basis elements span, $Γ$ and $Δ$ are mutually inverse isomorphisms. Hence $(1 \otimes a_{i})$ is an $S$ -basis of $S \otimes_{R} A$ , and an explicitly constructed basis isomorphism of a presented $R$ -algebra may be tensored with $S$ to transport the basis to every commutative specialization.

2.1step 1.1F1F3algebra

Inverse for part 1. The map $X \to S \otimes_{R} R ⟨ X ⟩$ , $x \mapsto 1 \otimes x$ , extends by the universal property of [F1] applied over $S$ to an $S$ -algebra homomorphism $Λ : S ⟨ X ⟩ \to S \otimes_{R} R ⟨ X ⟩$ , and for a word $w = x_{i_{1}} \dots x_{i_{n}}$ one has $Λ (ρ (w)) = Λ (x_{i_{1}} \dots x_{i_{n}}) = \prod_{k} Λ (x_{i_{k}}) = \prod_{k} (1 \otimes x_{i_{k}}) = 1 \otimes w$ by multiplicativity and [F3]; hence $Λ Θ (s \otimes w) = Λ (s w) = s (1 \otimes w) = s \otimes w$ for every $s \in S$ and word $w$ , so $Λ Θ = i d$ on the spanning elementary tensors $s \otimes w$ . Conversely $Θ Λ$ and the identity of $S ⟨ X ⟩$ are unital $S$ -algebra homomorphisms agreeing on the generators $x \in X$ , so by the uniqueness in [F1] they are equal. Thus $Θ$ is an $S$ -algebra isomorphism, unique because it is determined on the spanning elementary tensors, and $Λ$ sends a word to $1 \otimes w$ .

3.1step 1.1step 1.2step 1.3step 2.1∎

Collecting: step 1.1 constructs $Θ$ and step 2.1 proves it is the unique $S$ -algebra isomorphism of part 1 with the stated values; step 1.2 proves the quotient identification of part 2 with the image ideal generated by the relations and no flatness hypothesis; step 1.3 proves the free-basis transport of part 3.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-10-08Open item page →

Multivariate polynomial and Laurent rings over commutative rings, domains and fraction fields

Statement

Let $R$ be a commutative ring and $n \geq 0$ .

Multivariate polynomial ring. The ring $R [x_{1}, \dots, x_{n}]$ of Polynomial rings in finitely many commuting indeterminates by iteration is a commutative $R$ -algebra, free as an $R$ -module on the monomials $x^{α} = x_{1}^{α_{1}} \dots x_{n}^{α_{n}}$ ( $α \in N^{n}$ ), and for every commutative $R$ -algebra $A$ and elements $a_{1}, \dots, a_{n} \in A$ there is a unique $R$ -algebra homomorphism $R [x_{1}, \dots, x_{n}] \to A$ with $x_{i} \mapsto a_{i}$ . If $R$ is an integral domain, so is $R [x_{1}, \dots, x_{n}]$ .
Laurent polynomial ring. The set $Λ_{R, n}$ of finitely supported functions $Z^{n} \to R$ with coefficientwise addition and convolution $(a b)_{γ} = \sum_{α + β = γ} a_{α} b_{β}$ (summed only over the finite supports of $a$ and $b$ ) is a commutative $R$ -algebra, free as an $R$ -module on the monomials $x^{α}$ ( $α \in Z^{n}$ ); each $x_{i}$ is a unit. For every commutative $R$ -algebra $A$ and units $u_{1}, \dots, u_{n} \in A$ there is a unique $R$ -algebra homomorphism $Λ_{R, n} \to A$ with $x_{i} \mapsto u_{i}$ . If $R$ is an integral domain, so is $Λ_{R, n}$ ; no domain assertion is made over a ring with zero divisors. Only finitely many variables are used; in applications to Coxeter systems with a finite generator set, $W$ may nevertheless be infinite.
Fraction field. If $R$ is an integral domain, then on pairs $(f, g)$ with $f, g \in Λ_{R, n}$ and $g \neq 0$ , the relation $(f, g) \sim (f^{'}, g^{'})$ iff $f g^{'} = f^{'} g$ is an equivalence relation compatible with $(f, g) + (f^{'}, g^{'}) : = (f g^{'} + f^{'} g, g g^{'})$ and $(f, g) (f^{'}, g^{'}) : = (f f^{'}, g g^{'})$ , and the quotient $K (Λ_{R, n})$ is a field containing $Λ_{R, n}$ through $f \mapsto [(f, 1)]$ . The same construction gives the fraction field of $R [x_{1}, \dots, x_{n}]$ .

Facts & Assumptions

Given: A commutative ring $R$ , an integer $n \geq 0$ , and a commutative $R$ -algebra $A$ .

[F1]

The polynomial ring $R [x]$ is the set of finitely supported functions $N \to R$ with coefficientwise addition and convolution $(a b)_{k} = \sum_{i + j = k} a_{i} b_{j}$ ; its elements are written $\sum_{i} a_{i} x^{i}$ , and the constant embedding sends $r$ to the sequence supported at $0$ with coefficient $r$ (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution).

[F2]

These operations make $R [x]$ a commutative ring and the constant map $R \to R [x]$ an injective unital ring homomorphism (Polynomial convolution makes $R [x]$ a commutative ring containing $R$ as its constant subring).

[F3]

A coefficient homomorphism and the image of $x$ determine a unique unital ring homomorphism $R [x] \to S$ ; it is ${ev}_{φ, s} (\sum_{i} a_{i} x^{i}) = \sum_{i} φ (a_{i}) s^{i}$ (Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism).

[F4]

$R [x_{1}, \dots, x_{n}]$ is defined by $R [x_{1}, \dots, x_{0}] : = R$ and $R [x_{1}, \dots, x_{n + 1}] : = R [x_{1}, \dots, x_{n}] [x_{n + 1}]$ , so all indeterminates commute (Polynomial rings in finitely many commuting indeterminates by iteration).

[F5]

If $R$ is an integral domain then so is $R [x_{1}, \dots, x_{n}]$ , including $n = 0$ (A polynomial ring in finitely many indeterminates over an integral domain is an integral domain).

[F6]

The free module $R^{(X)} = ⨁_{x \in X} R$ has standard basis $e_{x}$ and every element is uniquely a finite sum $\sum_{x \in F} r_{x} e_{x}$ (The free module on a set and its standard basis).

[F7]

A set map from a basis into a module extends uniquely to an $R$ -linear map (Universal property of the free module on a set).

[F8]

Finite sums in a commutative monoid are invariant under reindexing, split over disjoint unions and satisfy Fubini (Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule).

[F9]

An integral domain is a commutative ring with $1 \neq 0$ and no zero divisors: $a b = 0$ implies $a = 0$ or $b = 0$ (Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors, Commutative ring).

[F10]

Equivalence relations, classes and quotient sets; a relation between pairs is an equivalence relation when reflexive, symmetric and transitive (Equivalence relation, equivalence class, and the quotient set $A / \sim$ ).

[F11]

An $R$ -algebra carries a central unital structure map (Algebras over a commutative ring, central structure maps, and algebra homomorphisms, Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ ).

[F12]

The integers form a totally ordered commutative ring; in particular their addition is associative and commutative and their order is translation-invariant (The integers form a totally ordered ring).

[F13]

The units of a ring form a group, with integer powers defined by natural powers and inverses and satisfying $u^{a + b} = u^{a} u^{b}$ (The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring, Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ , Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute).

Proof

technique · direct

1.1givenF1F2F3F4F6F11algebra

Multivariate basis and universal property, by induction on $n$ . For $n = 0$ , $R [x_{1}, \dots, x_{0}] = R$ by [F4], the single monomial $x^{(0)} = 1$ is a basis of the free rank-one module $R$ by [F6], and the structure map of $A$ is the unique $R$ -algebra homomorphism $R \to A$ by [F11]. For the step, put $S : = R [x_{1}, \dots, x_{n}]$ , free over $R$ on the monomials $x^{α}$ by induction, and $R [x_{1}, \dots, x_{n + 1}] = S [x_{n + 1}]$ by [F4]: by [F1] and [F2] (applied over the coefficient ring $S$ ) an element of $S [x_{n + 1}]$ is a finitely supported function $k \mapsto s_{k}$ with $s_{k} \in S$ , each $s_{k} = \sum_{α} a_{α, k} x^{α}$ a finite $R$ -linear combination, so substituting the unique coefficient expansions gives a unique finite $R$ -linear combination of the monomials $x^{α} x_{n + 1}^{k}$ , and these therefore form an $R$ -basis indexed by $N^{n} \times N ≅ N^{n + 1}$ ; applying [F3] twice, a unital $R$ -algebra homomorphism $S [x_{n + 1}] \to A$ is exactly a unital $R$ -algebra homomorphism $S \to A$ together with an element $a \in A$ (the image of $x_{n + 1}$ ), which by induction is exactly images $a_{1}, \dots, a_{n + 1} \in A$ of the generators.

1.2givenF5

If $R$ is an integral domain, $R [x_{1}, \dots, x_{n}]$ is one by [F5].

1.3givenF6F8F11F12algebra

Construction of $Λ_{R, n}$ : let $Λ_{R, n} : = R^{(Z^{n})}$ be the free $R$ -module with standard basis the monomials $x^{α}$ by [F6], and define multiplication on the basis by $x^{α} x^{β} : = x^{α + β}$ , extended $R$ -bilinearly, so that $(a b)_{γ} = \sum_{\begin{array}{c} α \in supp a, β \in supp b \\ α + β = γ \end{array}} a_{α} b_{β}$ is a finite sum by [F8], and $supp (a b) \subseteq supp a + supp b$ is finite. The rule is closed and associative because $Z^{n}$ has associative, commutative coordinatewise addition by [F12] and reindexing the finite triple sum gives $((a b) c)_{γ} = (a (b c))_{γ} = \sum_{α + β + δ = γ} a_{α} b_{β} c_{δ}$ by [F8]; it is commutative because $α + β = β + α$ and $R$ is commutative; it is distributive over the coefficientwise addition inherited from [F6]; and the basis vector $x^{0}$ , coefficient $1_{R}$ at $0$ and $0$ elsewhere, is a two-sided identity. Hence $Λ_{R, n}$ is a commutative ring, it is an $R$ -algebra through $r \mapsto r x^{0}$ by [F11], it is free on the monomials by construction, and each $x_{i}$ is a unit since the monomial $x^{- e_{i}}$ with coefficient $1_{R}$ satisfies $x_{i} x^{- e_{i}} = x^{e_{i} - e_{i}} = x^{0} = 1$ .

1.4givenF6F7F13algebra

Universal property of $Λ_{R, n}$ : let $u_{1}, \dots, u_{n} \in A$ be units and for $α \in Z^{n}$ put $u^{α} : = u_{1}^{α_{1}} \dots u_{n}^{α_{n}}$ , negative exponents denoting powers of the inverses. Every element of $Λ_{R, n}$ is a unique finite sum $\sum_{α} a_{α} x^{α}$ by [F6], so $x^{α} \mapsto u^{α}$ extends uniquely to an $R$ -linear map $Φ : Λ_{R, n} \to A$ by [F7]; it is a unital ring homomorphism because $u^{α + β} = u^{α} u^{β}$ and $u^{0} = 1$ in $A$ by [F13], applied in the unit group of $A$ , which is abelian because $A$ is commutative, and it is the unique $R$ -algebra homomorphism with $x_{i} \mapsto u_{i}$ because the monomials span.

1.5givenF8F9F12algebra

Domain of $Λ_{R, n}$ : suppose $R$ is an integral domain. Order $Z^{n}$ lexicographically: distinct tuples are compared at their first differing coordinate. By [F12] the integer order is total, and adding the same tuple preserves that first differing coordinate and its strict comparison; hence this is a translation-invariant total order (for $n = 0$ , there is just the empty tuple). For nonzero $a, b \in Λ_{R, n}$ the finite supports $supp a, supp b$ are nonempty and have greatest elements $α, β$ . A coefficient $(a b)_{γ}$ with $γ > α + β$ is a sum of products $a_{α^{'}} b_{β^{'}}$ with $α^{'} + β^{'} = γ$ , and each such term has $α^{'} > α$ (then $a_{α^{'}} = 0$ ) or $β^{'} > β$ (then $b_{β^{'}} = 0$ ), since $α^{'} \leq α$ and $β^{'} \leq β$ would give $α^{'} + β^{'} \leq α + β < γ$ ; so $(a b)_{γ} = 0$ for $γ > α + β$ . For $γ = α + β$ every term indexed within the supports with $α^{'} \neq α$ has $α^{'} < α$ and then $β^{'} > β$ by strict order invariance, hence vanishes, and the remaining term is $a_{α} b_{β} \neq 0$ because $R$ has no zero divisors [F9]; so $(a b)_{α + β} \neq 0$ and $a b \neq 0$ . Since $1_{R} \neq 0$ in $R$ , the unit of $Λ_{R, n}$ differs from $0$ , so $Λ_{R, n}$ is an integral domain.

1.6givenF9F10algebra

Fraction field for a commutative integral domain $D$ : on $P : = {(f, g) : f, g \in D, g \neq 0}$ define $(f, g) \sim (f^{'}, g^{'})$ iff $f g^{'} = f^{'} g$ . This is reflexive and symmetric, and transitive: from $f g^{'} = f^{'} g$ and $f^{'} g^{''} = f^{''} g^{'}$ one gets $g^{'} (f g^{''}) = (f g^{'}) g^{''} = (f^{'} g) g^{''} = f^{'} (g g^{''}) = f^{'} (g^{''} g) = (f^{'} g^{''}) g = (f^{''} g^{'}) g = f^{''} (g^{'} g) = g^{'} (f^{''} g)$ , so cancellation of the nonzero $g^{'}$ in the domain gives $f g^{''} = f^{''} g$ . Sums $(f, g) + (f^{'}, g^{'}) = (f g^{'} + f^{'} g, g g^{'})$ and products $(f, g) (f^{'}, g^{'}) = (f f^{'}, g g^{'})$ have nonzero second components since $D$ has no zero divisors, and they respect $\sim$ : if $f g^{'} = f^{'} g$ and $h k^{'} = h^{'} k$ then $(f k + h g) (g^{'} k^{'}) = f k g^{'} k^{'} + h g g^{'} k^{'} = f^{'} k^{'} (g k) + h^{'} g^{'} (g k) = (f^{'} k^{'} + h^{'} g^{'}) (g k)$ and $f h g^{'} k^{'} = f^{'} g h^{'} k = f^{'} h^{'} g k$ , so the operations are well defined on the quotient set $D^{'} = P / \sim$ of [F10]. Writing $f / g$ for $[(f, g)]$ , addition of $f / g, h / k, l / m$ in either order gives $(f k m + h g m + l g k) / (g k m)$ , multiplication in either order gives $f h l / (g k m)$ , and distributivity gives $f (h m + l k) / (g k m)$ on both sides. Commutativity follows from that in $D$ ; $0 / 1$ and $1 / 1$ are the identities, and $(- f) / g$ is the additive inverse of $f / g$ . Thus $D^{'}$ is a commutative ring, with $0 / 1 \neq 1 / 1$ because $0 \neq 1$ in $D$ , and $D^{'}$ is a field: for a class $[(f, g)] \neq [(0, 1)]$ one has $f \neq 0$ , so that $(f, g) (g, f) = (f g, f g) \sim (1, 1)$ since $f g \cdot 1 = 1 \cdot f g$ , and hence $[(g, f)]$ is an inverse. Finally $f \mapsto [(f, 1)]$ is a unital ring homomorphism with kernel ${f : [(f, 1)] = [(0, 1)]} = {f : f = 0}$ , so it injects $D$ into $D^{'}$ .

2.1step 1.1step 1.2step 1.3step 1.4step 1.5step 1.6F5∎

Collecting the clauses: step 1.1 gives the monomial basis, the universal property and, with [F5] as used in step 1.2, the domain assertion of the multivariate clause; step 1.3 gives the ring structure, freeness and the unit property of $Λ_{R, n}$ , step 1.4 its unit-substitution universal property, and step 1.5 its domain assertion. If $R$ is an integral domain then $D : = Λ_{R, n}$ is a domain by step 1.5 and $R [x_{1}, \dots, x_{n}]$ is a domain by step 1.2, so step 1.6 applies to both and yields the fraction field $K (Λ_{R, n})$ and the fraction field of $R [x_{1}, \dots, x_{n}]$ with the embedding $f \mapsto [(f, 1)]$ , which completes all three parts of the claim.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-10-08Open item page →

Finite matrix and module preliminaries: right inverses, rank invariance, finite length and nilpotent trace

Statement

Let $k$ be a field.

Let $R$ be a commutative ring, $M$ a free $R$ -module of finite rank $n$ with basis $(e_{0}, \dots, e_{n - 1})$ , and let $v_{0}, \dots, v_{n - 1} \in M$ span $M$ . Then the coordinate matrix $A$ of $(v_{0}, \dots, v_{n - 1})$ in the basis $(e_{i})$ has a right inverse, $\det (A)$ is a unit of $R$ , and $(v_{0}, \dots, v_{n - 1})$ is a basis of $M$ . The determinant is formed for $n \geq 1$ ; for $n = 0$ the module is the zero module and the empty family is its basis, with no determinant clause.
If $F \subseteq E$ is a field extension and $A \in M_{m \times n} (F)$ , then the rank of $A$ over $F$ equals its rank over $E$ .
Let $B$ be a finite-dimensional $k$ -algebra and $N$ a finite-dimensional left $B$ -module. A proper submodule has strictly smaller $k$ -dimension, a maximal proper submodule of a nonzero finite-dimensional module exists by maximal dimension (no arbitrary Choice), and iterating produces a finite composition series of $N$ (Composition series and length of a module); in particular $N$ has finite length.
If $S_{1}, \dots, S_{m}$ are simple $B$ -modules and $N \subseteq S_{1} \oplus \dots \oplus S_{m}$ is a submodule, then $N$ is isomorphic to a direct sum of a subfamily of the $S_{i}$ , and $N$ has a complement in $S_{1} \oplus \dots \oplus S_{m}$ .
If $T : N \to N$ is $k$ -linear with $T^{r} = 0$ for some $r \geq 1$ , then $tr (T) = 0$ (The basis-independent trace of an endomorphism of a finite-dimensional vector space).

Facts & Assumptions

Given: A field $k$ ; a commutative ring $R$ ; a free $R$ -module $M$ of finite rank $n$ with basis $(e_{0}, \dots, e_{n - 1})$ and vectors $v_{0}, \dots, v_{n - 1}$ spanning $M$ ; a field extension $F \subseteq E$ and a matrix $A \in M_{m \times n} (F)$ ; a finite-dimensional $k$ -algebra $B$ and a finite-dimensional left $B$ -module $N$ ; simple $B$ -modules $S_{1}, \dots, S_{m}$ and a submodule $N \subseteq S_{1} \oplus \dots \oplus S_{m}$ ; a $k$ -linear $T : N \to N$ with $T^{r} = 0$ for some $r \geq 1$ .

[F1]

A free module on a set has a standard basis with unique finite expansions, and a family $(b_{x})$ is a basis of a module when every element is uniquely a finite $R$ -linear combination of the $b_{x}$ (The free module on a set and its standard basis).

[F2]

Matrix product and identity over a commutative ring: $(A B)_{i k} = \sum_{j < n} a_{i j} b_{j k}$ , and $I_{n}$ has entry $1$ on the diagonal and $0$ elsewhere (Entrywise ring-matrix operations, rectangular matrix products, identity matrices and transpose).

[F3]

Invertibility over a commutative ring means the existence of $C$ with $A C = I_{n} = C A$ , and the inverse is unique (Invertible square matrices and similarity over a commutative ring).

[F4]

The Leibniz determinant is $\det (A) = \sum_{σ \in S_{n}} sgn (σ) \prod_{i < n} a_{σ (i), i}$ and is multiplicative, $\det (A B) = \det (A) \det (B)$ (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix, For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ ).

[F5]

A matrix $A \in M_{n} (R)$ over a commutative ring is invertible if and only if $\det (A)$ is a unit of $R$ (A positive-sized square matrix over a commutative ring is invertible if and only if its determinant is a unit).

[F6]

Every family of nonempty sets indexed by a natural number $n$ has a choice function (Every natural-number-indexed list of nonempty sets has a choice function on its family of values).

[F7]

The row space $Row (A) \subseteq F^{n}$ is the span of the rows, and $rank (A) = \dim_{F} Row (A)$ , defined because a finite list spans it (Row space, column space, nullspace, row rank, column rank and matrix rank).

[F8]

An $R$ -algebra has a central structure map, and a left module over a $k$ -algebra is a $k$ -vector space through that map (Algebras over a commutative ring, central structure maps, and algebra homomorphisms, Unital left and right modules over a ring; unqualified module means left module).

[F9]

A subspace of a finite-dimensional space is finite-dimensional with no larger dimension; $\dim_{F} U = \dim_{F} V$ holds for $U \subseteq V$ exactly when $U = V$ ; every independent subset of a finite-dimensional space is contained in a basis (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ ).

[F10]

Ordered bases give unique coordinates, and the matrix $[T]_{B}^{C}$ of a linear map has as its $j$ -th column the coordinate column of $T (b_{j})$ (A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i < n} λ_{i} v_{i}$ for exactly one $λ : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis, Coordinate columns $[v]_{B}$ and matrices $[T]_{B}^{C}$ of linear maps relative to ordered bases).

[F11]

The trace of a square matrix is the sum of its diagonal entries, and the trace of an endomorphism of a finite-dimensional space is the trace of any matrix of it (The trace $tr (A)$ as the sum of the diagonal entries, The basis-independent trace of an endomorphism of a finite-dimensional vector space).

[F12]

For a subfield $F$ of $E$ , every $E$ -vector space is an $F$ -vector space by restricting scalars, and $F^{X}$ , $F^{n}$ are the function spaces with pointwise operations (Subfield: a subring of a field closed under inverses of its nonzero elements, and therefore a field with the restricted operations, A field is a vector space over itself, and over any subfield $K \subseteq F$ every $F$ -vector space is a $K$ -vector space by restricting the scalars, The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = {0, 1, \dots, n - 1}$ ).

[F13]

Submodules, quotient modules, simple modules, composition series and length (Submodule of a module, Quotient module $M / N$ with scalar multiplication on additive cosets, Simple module: a nonzero module with no proper nonzero submodule, Composition series and length of a module).

[F14]

The direct sum $⨁_{i = 1}^{m} S_{i}$ is the submodule of the product of tuples with finite support, with coordinate inclusions (The direct sum of an indexed family of modules).

Proof

technique · direct

1.1givenF1F6algebrachoose

Write the unique coordinate expansion $v_{j} = \sum_{i < n} a_{i j} e_{i}$ of each $v_{j}$ in the basis $(e_{i})$ ; this is the coordinate matrix $A = (a_{i j})$ , and identifying $M$ with $R^{n}$ through coordinate expansions of [F1] carries $v_{j}$ to the $j$ -th column $a_{j}$ of $A$ and is an $R$ -module isomorphism. Since the $v_{j}$ span $M$ , the columns $a_{j}$ span $R^{n}$ , so for every $i < n$ the set ${b \in R^{n} : A b = e_{i}}$ is nonempty; by [F6] applied to the function $i \mapsto {b \in R^{n} : A b = e_{i}}$ on $n$ there are $b_{0}, \dots, b_{n - 1} \in R^{n}$ with $A b_{i} = e_{i}$ , and $B \in M_{n} (R)$ denotes the matrix with $i$ -th column $b_{i}$ .

1.2givenF7F12algebra

Let $U \subseteq F^{n}$ be the $F$ -row space of $A$ and $(u_{0}, \dots, u_{r - 1})$ an $F$ -basis of it, so that $r = {rank}_{F} (A)$ by [F7]; by [F12] the field extension makes $E^{n}$ an $F$ -vector space containing $F^{n}$ , and viewing rows and the $u_{i}$ in $E^{n}$ , every row of $A$ is an $F$ -combination of the $u_{i}$ and hence an $E$ -combination, so every $E$ -combination of the rows is an $E$ -combination of the $u_{i}$ : each $u_{i}$ is itself an $F$ -combination of the rows and therefore lies in $U_{E}$ , so the list $(u_{0}, \dots, u_{r - 1})$ $E$ -spans the $E$ -row space $U_{E}$ of the matrix $A$ read over $E$ .

1.3givenF8F9algebra

A submodule of $N$ is closed under $k$ -scalars through the structure map of the $k$ -algebra $B$ of [F8], hence is a $k$ -linear subspace of $N$ ; consequently, if $L ⊊ L^{'} \subseteq N$ are submodules then $\dim_{k} L < \dim_{k} L^{'}$ by [F9].

1.4givenF13F14algebra

Claim 4 is proved by induction on $m$ . For $m = 0$ the direct sum is $0$ , so $N = 0$ is the empty direct sum with complement $0$ ; for $m = 1$ the only submodules of the simple module $S_{1}$ are $0$ and $S_{1}$ , the direct sums over the empty and the full subfamily, with complements $S_{1}$ and $0$ ; for $m \geq 2$ , write $S : = S_{1}$ and $M : = S_{2} \oplus \dots \oplus S_{m}$ , a direct sum of $m - 1$ simple modules, so that $S_{1} \oplus \dots \oplus S_{m} = S \oplus M$ with the coordinate projections as in [F14], and let $N \subseteq S \oplus M$ be a submodule.

1.5givenF9F10algebra

If $T : N \to N$ is $k$ -linear with $T^{r} = 0$ , put $K_{j} : = \ker (T^{j})$ for $j \geq 0$ , so that $K_{0} = 0$ and $0 = K_{0} \subseteq K_{1} \subseteq \dots \subseteq K_{r} = N$ is a chain of $k$ -subspaces of the finite-dimensional space $N$ , with $K_{j}$ finite-dimensional for every $j$ by [F9]; extend a basis of $K_{j}$ successively to a basis of $K_{j + 1}$ using the extension clause of [F9] and concatenate the successive blocks to an ordered basis $(w_{1}, \dots, w_{d})$ of $N$ in the sense of [F10].

2.1step 1.1F2F4algebra

By [F2] the $i$ -th column of $A B$ is $A b_{i} = e_{i}$ , so $A B = I_{n}$ ; by [F4] $\det (A) \det (B) = \det (A B) = \det (I_{n})$ , and the Leibniz expansion of $\det (I_{n})$ has vanishing products $\prod_{i < n} (I_{n})_{σ (i), i}$ unless $σ = id$ , so $\det (I_{n}) = 1$ for $n \geq 1$ and $\det (A) \det (B) = 1$ : the determinant $\det (A)$ is a unit of $R$ with inverse $\det (B)$ .

2.2step 1.2F7F12algebra

Suppose $\sum_{i < r} λ_{i} u_{i} = 0$ in $E^{n}$ with all $λ_{i} \in E$ . The coefficients span an $F$ -subspace $L \subseteq E$ that is spanned by the finitely many $λ_{i}$ ; discarding from that finite spanning list each vector lying in the span of those retained leaves an $F$ -basis $(μ_{1}, \dots, μ_{t})$ of $L$ , and $λ_{i} = \sum_{t} c_{i t} μ_{t}$ with $c_{i t} \in F$ . For each coordinate $k < n$ the equality gives $0 = \sum_{i < r} λ_{i} u_{i} (k) = \sum_{t} μ_{t} (\sum_{i < r} c_{i t} u_{i} (k))$ in $E$ ; the $μ_{t}$ are $F$ -independent in $E$ and the coefficients $\sum_{i < r} c_{i t} u_{i} (k)$ lie in $F$ , so they all vanish, that is $\sum_{i < r} c_{i t} u_{i} = 0$ in $F^{n}$ for every $t$ ; the $u_{i}$ are $F$ -independent, so every $c_{i t} = 0$ and hence every $λ_{i} = 0$ : the list $(u_{0}, \dots, u_{r - 1})$ is $E$ -independent.

2.3step 1.3F9choosealgebra

Let $N \neq 0$ . The $k$ -dimensions $\dim_{k} L$ of the proper submodules $L ⊊ N$ form a nonempty set of natural numbers, since the zero submodule is proper and has dimension $0$ , bounded above by $\dim_{k} N - 1$ by [F9]; let $d$ be its largest element and let $L$ be a proper submodule with $\dim_{k} L = d$ (one selection from a nonempty set of submodules). Then $L$ is maximal proper: if $L ⊊ L^{'} \subseteq N$ , step 1.3 gives $\dim_{k} L^{'} > \dim_{k} L = d$ , so $L^{'}$ is not a proper submodule with dimension at most $d$ , whence $L^{'} = N$ .

2.4step 1.4F13F14algebra

Case $N \cap S \neq 0$ of claim 4. Since $S$ is simple and $N \cap S$ is a nonzero submodule of it, $N \cap S = S$ , and for $n = s + u \in N$ with $s \in S$ , $u \in M$ one has $u = n - s \in N \cap M$ ; hence $N = S \oplus (N \cap M)$ with the sum direct because $S \cap M = 0$ . By the induction hypothesis applied to $N \cap M \subseteq M$ there are a subfamily $(S_{i})_{i \in J}$ , $J \subseteq {2, \dots, m}$ , with $N \cap M ≅ ⨁_{i \in J} S_{i}$ and a submodule $K^{'} \subseteq M$ with $(N \cap M) \cap K^{'} = 0$ and $(N \cap M) + K^{'} = M$ ; then $N ≅ S \oplus ⨁_{i \in J} S_{i}$ , and $K^{'}$ is a complement of $N$ in $S \oplus M$ , because $N + K^{'} = S + (N \cap M) + K^{'} = S \oplus M$ and an element of $N \cap K^{'}$ lies in $M$ , hence has zero $S$ -component and lies in $(N \cap M) \cap K^{'} = 0$ .

3.1step 2.1F1F3F5algebra

By [F5] the unit determinant of step 2.1 makes $A$ invertible, so [F3] gives $C$ with $C A = A C = I_{n}$ ; if $\sum_{j < n} λ_{j} a_{j} = 0$ in $R^{n}$ , then $0 = C (0) = C A λ = λ$ , so the columns $a_{j}$ of $A$ are linearly independent and, being spanning, they form a basis of $R^{n}$ ; applying the coordinate isomorphism to the preimages shows $(v_{0}, \dots, v_{n - 1})$ is a basis of $M$ , which proves claim 1 for $n \geq 1$ ; for $n = 0$ the module is $0$ with empty basis by [F1] and there is no determinant clause.

3.2step 1.2step 2.2F7algebra

The list $(u_{0}, \dots, u_{r - 1})$ is $E$ -independent by step 2.2 and $E$ -spans $U_{E}$ by step 1.2, so it is an $E$ -basis of $U_{E}$ ; therefore ${rank}_{E} (A) = \dim_{E} U_{E} = r = \dim_{F} U = {rank}_{F} (A)$ by [F7], which is claim 2.

3.3step 2.3F9F13algebra

Iteration for claim 3: starting from $N$ and repeatedly replacing a nonzero module by a maximal proper submodule, which exists by step 2.3, produces a strictly decreasing chain $N = N_{0} ⊋ N_{1} ⊋ \dots ⊋ N_{t} = 0$ of finite-dimensional submodules, because each $N_{i}$ is finite-dimensional by [F9] and each step strictly lowers the $k$ -dimension, so the process terminates after at most $\dim_{k} N$ steps. Reading the chain upwards gives $0 = N_{t} ⊊ \dots ⊊ N_{0} = N$ , and each factor $N_{i - 1} / N_{i}$ is simple: if it had a nonzero proper submodule $Q$ , the preimage ${x \in N_{i - 1} : x + N_{i} \in Q}$ would be a submodule strictly between $N_{i}$ and $N_{i - 1}$ , contradicting maximality of $N_{i}$ in $N_{i - 1}$ ; this is a finite composition series of $N$ in the sense of [F13], so $N$ has finite length.

3.4step 2.4F13F14algebra

Case $N \cap S = 0$ of claim 4. The projection $π : N \to M$ is then injective, and $π (N) \subseteq M$ is a submodule to which the induction hypothesis applies: there are a subfamily $(S_{i})_{i \in J}$ , $J \subseteq {2, \dots, m}$ , with $π (N) ≅ ⨁_{i \in J} S_{i}$ and a submodule $K \subseteq M$ with $π (N) \cap K = 0$ and $π (N) + K = M$ . For $u \in π (N)$ let $f (u) \in S$ be the $S$ -component of the unique $n \in N$ with $π (n) = u$ , so that this $n$ is $f (u) + u$ and $N = {f (u) + u : u \in π (N)}$ ; the projection and the $S$ -component map are $B$ -module homomorphisms; the inverse of the restricted projection respects addition and $B$ -scalars, so $f$ is a $B$ -module homomorphism. Every $x = s + u \in S \oplus M$ with $u = u_{1} + u_{2}$ , $u_{1} \in π (N)$ , $u_{2} \in K$ , equals $(f (u_{1}) + u_{1}) + ((s - f (u_{1})) + u_{2}) \in N + (S \oplus K)$ , and if $n = f (u) + u \in N \cap (S \oplus K)$ with $n = s^{'} + k$ , $k \in K$ , then projecting to $M$ gives $u = k$ , so $u \in π (N) \cap K = 0$ , $u = 0$ and $n = f (0) = 0$ by linearity of $f$ , whence $N \cap (S \oplus K) = 0$ ; therefore $S \oplus K$ is a complement of $N$ in $S \oplus M$ , and $N ≅ π (N) ≅ ⨁_{i \in J} S_{i}$ through the injective projection.

4.1step 1.4step 2.4step 3.4algebra

Cases 2.4 and 3.4 cover every submodule $N \subseteq S \oplus M$ according to whether $N \cap S$ is zero, and in both cases $N$ is isomorphic to a direct sum of a subfamily of $S_{1}, \dots, S_{m}$ and has a complement in $S_{1} \oplus \dots \oplus S_{m}$ ; with the case $m \leq 1$ of step 1.4 this proves claim 4 by induction on $m$ .

5.1step 1.5step 3.1step 3.2step 3.3step 4.1F10F11algebra∎

In the ordered basis of step 1.5, a basis vector $w \in K_{j}$ coming from the $j$ -th block satisfies $T (w) \in K_{j - 1}$ , which is the span of the blocks preceding $w$ ; the matrix $[T]$ of $T$ in this ordered basis of [F10] therefore has zeros on its diagonal, and $tr (T)$ is the trace of that matrix by [F11] and thus the sum of its diagonal entries, which is $0$ . This proves claim 5, and with steps 3.1, 3.2, 3.3, 4.1 all five claims are proved.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

The left regular module and its tensor powers detect linear and tensor identities

Statement

Let $R$ be a commutative ring and let $A$ be a unital $R$ -algebra (Algebras over a commutative ring, central structure maps, and algebra homomorphisms), with $A$ regarded as its left regular module.

For $a, b \in A$ one has $a = b$ if and only if $a \cdot 1_{A} = b \cdot 1_{A}$ .
For $n \geq 1$ the tensor power $A^{\otimes n}$ is a left $A$ -module via $a \cdot (a_{1} \otimes \dots \otimes a_{n}) = (a a_{1}) \otimes \dots \otimes a_{n}$ . If $a \cdot z = b \cdot z$ for all $z \in A^{\otimes n}$ , then $a = b$ ; in particular the single element $1_{A}^{\otimes n}$ detects equality. The analogous statement holds for the right regular structure. The empty tensor power $A^{\otimes 0} = R$ carries no canonical left $A$ -module structure here, so no $n = 0$ case is claimed.
For $n \geq 1$ , if $f, g : A^{\times n} \to M$ are $R$ -multilinear and agree on every $n$ -tuple, then the $R$ -linear maps $A^{\otimes n} \to M$ they induce (Finite iterated tensor products represent multilinear maps independently of parenthesization) agree; conversely, agreement of the induced maps gives agreement on all pure tensors $a_{1} \otimes \dots \otimes a_{n}$ .
Descent warning: if $π : A \to A / I$ is a quotient of $R$ -algebras (A ring homomorphism whose kernel contains a two-sided ideal factors uniquely through the quotient ring), the induced map $A^{\otimes n} \to (A / I)^{\otimes n}$ is surjective, but an identity between $R$ -linear maps out of $(A / I)^{\otimes n}$ may be checked on images of pure tensors of $A$ only after those maps have been shown to be well defined on the quotient; surjectivity alone is not descent.

Facts & Assumptions

Given: A commutative ring $R$ , a unital $R$ -algebra $A$ , an $R$ -module $M$ for claim 3, a two-sided ideal $I \subseteq A$ , and an integer $n \geq 1$ .

[F1]

A unital $R$ -algebra $A$ has a central unital structure map and is a left and right module over itself; for a left $A$ -module the action satisfies $1_{A} \cdot x = x$ and $a \cdot (b \cdot x) = (a b) \cdot x$ (Algebras over a commutative ring, central structure maps, and algebra homomorphisms, Unital left and right modules over a ring; unqualified module means left module).

[F2]

The tensor product $A \otimes_{R} \dots \otimes_{R} A$ exists with its balanced universal property, every element is a finite sum of elementary tensors, and an $R$ -multilinear map out of $A^{\times n}$ induces a unique $R$ -linear map out of the $n$ -fold tensor power, independently of parenthesization (The tensor product $M \otimes_{R} N$ from the additive group underlying the free $Z$ -module on $M \times N$ , elementary tensors, and finite tensor sums, Universal property of the tensor product for balanced maps into abelian groups, Finite iterated tensor products represent multilinear maps independently of parenthesization).

[F3]

Tensoring a surjective homomorphism is surjective (Tensoring is right exact).

[F4]

A ring homomorphism whose kernel contains a two-sided ideal factors uniquely through the quotient ring (A ring homomorphism whose kernel contains a two-sided ideal factors uniquely through the quotient ring, The quotient ring $R / I$ with $(r + I) (s + I) = r s + I$ ).

Proof

technique · direct

1.1givenF1

Claim 1: $1_{A}$ is a two-sided identity of the regular module, so $a = a \cdot 1_{A}$ and $b = b \cdot 1_{A}$ by [F1]; conversely $a \cdot 1_{A} = b \cdot 1_{A}$ is $a = b$ .

1.2givenF1F2algebra

Claim 2: for $a \in A$ the map $A^{\times n} \to A^{\otimes n}$ , $(a_{1}, \dots, a_{n}) \mapsto (a a_{1}) \otimes a_{2} \otimes \dots \otimes a_{n}$ , is $R$ -multilinear, so it induces an $R$ -linear map $λ_{a} : A^{\otimes n} \to A^{\otimes n}$ with $λ_{a} (a_{1} \otimes \dots \otimes a_{n}) = (a a_{1}) \otimes a_{2} \otimes \dots \otimes a_{n}$ by [F2]. On elementary tensors $λ_{a} λ_{b} = λ_{a b}$ , $λ_{a + b} = λ_{a} + λ_{b}$ and $λ_{1_{A}} = id$ , and since elementary tensors span [F2], these identities extend to all of $A^{\otimes n}$ , so $a \cdot z : = λ_{a} (z)$ makes $A^{\otimes n}$ a left $A$ -module [F1]. The $n$ -fold multiplication $a_{1} \otimes \dots \otimes a_{n} \mapsto a_{1} a_{2} \dots a_{n}$ is the $R$ -linear map $μ : A^{\otimes n} \to A$ induced by the $R$ -multilinear product [F1, F2], and $μ (λ_{a} (z)) = a μ (z)$ for elementary tensors and hence for all $z$ by linearity. If $a \cdot z = b \cdot z$ for all $z$ , then in particular $λ_{a} (1_{A}^{\otimes n}) = λ_{b} (1_{A}^{\otimes n})$ , that is $a \otimes 1_{A}^{\otimes (n - 1)} = b \otimes 1_{A}^{\otimes (n - 1)}$ ; applying $μ$ gives $a \cdot 1_{A} = b \cdot 1_{A}$ , and $a \cdot 1_{A} = a$ while $b \cdot 1_{A} = b$ by [F1], so $a = b$ : the single element $1_{A}^{\otimes n}$ detects equality. The right-handed statement is the mirror computation with $ρ_{a} (a_{1} \otimes \dots \otimes a_{n}) = a_{1} \otimes \dots \otimes (a_{n} a)$ .

1.3givenF2algebra

Claim 3: by [F2] the $R$ -multilinear map $f$ induces the unique $R$ -linear map $\overline{f} : A^{\otimes n} \to M$ with $\overline{f} (a_{1} \otimes \dots \otimes a_{n}) = f (a_{1}, \dots, a_{n})$ , and likewise for $g$ ; if $f = g$ as functions then $\overline{f}$ and $\overline{g}$ agree on every elementary tensor, hence on the whole tensor power, so $\overline{f} = \overline{g}$ . Conversely, if $\overline{f} = \overline{g}$ , then for every tuple $f (a_{1}, \dots, a_{n}) = \overline{f} (a_{1} \otimes \dots \otimes a_{n}) = \overline{g} (a_{1} \otimes \dots \otimes a_{n}) = g (a_{1}, \dots, a_{n})$ , so the multilinear maps agree on all tuples and the induced maps agree on all pure tensors.

1.4givenF3F4algebra

Claim 4: the quotient map $π : A \to A / I$ is a surjective $R$ -algebra homomorphism, and $π^{\otimes n} : = π \otimes \dots \otimes π : A^{\otimes n} \to (A / I)^{\otimes n}$ is surjective by iterated [F3]; it sends an elementary tensor $a_{1} \otimes \dots \otimes a_{n}$ to $π (a_{1}) \otimes \dots \otimes π (a_{n})$ . If $\overline{φ} : (A / I)^{\otimes n} \to Q$ is an $R$ -linear map out of the quotient tensor power, then to check an identity between two such maps on images of pure tensors of $A$ one must first know that each map is well defined on the quotient, equivalently, any proposed map $φ : A^{\otimes n} \to Q$ must annihilate $\ker (π^{\otimes n})$ , so that $φ = \overline{φ} \circ π^{\otimes n}$ for a well-defined map $\overline{φ}$ ; that is the descent obligation, and surjectivity of $π^{\otimes n}$ alone equips no map out of $A^{\otimes n}$ with a well-defined value on a class. For ring-level descent the quotient universal property [F4] is the tool: a homomorphism killing $I$ factors uniquely through $π$ .

2.1step 1.1step 1.2step 1.3step 1.4∎

Collecting: step 1.1 proves claim 1, step 1.2 proves both the module structure and the detection statement of claim 2 together with its right-handed mirror, step 1.3 proves both directions of claim 3, and step 1.4 records the descent warning of claim 4.

5 · Examples, counterexamples and false statements

None yet.

Tensor Coherence and Algebraic Descent

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Scalars, tensor powers, the empty tensor, opposite algebras and finite sums

Definition

Associator naturality, pentagon, unit triangle and symmetry hexagons on elementary tensors

Statement

Facts & Assumptions

Proof

Tensoring injections and the kernel of a tensor product of quotient maps over a field

Statement

Facts & Assumptions

Proof

Coefficient separation for an independent family of vectors, with explicit Choice assumptions

Statement

Facts & Assumptions

Proof

Remarks

Finite tensor duality and basis-independent coevaluation

Statement

Facts & Assumptions

Proof

Remarks

The free associative R-algebra on a set and descent of relations

Statement

Facts & Assumptions

Proof

Presentation base change and transport of explicit bases to commutative specializations

Statement

Facts & Assumptions

Proof

Multivariate polynomial and Laurent rings over commutative rings, domains and fraction fields

Statement

Facts & Assumptions

Proof

Finite matrix and module preliminaries: right inverses, rank invariance, finite length and nilpotent trace

Statement

Facts & Assumptions

Proof

The left regular module and its tensor powers detect linear and tensor identities

Statement

Facts & Assumptions

Proof

5 · Examples, counterexamples and false statements

Sources