Pipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

✓ 14 results · all verified · 13 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 1 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Hochschild Homology and Diagonal Koszul Resolutions

1 · Prerequisites

2 · Summary

This page develops Hochschild homology of unital algebras over a field. It constructs the two-sided bar resolution, identifies its tensor product over the enveloping algebra with Hochschild chains, and obtains the Tor description, coefficient functoriality, long exact sequences, and degree-zero coinvariants. For polynomial algebras, it proves that the diagonal Koszul complex resolves the regular bimodule and computes Hochschild homology with central bimodule coefficients under AC. The regular coefficient case gives the exterior-power formula and its internal grading.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableprecheck passjudge pass (gpt-6-sol)audited 2026-09-30Open item page →

Enveloping algebra and the bimodule–module dictionary

Definition

Let $k$ be a field and let $A$ be a unital associative $k$ -algebra (Algebras over a commutative ring, central structure maps, and algebra homomorphisms). Its enveloping algebra is

$A^{e} : = A \otimes_{k} A^{o p} .$

The multiplication from the tensor-product algebra structure (The tensor product of $R$ -algebras has multiplication $(a \otimes b) (a^{'} \otimes b^{'}) = a a^{'} \otimes b b^{'}$ ) and opposite-ring multiplication (The opposite ring $R^{o p}$ ) is

$(a \otimes b^{o p}) (c \otimes d^{o p}) = a c \otimes (d b)^{o p},$

with unit $1 \otimes 1^{o p}$ . A $k$ -central $A$ -bimodule is a bimodule with commuting actions ( $(S, R)$ -bimodules and commuting left and right scalar actions) whose induced scalar actions agree as required by Associative graded algebras, bimodules, and internal shifts.

For such a bimodule $M$ , the formula

$(a \otimes b^{o p}) m : = a m b$

defines a unital left $A^{e}$ -module. Indeed,

$((a \otimes b^{o p}) (c \otimes d^{o p})) m = (a c) m (d b) = a (c m d) b = (a \otimes b^{o p}) ((c \otimes d^{o p}) m),$

and the unit acts as $1_{A} m 1_{A} = m$ . Conversely, if $M$ is a left $A^{e}$ -module, set $a m : = (a \otimes 1) m$ and $m b : = (1 \otimes b^{o p}) m$ . The two actions are unital, associative, commute because the two tensor factors commute in $A^{e}$ , and are $k$ -central because the two copies of a scalar $λ \in k$ define the same element of $A^{e}$ . These constructions are inverse: $(a \otimes b^{o p}) m = a (m b) = a m b$ .

The same bimodule is a unital right $A^{e}$ -module by

$m (a \otimes b^{o p}) : = b m a .$

Indeed, for $x = a \otimes b^{o p}$ and $y = c \otimes d^{o p}$ ,

$(m x) y = d (b m a) c = (d b) m (a c) = m ((a c) \otimes (d b)^{o p}) = m (x y),$

and $m (1 \otimes 1^{o p}) = m$ .

For the right-module converse, define $a m : = m (1 \otimes a^{o p})$ and $m b : = m (b \otimes 1)$ . Right associativity gives the left and right $A$ -module laws, and the two actions commute because $(1 \otimes a^{o p})$ commutes with $(b \otimes 1)$ . They are $k$ -central for the same scalar-balancing reason. The constructions are inverse since

$m ((a \otimes 1) (1 \otimes b^{o p})) = (m a) (1 \otimes b^{o p}) = b (m a) = b m a .$

Thus $k$ -central $A$ -bimodules, left $A^{e}$ -modules, and right $A^{e}$ -modules have the same objects and morphisms under these dictionaries. In particular, the regular bimodule $A$ corresponds to $A$ with left action $(a \otimes b^{o p}) c = a c b$ and right action $c (a \otimes b^{o p}) = b c a$ .

DefinitionDefinition: Literature-sourcedProof: Not applicableprecheck passjudge pass (gpt-6-sol)audited 2026-09-30Open item page →

The augmented two-sided bar complex

Definition

Let $k$ be a field and $A$ a unital associative $k$ -algebra. Set $A^{\otimes_{k} 0} : = k$ and, for every $n \geq 0$ , set

${Bar}_{n} (A) : = A \otimes_{k} A^{\otimes_{k} n} \otimes_{k} A .$

Write a pure tensor as $a_{0} \otimes \dots \otimes a_{n + 1}$ . For $n \geq 1$ define

$d_{n} : = \sum_{r = 0}^{n} (- 1)^{r} μ_{r, r + 1} : {Bar}_{n} (A) \to {Bar}_{n - 1} (A),$

where $μ_{r, r + 1}$ multiplies slots $r$ and $r + 1$ and leaves the other slots in order. The augmentation is

$ε : = μ : {Bar}_{0} (A) = A \otimes_{k} A \to A, a_{0} \otimes a_{1} \mapsto a_{0} a_{1} .$

In degree one, $d_{1} (a_{0} \otimes a_{1} \otimes a_{2}) = a_{0} a_{1} \otimes a_{2} - a_{0} \otimes a_{1} a_{2} .$

The empty middle tensor convention makes ${Bar}_{0} (A) = A \otimes_{k} A$ ; there is no unaugmented differential out of degree zero.

Using Enveloping algebra and the bimodule–module dictionary, each term has the following left and right $A^{e}$ -module structures, considered separately:

$(c \otimes d^{o p}) \cdot (a_{0} \otimes \dots \otimes a_{n + 1}) = c a_{0} \otimes a_{1} \otimes \dots \otimes a_{n + 1} d,$

$(a_{0} \otimes \dots \otimes a_{n + 1}) \cdot (c \otimes d^{o p}) = d a_{0} \otimes a_{1} \otimes \dots \otimes a_{n + 1} c .$

Every adjacent-multiplication face is linear for each of these module structures: at the first and last faces this is associativity, and at internal faces the outer factors are unchanged. The augmentation is linear on both sides, since $ε (c a_{0} \otimes a_{1} d) = c (a_{0} a_{1}) d$ and $ε (d a_{0} \otimes a_{1} c) = d (a_{0} a_{1}) c$ . This item specifies the bar terms, maps, and outer actions; the asserted zero-composite identities are addressed by the following bar-boundary lemma.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30Open item page →

The bar boundary squares to zero and is augmented

Statement

For the bar maps of The augmented two-sided bar complex, $d_{n - 1} d_{n} = 0$ for $n \geq 2$ and $ε d_{1} = 0$ . Thus the augmented bar sequence is a chain complex of both left and right $A^{e}$ -modules.

Facts & Assumptions

Given: A unital associative algebra $A$ over a field $k$ , with the bar terms, faces, augmentation and outer $A^{e}$ -actions defined in the cited item.

[F1]

The maps are the alternating sum of adjacent-slot multiplication faces (The augmented two-sided bar complex).

[F2]

Every adjacent-multiplication face is linear on both $A^{e}$ sides (The augmented two-sided bar complex).

[F3]

The augmentation is multiplication and is linear on both $A^{e}$ sides (The augmented two-sided bar complex).

Proof

technique · direct

For $n \geq 1$ , write $\partial_{i}^{(n)}$ for the face that multiplies slots $i$ and $i + 1$ , so $d_{n} = \sum_{i = 0}^{n} (- 1)^{i} \partial_{i}^{(n)}$ .

1.1F1givenalgebra

If $0 \leq i < j - 1$ , the two faces multiply disjoint pairs of slots; doing the later one first and reindexing it by one gives $\partial_{i}^{(n - 1)} \partial_{j}^{(n)} = \partial_{j - 1}^{(n - 1)} \partial_{i}^{(n)}$ . The products are independent and retain their order, so the identity holds on the tensor terms.

1.2F1givenalgebra

If $j = i + 1$ , both composites multiply the consecutive triple $a_{i}, a_{i + 1}, a_{i + 2}$ into one slot, giving $a_{i} (a_{i + 1} a_{i + 2}) = (a_{i} a_{i + 1}) a_{i + 2}$ by associativity; thus the same face identity holds for every $i < j$ .

1.3F3givenalgebra

In degree one, $ε d_{1} (a_{0} \otimes a_{1} \otimes a_{2}) = (a_{0} a_{1}) a_{2} - a_{0} (a_{1} a_{2}) = 0$ by associativity, so $ε d_{1} = 0$ .

2.1step 1.1step 1.2F1algebra

In the double sum for $d_{n - 1} d_{n}$ , terms indexed by $i < j$ pair with $(j - 1, i)$ , exactly the terms with first index at least the second. Steps 1.1 and 1.2 identify the composites, and $(- 1)^{i + j} = - (- 1)^{(j - 1) + i}$ , so every term cancels and $d_{n - 1} d_{n} = 0$ for every $n \geq 2$ .

3.1step 2.1step 1.3F2F3algebra∎

By [F2] and [F3], all these maps are linear for both outer $A^{e}$ actions; steps 2.1 and 1.3 therefore give the augmented chain-complex identities in both module categories.

Remark

For $0 \leq i < j - 1$ , the bar faces multiply disjoint adjacent pairs; their composites agree after the later face is reindexed by one. For $j = i + 1$ , associativity on the overlapping triple gives the same face identity. These are the internal adjacent-multiplication cases used in the proof above.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30Open item page →

The two-sided bar complex is a projective $A^{e}$ -resolution

Statement

Assume the Axiom of Choice (AC). Let $k$ be a field and $A$ a unital associative $k$ -algebra. The augmented two-sided bar complex $\dots ⟶ {Bar}_{1} (A) ⟶ {Bar}_{0} (A) \overset{ε}{\to} A$ is a projective resolution of $A$ both as a right and as a left $A^{e} = A \otimes_{k} A^{o p}$ -module. The contraction below is $k$ -linear; it is not asserted to be $A^{e}$ -linear.

Facts & Assumptions

Given: AC, a field $k$ , and a unital associative $k$ -algebra $A$ .

[F1]

The bar terms, adjacent-multiplication differential, and multiplication augmentation are as defined in The augmented two-sided bar complex.

[F2]

Each bar term has the separate outer left and right $A^{e}$ -actions specified in The augmented two-sided bar complex.

[F3]

The maps are linear for both outer actions and satisfy $d_{n - 1} d_{n} = 0$ and $ε d_{1} = 0$ (The bar boundary squares to zero and is augmented).

[F4]

The regular bimodule $A$ has the left and right $A^{e}$ -actions specified by the enveloping-algebra dictionary (Enveloping algebra and the bimodule–module dictionary).

[F5]

Under AC every vector space has a basis (Every vector space has a basis).

[F6]

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

[F7]

Tensor products commute with arbitrary direct sums in either variable (Tensor products commute with arbitrary direct sums).

[F8]

$k \otimes_{k} V ≅ V ≅ V \otimes_{k} k$ (The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ ).

[F9]

A module with a basis indexed by $X$ is isomorphic to the free module $R^{(X)}$ (The free module on a set and its standard basis).

[F10]

Under AC every free module is projective (Free modules are projective, with the exact choice boundary).

[F11]

AC means every family of nonempty sets has a choice function (The Axiom of Choice).

[F12]

A right $E$ -module is regarded as a left $E^{o p}$ -module by $r^{o p} m : = m r$ ; conversely a left $E^{o p}$ -module gives a right $E$ -module (Unital left and right modules over a ring; unqualified module means left module, The opposite ring $R^{o p}$ ).

[F13]

For every ring $E$ , the category of left $E$ -modules is abelian (Modules over a ring form an abelian category).

[F14]

A projective resolution is an exact augmented complex whose terms are projective (Projective resolutions in an abelian category).

[F15]

A multilinear prescription on finitely many tensor factors induces a linear map from their tensor product (Finite iterated tensor products represent multilinear maps independently of parenthesization).

[F16]

Kernels and images of module homomorphisms are the usual kernel and image submodules (Module homomorphism and isomorphism, kernel, image and cokernel).

Proof

technique · direct

1.1F1F8F15givenalgebra

Define $h_{- 1} : A \to {Bar}_{0} (A)$ by $h_{- 1} (a) = 1 \otimes a$ , and for $n \geq 0$ define $h_{n} (a_{0} \otimes \dots \otimes a_{n + 1}) = 1 \otimes a_{0} \otimes \dots \otimes a_{n + 1} .$ The formulas are $k$ -multilinear, so [F15] makes them well-defined $k$ -linear maps. In $d_{n + 1} h_{n}$ , the first face is the identity term $a_{0} \otimes \dots \otimes a_{n + 1}$ . Every later face, with index $r \geq 1$ , is $- h_{n - 1}$ applied to the face of index $r - 1$ in $d_{n}$ , since $(- 1)^{r} = - (- 1)^{r - 1}$ . For $n = 0$ , this gives $d_{1} h_{0} (a_{0} \otimes a_{1}) = a_{0} \otimes a_{1} - 1 \otimes a_{0} a_{1}$ , while $h_{- 1} ε (a_{0} \otimes a_{1}) = 1 \otimes a_{0} a_{1}$ . For $n = 1$ , $d_{2} h_{1} (a_{0} \otimes a_{1} \otimes a_{2}) = a_{0} \otimes a_{1} \otimes a_{2} - 1 \otimes a_{0} a_{1} \otimes a_{2} + 1 \otimes a_{0} \otimes a_{1} a_{2}$ , and $h_{0} d_{1} (a_{0} \otimes a_{1} \otimes a_{2}) = 1 \otimes a_{0} a_{1} \otimes a_{2} - 1 \otimes a_{0} \otimes a_{1} a_{2}$ . In general the same opposite-sign pairing yields $d_{n + 1} h_{n} + h_{n - 1} d_{n} = 1 (n \geq 0),$ where $d_{0} : = ε$ , and $ε h_{- 1} = 1_{A}$ . If $A = k$ , [F8] identifies every bar term with $k$ and every face with the identity, so $d_{n} = (\sum_{r = 0}^{n} (- 1)^{r}) 1_{k}$ is zero for odd $n$ and the identity for even $n$ .

1.2F1F2F5F6F7F8F9F11F12F15givenalgebra

By the assumed AC [F11] and the basis theorem [F5], choose a $k$ -basis $B$ of $A$ . For every $n \geq 1$ , [F6] applied inductively gives the basis of $A^{\otimes_{k} n}$ consisting of tensors $b_{1} \otimes \dots \otimes b_{n}$ with $b_{i} \in B$ . For $n = 0$ , use the basis ${1_{k}}$ of $A^{\otimes_{k} 0} = k$ . Let $B_{n}$ denote these basis index sets. Using [F7]–[F9], $A^{\otimes_{k} n} \otimes_{k} A^{e} ≅ ⨁_{B_{n}} (k \otimes_{k} A^{e}) ≅ (A^{e})^{(B_{n})}$ as right $A^{e}$ -modules, where the canonical map on pure tensors is $((a_{1} \otimes \dots \otimes a_{n}) \otimes (a \otimes b^{o p})) ⟼ b \otimes a_{1} \otimes \dots \otimes a_{n} \otimes a .$ For $n = 0$ , it sends $1_{k} \otimes (a \otimes b^{o p})$ to $b \otimes a$ . The inverse extracts the middle tensor and the two outer factors; multilinearity makes both maps well-defined by [F15]. By [F2], multiplication by $c \otimes d^{o p}$ sends the outer factors to $d b$ and $a c$ on each side, so this isomorphism respects the right action. By [F12], this right free module is a free left $(A^{e})^{o p}$ -module.

2.1step 1.2F1F2F6F7F8F9F15givenalgebra

Similarly, $A^{e} \otimes_{k} A^{\otimes_{k} n} ≅ ⨁_{B_{n}} (A^{e} \otimes_{k} k) ≅ (A^{e})^{(B_{n})}$ as left $A^{e}$ -modules, by the map $((a \otimes b^{o p}) \otimes (a_{1} \otimes \dots \otimes a_{n})) ⟼ a \otimes a_{1} \otimes \dots \otimes a_{n} \otimes b .$ For $n = 0$ , it sends $(a \otimes b^{o p}) \otimes 1_{k}$ to $a \otimes b$ . The inverse extracts the two outer factors and the middle tensor. Left multiplication by $c \otimes d^{o p}$ sends those outer factors to $c a$ and $b d$ on both sides, by [F2], and [F15] makes the maps well-defined. Thus, using the separate outer actions in [F1]–[F2] and the basis in step 1.2, every bar term is free on both sides.

2.2step 1.1F3F4F12F16givenalgebra

If $z \in \ker ε$ , step 1.1 gives $z = d_{1} h_{0} (z)$ . If $n \geq 1$ and $z \in \ker d_{n}$ , it gives $z = d_{n + 1} h_{n} (z)$ . Conversely, each image lies in the next kernel by [F3]. Also $ε h_{- 1} = 1_{A}$ , so $ε$ is onto. The augmented bar complex is therefore exact as a complex of $k$ -vector spaces. Since each differential and the augmentation are $A^{e}$ -linear by [F3] and the target actions are those of [F4], these elementwise kernel-image equalities are exactness in both module categories by [F12] and [F16]. The contraction need not be $A^{e}$ -linear.

3.1step 1.2step 2.1F10F11F12given

By the assumed AC [F11], each free module in steps 1.2 and 2.1 is projective by [F10], applying that theorem to the ring $A^{e}$ for left modules and $(A^{e})^{o p}$ for right modules via [F12]. Therefore every bar term is projective in both module categories.

4.1

By [F13], the left $A^{e}$ -module category and the left $(A^{e})^{o p}$ -module category are abelian; by [F12] the latter is the right $A^{e}$ -module category. Steps 2.2 and 3.1 give exactness and termwise projectivity in each category. Thus [F14] makes $Bar (A) \to A$ a projective resolution on both sides. AC is used to obtain a basis of $A$ and for projectivity of free modules with arbitrary basis; the contracting homotopy and exactness calculation are choice-free. [step 2.2, step 3.1, F11, F12, F13, F14, given] $□$

DefinitionDefinition: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-30Open item page →

Hochschild chains and Hochschild homology with coefficients

Definition

Let $k$ be a field, $A$ a unital associative $k$ -algebra, and $M$ a $k$ -central $A$ -bimodule (Enveloping algebra and the bimodule–module dictionary). For $n \geq 1$ put

$C_{n} (A, M) : = M \otimes_{k} A^{\otimes_{k} n},$

and put $C_{0} (A, M) : = M$ , using the canonical tensor-unit identification $M \otimes_{k} k ≅ M$ . For $n \geq 1$ and $0 \leq i \leq n$ , define the faces on elementary tensors by

$δ_{i}^{(n)} (m \otimes a_{1} \otimes \dots \otimes a_{n}) = {\begin{cases} (m a_{1}) \otimes a_{2} \otimes \dots \otimes a_{n}, & i = 0, \\ m \otimes a_{1} \otimes \dots \otimes (a_{i} a_{i + 1}) \otimes \dots \otimes a_{n}, & 0 < i < n, \\ (a_{n} m) \otimes a_{1} \otimes \dots \otimes a_{n - 1}, & i = n . \end{cases}$

These are well-defined $k$ -linear maps by the multilinear universal property of the finite tensor product. Set $C_{- 1} (A, M) = 0$ and $b_{0} = 0$ . The Hochschild boundary is

$b_{n} : = \sum_{i = 0}^{n} (- 1)^{i} δ_{i}^{(n)} : C_{n} (A, M) ⟶ C_{n - 1} (A, M) (n \geq 1) .$

In particular, $b_{1} (m \otimes a) = m a - a m$ . For $n \geq 2$ the faces satisfy

$δ_{i}^{(n - 1)} δ_{j}^{(n)} = δ_{j - 1}^{(n - 1)} δ_{i}^{(n)} (0 \leq i < j \leq n),$

so the alternating-sum boundaries satisfy $b_{n - 1} b_{n} = 0$ for $n \geq 2$ ; $b_{0} b_{1} = 0$ because $b_{0} = 0$ . Thus $C_{∙} (A, M)$ is a chain complex of $k$ -modules. Its $n$ th homology object is the Hochschild homology with coefficients in $M$ ,

$H H_{n} (A, M) : = H_{n} (C_{∙} (A, M)) .$

Facts & Assumptions

Given: A field $k$ , a unital associative $k$ -algebra $A$ , and a $k$ -central $A$ -bimodule $M$ .

[F1]

The left and right $A$ -actions on $M$ commute, and their scalar actions agree because $M$ is $k$ -central (Enveloping algebra and the bimodule–module dictionary).

[F2]

A finite tensor product over a commutative ring represents multilinear maps into a module, so a multilinear face formula induces a unique linear map on the tensor product (Finite iterated tensor products represent multilinear maps independently of parenthesization).

[F3]

The canonical map $M \otimes_{k} k \to M$ , $m \otimes λ \mapsto m λ$ , is an isomorphism with inverse $m \mapsto m \otimes 1$ (The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ ).

[F4]

Disjoint adjacent multiplication faces satisfy the reindexed face identity (The bar boundary squares to zero and is augmented).

[F5]

Overlapping adjacent multiplication faces satisfy the face identity by associativity (The bar boundary squares to zero and is augmented).

[F6]

The homology object $H_{n} (C)$ is defined when $C_{∙}$ is a chain complex (Homology object of a chain complex).

Proof

technique · direct

1.1F2F3givenalgebra

Each endpoint formula is multilinear because the bimodule actions are $k$ -bilinear, and each interior formula is multilinear because multiplication in $A$ is $k$ -bilinear; hence [F2] induces the displayed $k$ -linear face maps and every face sends a zero input to zero. The tensor-unit isomorphism [F3] identifies the degree-zero term with $M$ .

1.2givenalgebra

The degree-zero boundary is zero by definition, while in degree one the two faces are $δ_{0}^{(1)} (m \otimes a) = m a$ and $δ_{1}^{(1)} (m \otimes a) = a m$ , so $b_{1} (m \otimes a) = m a - a m$ and $b_{0} b_{1} = 0$ .

1.3F4F5givenalgebra

If $1 \leq i < j \leq n - 1$ , both faces are internal adjacent multiplications among the $A$ -slots. For disjoint slots their operations commute and reindex as in [F4]; for overlapping slots the two composites multiply the same triple, and associativity gives [F5]. Thus $δ_{i}^{(n - 1)} δ_{j}^{(n)} = δ_{j - 1}^{(n - 1)} δ_{i}^{(n)}$ for all such internal pairs.

1.4F1givenalgebra

For the adjacent first pair $i = 0, j = 1$ , the two composites have first coefficients $(m a_{1}) a_{2}$ and $m (a_{1} a_{2})$ , which agree by the right module law. For $i = n - 1, j = n$ , they have first coefficients $a_{n - 1} (a_{n} m)$ and $(a_{n - 1} a_{n}) m$ , which agree by the left module law. For the pair $i = 0, j = n$ , they have first coefficients $(a_{n} m) a_{1}$ and $a_{n} (m a_{1})$ , which agree because the left and right actions commute by [F1]. The untouched slots agree in their original order in all three cases.

1.5givenalgebra

The remaining pairs with one endpoint face and one internal face act on disjoint data: for $i = 0$ and $2 \leq j < n$ , the first face acts on $m, a_{1}$ while the other multiplies $a_{j}, a_{j + 1}$ ; for $1 \leq i < n - 1$ and $j = n$ , the internal face multiplies $a_{i}, a_{i + 1}$ while the last face acts by $a_{n}$ on $m$ . In either order the same multiplication/action is applied to each of these disjoint slots, with the same remaining tensor factors and order, proving the face identity. Together with 1.3 and 1.4 this covers every $0 \leq i < j \leq n$ .

1.6F2F3givenalgebra

In the degenerate case $A = k$ , the canonical tensor-unit identifications turn each face into the identity on $M$ ; hence $b_{n} = \sum_{i = 0}^{n} (- 1)^{i} {id}_{M}$ , which is the identity for even $n$ and zero for odd $n$ . Consecutive composites therefore vanish in this case too.

2.1step 1.3step 1.4step 1.5algebra

Expand $b_{n - 1} b_{n}$ as the sum of terms $(- 1)^{i + j} δ_{i}^{(n - 1)} δ_{j}^{(n)}$ . For every $i < j$ , the face identity from 1.3--1.5 pairs this term with the term indexed by $(j - 1, i)$ ; their composites agree and their signs are opposite because $(- 1)^{i + j} = - (- 1)^{(j - 1) + i}$ . Every term with first index $r \geq s$ is uniquely such a partner, obtained from $i = s, j = r + 1$ ; thus every term occurs in exactly one pair. Hence $b_{n - 1} b_{n} = 0$ for $n \geq 2$ ; the case $n = 1$ was checked in 1.2.

3.1step 1.1step 1.2step 2.1F6∎

The maps are $k$ -linear by 1.1 and their consecutive composites vanish by 1.2 and 2.1, so they form a chain complex. The definition of homology applies by [F6], giving $H H_{n} (A, M) = H_{n} (C_{∙} (A, M))$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30Open item page →

Hochschild chains are bar tensor chains

Statement

Let $k$ be a field, $A$ a unital associative $k$ -algebra, and $M$ a $k$ -central $A$ -bimodule. For every $n \geq 0$ , define $Φ_{n} : {Bar}_{n} (A) \otimes_{A^{e}} M ⟶ C_{n} (A, M), (a_{0} \otimes \dots \otimes a_{n + 1}) \otimes m ⟼ (a_{n + 1} m a_{0}) \otimes a_{1} \otimes \dots \otimes a_{n}$ . The family $(Φ_{n})_{n \geq 0}$ is an isomorphism of chain complexes, natural in the $k$ -central bimodule $M$ , from ${Bar}_{∙} (A) \otimes_{A^{e}} M$ to the Hochschild chain complex $C_{∙} (A, M)$ . For $n = 0$ , the target is $C_{0} (A, M) = M$ and the formula is $Φ_{0} ((a_{0} \otimes a_{1}) \otimes m) = a_{1} m a_{0}$ . No projectivity assumption on $M$ is needed.

Facts & Assumptions

Given: A field $k$ , a unital associative $k$ -algebra $A$ , and a $k$ -central $A$ -bimodule $M$ .

[F1]

The two-sided bar term is ${Bar}_{n} (A) = A \otimes_{k} A^{\otimes_{k} n} \otimes_{k} A$ , with differential the alternating sum of adjacent-multiplication faces (The augmented two-sided bar complex).

[F2]

Its right $A^{e}$ -action is $(a_{0} \otimes \dots \otimes a_{n + 1}) \cdot (c \otimes d^{o p}) = d a_{0} \otimes a_{1} \otimes \dots \otimes a_{n + 1} c$ (The augmented two-sided bar complex).

[F3]

The $k$ -central bimodule $M$ is a left $A^{e}$ -module by $(c \otimes d^{o p}) m = c m d$ (Enveloping algebra and the bimodule–module dictionary).

[F4]

The Hochschild chain terms are $C_{n} (A, M) = M \otimes_{k} A^{\otimes_{k} n}$ for $n \geq 1$ , and $C_{0} (A, M) = M$ (Hochschild chains and Hochschild homology with coefficients).

[F5]

The Hochschild face maps have first and last module-action faces and internal adjacent-multiplication faces (Hochschild chains and Hochschild homology with coefficients).

[F6]

A balanced map from a right module and a left module into an abelian group induces a unique homomorphism from their tensor product (Universal property of the tensor product for balanced maps into abelian groups).

[F7]

A multilinear map on finitely many $k$ -module factors induces a unique linear map from their iterated tensor product (Finite iterated tensor products represent multilinear maps independently of parenthesization).

[F8]

The tensor unit maps $k \otimes_{k} V \to V$ and $V \otimes_{k} k \to V$ are isomorphisms (The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ ).

[F9]

The Hochschild boundary $b_{n}$ is the alternating sum of its face maps (Hochschild chains and Hochschild homology with coefficients).

[F10]

The enveloping algebra is $A^{e} = A \otimes_{k} A^{o p}$ (Enveloping algebra and the bimodule–module dictionary).

Proof

technique · direct

1.1F1F2F3F4F6F7givenalgebra

For $n \geq 0$ , define on pure tensors $f_{n} (a_{0} \otimes \dots \otimes a_{n + 1}, m) = (a_{n + 1} m a_{0}) \otimes a_{1} \otimes \dots \otimes a_{n},$ where at $n = 0$ the value is $a_{1} m a_{0} \in M$ . The formula is $k$ -multilinear in the $n + 2$ algebra slots and $m$ , so [F7] gives a bilinear map $f_{n} : {Bar}_{n} (A) \times M \to C_{n} (A, M)$ . It is balanced over $A^{e}$ : for $z = a_{0} \otimes \dots \otimes a_{n + 1}$ , $f_{n} (z \cdot (c \otimes d^{o p}), m) = (a_{n + 1} c) m d a_{0} \otimes a_{1} \otimes \dots \otimes a_{n} = a_{n + 1} (c m d) a_{0} \otimes a_{1} \otimes \dots \otimes a_{n} = f_{n} (z, (c \otimes d^{o p}) m),$ using [F2], [F3], and associativity. By [F6] it induces $Φ_{n}$ with the stated formula. In degree zero this is precisely $Φ_{0} ((a_{0} \otimes a_{1}) \otimes m) = a_{1} m a_{0}$ .

1.2F1F4F7givenalgebra

Define $Ψ_{n}$ on pure Hochschild tensors by $Ψ_{n} (m \otimes a_{1} \otimes \dots \otimes a_{n}) = (1 \otimes a_{1} \otimes \dots \otimes a_{n} \otimes 1) \otimes_{A^{e}} m .$ This prescription is $k$ -multilinear and hence defines a linear map by [F7]. At $n = 0$ , set $Ψ_{0} (m) = (1 \otimes 1) \otimes_{A^{e}} m$ , consistent with $C_{0} (A, M) = M$ .

1.3F1F4F5F9givenalgebra

For $n \geq 1$ , the bar face $r = 0$ becomes the first Hochschild face, because $a_{n + 1} m (a_{0} a_{1}) = (a_{n + 1} m a_{0}) a_{1}$ . Each internal face $1 \leq r < n$ keeps the coefficient $a_{n + 1} m a_{0}$ and multiplies the same adjacent pair $a_{r}, a_{r + 1}$ . The last bar face $r = n$ becomes the cyclic face because $(a_{n} a_{n + 1}) m a_{0} = a_{n} (a_{n + 1} m a_{0})$ . These faces have the same alternating sign $(- 1)^{r}$ , so $Φ_{n - 1} (d_{n} \otimes 1_{M}) = b_{n} Φ_{n}$ . When $n = 1$ , there are no internal faces: applying $Φ_{0}$ to $d_{1} \otimes 1_{M}$ gives $a_{2} m (a_{0} a_{1}) - (a_{1} a_{2}) m a_{0}$ , equal to $(a_{2} m a_{0}) a_{1} - a_{1} (a_{2} m a_{0}) = b_{1} Φ_{1}$ by associativity. At $n = 0$ both outgoing differentials are zero.

2.1step 1.1step 1.2F2F3F6givenalgebra

On a pure Hochschild tensor, $Φ_{n} Ψ_{n}$ is the identity because the outer units act trivially on $M$ . Conversely, for $z = a_{0} \otimes \dots \otimes a_{n + 1}$ , [F2] gives $(1 \otimes a_{1} \otimes \dots \otimes a_{n} \otimes 1) \cdot (a_{n + 1} \otimes a_{0}^{o p}) = z$ , and [F3] gives $(a_{n + 1} \otimes a_{0}^{o p}) m = a_{n + 1} m a_{0}$ . The balancing relation in $\otimes_{A^{e}}$ therefore gives $Ψ_{n} Φ_{n} (z \otimes m) = z \otimes m$ . The same calculation at $n = 0$ uses the empty middle tensor. Thus $Φ_{n}$ and $Ψ_{n}$ are inverse in every degree.

3.1

If $g : M \to N$ is an $A$ -bimodule map, then $Φ_{n}^{N} (z \otimes g (m)) = a_{n + 1} g (m) a_{0} \otimes a_{1} \otimes \dots \otimes a_{n} = g (a_{n + 1} m a_{0}) \otimes a_{1} \otimes \dots \otimes a_{n}$ , so the isomorphisms are natural in the coefficient bimodule. If $A = k$ , the multiplication map $k \otimes_{k} k^{o p} \to k$ has inverse $λ \mapsto λ \otimes 1$ , since $c \otimes d = c d \otimes 1$ in the tensor product; [F10] identifies $A^{e} = k \otimes_{k} k^{o p}$ with $k$ . Under this identification, the right action in [F2] on ${Bar}_{n} (k) ≅ k$ is scalar multiplication by $c d$ , and the left action in [F3] on $M$ is also multiplication by $c d$ . Thus ${Bar}_{n} (k) \otimes_{k^{e}} M ≅ k \otimes_{k} M ≅ M$ by [F8]. The Hochschild term $C_{n} (k, M) ≅ M$ by the tensor-unit maps, since the $n$ scalar factors multiply into the coefficient. Every bar and Hochschild face preserves this total scalar, so each face identifies with ${id}_{M}$ , and the formula for $Φ_{n}$ also identifies with ${id}_{M}$ . This checks the degenerate ground-field case directly. The displayed isomorphisms use no projectivity or choice. [step 1.1, step 2.1, step 1.3, F1, F2, F3, F4, F5, F8, F10, given, algebra] $□$

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30Open item page →

Hochschild homology is Tor over the enveloping algebra

Statement

Assume the Axiom of Choice (AC). Let $k$ be a field, let $A$ be a unital associative $k$ -algebra, and let $M$ be a $k$ -central $A$ -bimodule. Regard $A$ as a right $A^{e} = A \otimes_{k} A^{o p}$ -module by $a (c \otimes d^{o p}) = d a c$ , and regard $M$ as a left $A^{e}$ -module by $(c \otimes d^{o p}) m = c m d$ . For every $n \geq 0$ , there is a canonical isomorphism

$H H_{n} (A, M) ≅ {Tor}_{n}^{A^{e}} (A, M),$

natural in the coefficient bimodule $M$ . The Axiom of Choice is assumed; no $A^{e}$ -projectivity of $M$ is assumed.

Facts & Assumptions

Given: AC, a field $k$ , a unital associative $k$ -algebra $A$ , and a $k$ -central $A$ -bimodule $M$ .

[F1]

A $k$ -central $A$ -bimodule $M$ is a left $A^{e}$ -module by $(c \otimes d^{o p}) m = c m d$ (Enveloping algebra and the bimodule–module dictionary).

[F2]

The regular bimodule $A$ is a right $A^{e}$ -module by $a (c \otimes d^{o p}) = d a c$ (Enveloping algebra and the bimodule–module dictionary).

[F3]

The augmented two-sided bar complex has terms ${Bar}_{n} (A) = A \otimes_{k} A^{\otimes_{k} n} \otimes_{k} A$ and the specified alternating adjacent-multiplication differential, including the empty middle tensor in degree zero (The augmented two-sided bar complex).

[F4]

Under AC, ${Bar}_{∙} (A) \to A$ is a projective resolution of the regular right $A^{e}$ -module $A$ (The two-sided bar complex is a projective $A^{e}$ -resolution).

[F5]

The Hochschild complex has $C_{0} (A, M) = M$ , $C_{n} (A, M) = M \otimes_{k} A^{\otimes_{k} n}$ for $n \geq 1$ , and $H H_{n} (A, M) = H_{n} (C_{∙} (A, M))$ (Hochschild chains and Hochschild homology with coefficients).

[F6]

The maps $(a_{0} \otimes \dots \otimes a_{n + 1}) \otimes m \mapsto (a_{n + 1} m a_{0}) \otimes a_{1} \otimes \dots \otimes a_{n}$ give a chain isomorphism ${Bar}_{∙} (A) \otimes_{A^{e}} M ≅ C_{∙} (A, M)$ , natural in $M$ , with no projectivity assumption on $M$ (Hochschild chains are bar tensor chains).

[F7]

If $Q_{∙} ↠ N$ is a specified projective resolution of a right $R$ -module, the right-resolution construction is ${Tor}_{n}^{R, Q} (N, L) : = H_{n} (Q_{∙} \otimes_{R} L)$ for a left $R$ -module $L$ (Tor from a projective resolution of the right module).

[F8]

Under DC and with projective resolutions supplied for both modules, balanced ${Tor}_{n}^{R} (N, L)$ is identified from either resolution; the identifications are canonical under change of resolution and define a covariant bifunctor up to those canonical isomorphisms (The balanced Tor bifunctor).

[F9]

Under AC, every left module over a unital ring admits a projective resolution (Under the Axiom of Choice, every module admits a projective resolution).

[F10]

AC means every family of nonempty sets has a choice function (The Axiom of Choice).

[F11]

In ZF, AC implies DC (AC implies DC implies countable choice).

Proof

technique · direct

1.1F1F2given

By [F1] and [F2], the right module in the Tor expression is the regular right $A^{e}$ -module $A$ , and the coefficient bimodule is a left $A^{e}$ -module. Thus the tensor products and resolution statements below use the stated sides of the enveloping algebra.

1.2F3F4given

Put $P_{∙} = {Bar}_{∙} (A)$ . By [F4], under the assumed AC this is a projective resolution of the right $A^{e}$ -module $A$ . The augmentation endpoint is ${Bar}_{0} (A) \to A$ by multiplication.

2.1F7step 1.2

Applying the right-resolution definition [F7] to this specified resolution gives ${Tor}_{n}^{A^{e}, P} (A, M) = H_{n} (P_{∙} \otimes_{A^{e}} M)$ . This construction does not require $M$ itself to be projective.

2.2F4F8F9step 1.2

Applying [F9] to the unital ring $A^{e}$ supplies a left projective resolution of $M$ . Together with the right projective resolution $P_{∙}$ of $A$ from step 1.2, this supplies both resolutions required in [F8]. The corollary does not assert that $M$ is itself projective.

3.1F5F6step 2.1

The chain isomorphism [F6], followed by the homology definition [F5], gives for every $n \geq 0$ the identity $H H_{n} (A, M) = H_{n} (C_{∙} (A, M)) ≅ H_{n} (P_{∙} \otimes_{A^{e}} M) = {Tor}_{n}^{A^{e}, P} (A, M)$ . In degree zero, [F6] maps $(a_{0} \otimes a_{1}) \otimes m$ to $a_{1} m a_{0}$ ; in degree one its chain-map identity uses $b_{1} (m \otimes a) = m a - a m$ , so the first nonzero boundary is included.

4.1F4F5F6F8F10F11step 3.1step 2.2

By [F10] AC is the assumed choice principle, and [F11] gives the DC hypothesis of [F8]. Hence the right-resolution group in step 3.1 is canonically identified with the balanced ${Tor}_{n}^{A^{e}} (A, M)$ , independently of the chosen projective resolutions. For a bimodule map $f : M \to M^{'}$ , the map $1_{P_{∙}} \otimes f$ induces the right-resolution homology map; [F5] commutes with $f$ , and [F8] makes the balanced Tor identifications natural. Thus the isomorphism in the statement is natural in $M$ . AC is used in [F4] to choose a $k$ -basis of $A$ and make the resulting free bar terms projective, in [F9] to make the canonical free resolution of $M$ projective, and through AC $\Rightarrow$ DC for balanced Tor comparison and naturality; no choice is used in the chain isomorphism itself.

5.1F3F4F6F7F8step 3.1∎

If $M = 0$ , both chain complexes in step 3.1 vanish and both sides are zero. If $A = k$ , every bar term identifies with $k$ and every adjacent-multiplication face identifies with $1_{k}$ , so $d_{n} = \sum_{r = 0}^{n} (- 1)^{r} 1_{k}$ : it is the identity for even $n$ and zero for odd $n$ . After tensoring with $M$ , this augmented complex has homology $M$ in degree zero and zero in positive degrees; the length-zero resolution of the projective right $k$ -module $k$ gives the same Tor groups. The degree-zero and degree-one endpoints are those already checked in step 3.1.

Source comparison

Weibel, An Introduction to Homological Algebra, §9.1.3, Lemma 9.1.3, printed pp. 302–303 (PDF pp. 2–3), identifies Hochschild homology with relative Tor over $k \to A^{e}$ and gives the bar-tensor chain isomorphism. Section 9.1.4 and Corollary 9.1.5, printed p. 303 (PDF p. 3), explain that when $A$ is projective over $k$ , the bar terms are projective over $A^{e}$ and the relative Tor computation agrees with absolute Tor. These passages corroborate the relative/absolute distinction; the proof above obtains the absolute Tor resolution directly from the AC-qualified projectivity theorem [F3].

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30Open item page →

Functoriality and coefficient long exact sequences for Hochschild homology

Statement

Assume the Axiom of Choice (AC). Let $k$ be a field and $A$ a unital associative $k$ -algebra. Bimodule maps induce natural maps on $H H_{n} (A, -)$ . Each short exact sequence

$0 ⟶ M^{'} ⟶ M ⟶ M^{''} ⟶ 0$

of $k$ -central $A$ -bimodules yields the natural long exact sequence in Hochschild homology, with connecting maps $H H_{n} (A, M^{''}) ⟶ H H_{n - 1} (A, M^{'})$ for $n \geq 1$ . In particular, its bottom endpoint is

$H H_{0} (A, M^{'}) ⟶ H H_{0} (A, M) ⟶ H H_{0} (A, M^{''}) ⟶ 0.$

Facts & Assumptions

Given: AC, a field $k$ , a unital associative $k$ -algebra $A$ , and $k$ -central $A$ -bimodules.

[F1]

The Hochschild chain terms are $C_{0} (A, M) = M$ and $C_{n} (A, M) = M \otimes_{k} A^{\otimes_{k} n}$ for $n \geq 1$ (Hochschild chains and Hochschild homology with coefficients).

[F2]

The first and last Hochschild faces use the right and left bimodule actions, and the internal faces multiply adjacent algebra factors (Hochschild chains and Hochschild homology with coefficients).

[F3]

AC says that every family of nonempty sets has a choice function (The Axiom of Choice).

[F4]

Assuming AC, every vector space over a field has a basis, including the zero space with empty basis (Every vector space has a basis).

[F5]

Every free module over a commutative ring is flat, without an additional choice assumption (Under the stated choice boundary, free modules are projective and hence flat).

[F6]

A chain map induces a unique map on homology compatible with the quotient from cycles (A chain map induces a well-defined map on homology).

[F7]

The category of modules over a ring is abelian, hence so is the category of $k$ -modules (Modules over a ring form an abelian category).

[F8]

A short exact sequence of complexes is a sequence of chain maps that is exact in each degree in the ambient abelian category (Short exact sequence of complexes).

[F9]

A morphism of short exact sequences of complexes is a commutative ladder whose rows are short exact sequences of complexes and whose vertical maps are chain maps (A morphism of short exact sequences of complexes).

[F10]

A short exact sequence of chain complexes in an abelian category gives the long exact sequence in homology (The long exact sequence in homology).

[F11]

A morphism of short exact sequences of complexes induces a commutative square between their homology connecting morphisms (Naturality of the homology connecting morphism).

[F12]

Under AC, the canonical isomorphism $H H_{n} (A, M) ≅ {Tor}_{n}^{A^{e}} (A, M)$ is natural in the coefficient bimodule (Hochschild homology is Tor over the enveloping algebra).

[F13]

For every $k$ -module $M$ , the tensor-unit maps $k \otimes_{k} M \to M$ and $M \otimes_{k} k \to M$ are isomorphisms (The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ ).

Proof

technique · direct

1.1F1F2F6givenalgebra

Let $f : M \to N$ be a $k$ -central $A$ -bimodule map. In degree $n \geq 1$ set $C_{n} (f) = f \otimes 1_{A^{\otimes_{k} n}}$ , and set $C_{0} (f) = f$ . For the first face, $f (m a_{1}) = f (m) a_{1}$ ; for each internal face the map on the coefficient factor does not alter the multiplied algebra entries; for the last face, $f (a_{n} m) = a_{n} f (m)$ . Thus $C (f)$ commutes with every face and with every boundary, including $b_{0} = 0$ , so it is a chain map. The identity bimodule map gives the identity chain map, and $C (g \circ f) = C (g) \circ C (f)$ . By [F6] the induced homology maps obey the same identities. Hence $M \mapsto H H_{n} (A, M)$ is a covariant functor.

1.2F1F2F12givenalgebra

The maps just defined agree with the coefficient maps under the preceding Tor comparison. On an elementary bar tensor, $C_{n} (f) ((a_{n + 1} m a_{0}) \otimes a_{1} \otimes \dots \otimes a_{n}) = (a_{n + 1} f (m) a_{0}) \otimes a_{1} \otimes \dots \otimes a_{n},$ which is the image of $(a_{0} \otimes \dots \otimes a_{n + 1}) \otimes f (m)$ under the comparison for $N$ . The equality uses that $f$ is a bimodule map. Since elementary tensors span, the comparison square commutes; [F12] therefore identifies the induced Hochschild map with the natural map on Tor. This compatibility uses the completed preceding theorem and adds no projectivity hypothesis on $M$ .

1.3F1F3F4F5givenalgebra

For $n \geq 0$ put $V_{n} = A^{\otimes_{k} n}$ , with $V_{0} = k$ . By [F3] and [F4], choose bases $B_{n}$ for this set-indexed family of vector spaces; take $B_{0} = {1}$ . Each $V_{n}$ is then a free, hence flat, $k$ -module by [F5]. For any exact sequence of $k$ -modules $0 \to X^{'} \to X \to X^{''} \to 0$ , tensoring with $V_{n}$ is exact: using the chosen basis, the tensor sequence identifies with the direct sum over $B_{n}$ of copies of the original sequence. In particular, for each $n \geq 0$ , $0 ⟶ C_{n} (A, M^{'}) ⟶ C_{n} (A, M) ⟶ C_{n} (A, M^{''}) ⟶ 0$ is exact. At $n = 0$ , this is the original coefficient sequence under $C_{0} (A, M) = M$ . For $n < 0$ all three chain groups are zero.

2.1F1F8step 1.1step 1.3givenconstruct

The inclusions and quotient map in the coefficient sequence are $A$ -bimodule maps. By step 1.1, their maps on every chain degree commute with the Hochschild boundaries. By [F8], the degreewise exact sequences in step 1.3, with the chain maps checked in step 1.1, form a short exact sequence of chain complexes.

3.1F7F8F10step 2.1givenalgebra

By [F7] the category of $k$ -modules is abelian; apply [F10] to the short exact sequence of complexes from step 2.1, which qualifies by [F8]. This gives, in each degree $n \geq 1$ , $\dots \to H H_{n} (A, M^{'}) \to H H_{n} (A, M) \to H H_{n} (A, M^{''}) \overset{\partial_{n}}{\to} H H_{n - 1} (A, M^{'}) \to H H_{n - 1} (A, M) \to \dots .$ At the lower endpoint $C_{- 1} = 0$ , so the sequence ends as $H H_{0} (A, M^{'}) \to H H_{0} (A, M) \to H H_{0} (A, M^{''}) \to 0.$ This is the asserted long exact sequence.

3.2F6F9F11step 1.1step 2.1givenconstruct

A morphism between two short exact sequences of $k$ -central $A$ -bimodules induces in each degree the corresponding morphism between the short exact sequences of Hochschild chains: the vertical maps are the tensor maps of step 1.1, and commute with the differentials there. By [F9] this is a morphism of short exact sequences of complexes; [F11] makes the square for the homology connecting maps commute. The maps at all other positions are the functorial homology maps of step 1.1, so the entire long exact sequence is natural in the coefficient sequence.

4.1F1F2F3F4F12F13step 1.1step 1.2step 1.3step 3.1givenalgebra∎

If a coefficient module is zero, all its chain groups and homology groups are zero. A zero bimodule map induces the zero chain and homology maps, while an identity map induces identities; the composition check in step 1.1 covers all composites. As a unit-case check, when $A = k$ the maps in [F13] identify $C_{n} (k, M) ≅ M$ and every face is the identity, so $b_{n} = 0$ for odd $n$ and $b_{n} = 1_{M}$ for positive even $n$ . Thus $H H_{0} (k, M) = M$ and $H H_{j} (k, M) = 0$ for $j > 0$ ; the coefficient long exact sequence reduces to the original short exact sequence in degree zero and zeros in positive degrees. No iff claim occurs. AC is used through [F12] for the Tor comparison in step 1.2 and in step 1.3 to supply bases for the tensor powers; the chain-map and connecting-map constructions are choice-free.

Source notes

Weibel, An Introduction to Homological Algebra, §9.1.2, Exercise 9.1.2, printed p.301/PDF p.1, lines 40–43, asks for the coefficient long exact sequence when the short exact sequence of bimodules is $k$ -split. It states the result but leaves the proof as an exercise. Under the stated AC assumption the local basis argument above proves degreewise exactness for every short exact sequence of $k$ -central bimodules, rather than relying on the exercise as proof text.

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30Open item page →

Degree-zero Hochschild homology is bimodule coinvariants

Statement

Let $k$ be a field, $A$ a unital associative $k$ -algebra, and $M$ a $k$ -central $A$ -bimodule. Then there is a canonical $k$ -module isomorphism

$H H_{0} (A, M) ≅ M / D (A, M), D (A, M) : = {span}_{k} {a m - m a : a \in A, m \in M} .$

The denominator $D (A, M)$ is a $k$ -subspace of $M$ . It need not be a two-sided ideal or a sub-bimodule: this failure occurs for the regular bimodule of $M_{2} (k)$ .

Facts & Assumptions

Given: A field $k$ , a unital associative $k$ -algebra $A$ , and a $k$ -central $A$ -bimodule $M$ .

[F1]

The Hochschild chain definition sets $C_{0} (A, M) = M$ (Hochschild chains and Hochschild homology with coefficients).

[F2]

The Hochschild chain definition sets $b_{0} = 0$ and $b_{1} (m \otimes a) = m a - a m$ (Hochschild chains and Hochschild homology with coefficients).

[F3]

The Hochschild homology definition sets $H H_{n} (A, M) = H_{n} (C_{∙} (A, M))$ (Hochschild chains and Hochschild homology with coefficients).

[F4]

The degree- $n$ cycle and boundary subobjects are respectively $\ker (d_{n})$ and $im (d_{n + 1})$ (Cycle and boundary subobjects of a complex).

[F5]

Homology is the cokernel of the boundary-to-cycle map, equivalently the quotient of cycles by boundaries (Homology object of a chain complex).

[F6]

Every element of $M \otimes_{k} A$ is a finite sum of elementary tensors (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).

[F7]

The span of a subset of a vector space is a linear subspace and consists of its finite linear combinations, including the empty sum (Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ , $span (S)$ is exactly the set of linear combinations of finite lists of elements of $S$ , and $span (\emptyset) = {0_{V}}$ ).

[F8]

The cokernel of a module homomorphism is the quotient by its image (Module homomorphism and isomorphism, kernel, image and cokernel).

[F9]

The image of a module homomorphism is a submodule (Kernels and images of module homomorphisms are submodules, and injectivity is equivalent to trivial kernel).

[F10]

A quotient by a submodule has the induced module structure (Quotient module $M / N$ with scalar multiplication on additive cosets).

[F11]

The quotient action is well-defined and satisfies the module laws (The quotient action is well defined and makes $M / N$ a module).

[F12]

For every ring $R$ , the category of left $R$ -modules is abelian (Modules over a ring form an abelian category).

[F13]

$M_{n} (k)$ is the vector space of $n$ by $n$ matrices with entrywise addition and scalar multiplication (The vector space $M_{m \times n} (F) : = F^{m \times n}$ of $m$ by $n$ matrices over a field, with entrywise operations).

[F14]

$M_{n} (k)$ is a unital ring with entrywise addition and matrix multiplication ( $M_{n} (F)$ is a ring under entrywise addition and matrix multiplication, including the zero ring $M_{0} (F)$ ).

[F15]

Matrix multiplication is associative and unital, distributes over addition, and is compatible with scalar multiplication (Matrix multiplication is associative, unital, distributive, and compatible with scalar multiplication).

[F16]

A $k$ -algebra is a unital ring with a unital ring map from $k$ whose image is central (Algebras over a commutative ring, central structure maps, and algebra homomorphisms).

[F17]

Matrix products are defined by row-by-column sums and $I_{n}$ is the identity matrix (Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes).

[F18]

The matrix unit $E_{i j}$ has a single $1$ in entry $(i, j)$ and zeros elsewhere (Matrix units $E_{i j}$ and the Kronecker delta).

[F19]

Matrix units multiply by $E_{i j} E_{r s} = δ_{j r} E_{i s}$ ( $E_{i j} E_{k ℓ} = δ_{j k} E_{i ℓ}$ ).

[F20]

The trace of a square matrix is the sum of its diagonal entries (The trace $tr (A)$ as the sum of the diagonal entries).

[F21]

For square matrices $X, Y$ over $k$ , $tr (X Y) = tr (Y X)$ (For $A \in M_{m \times n} (F)$ and $B \in M_{n \times m} (F)$ , $tr (A B) = tr (B A)$ ).

[F22]

Trace is $k$ -linear (Trace is a linear functional on $M_{n} (F)$ ).

[F23]

The regular bimodule has left and right actions given by multiplication (Enveloping algebra and the bimodule–module dictionary).

Source notes

Weibel, An Introduction to Homological Algebra, §9.1.1, printed p.300/PDF p.0, lines 17–19, identifies the image of the degree-one face difference with the commutator submodule and gives the degree-zero quotient. Khovanov, “Triply-graded link homology and Hochschild homology of Soergel bimodules,” “Hochschild homology,” PDF p.1, lines 9–13, defines the coinvariant quotient by the span of commutators and identifies it with $R \otimes_{R^{e}} M$ . These passages confirm the convention up to the harmless sign reversal in our $b_{1}$ ; the equality $im b_{1} = D (A, M)$ is proved in steps 2.1–2.2.

Proof

technique · direct

1.1F1F2F3F4F5F12given

By [F1], [F2], and [F4], $Z_{0} (C) = \ker b_{0} = M$ and $B_{0} (C) = im b_{1}$ . Thus [F5] identifies $H H_{0} (A, M)$ with the cokernel of the inclusion $im b_{1} ↪ M$ .

1.2F2F6F7given

Let $z \in C_{1} = M \otimes_{k} A$ . By [F6], write $z = \sum_{i < r} m_{i} \otimes a_{i}$ . Then [F2] gives $b_{1} (z) = \sum_{i < r} (m_{i} a_{i} - a_{i} m_{i}) = - \sum_{i < r} (a_{i} m_{i} - m_{i} a_{i}) \in D (A, M) .$ So $im b_{1} \subseteq D (A, M)$ .

1.3F13F14F15F16F17F23given

To verify the asserted failure of ideal and sub-bimodule closure, take $A = M = M_{2} (k)$ with the regular bimodule. By [F13] and [F14] this is a vector space and a unital ring. Define $η : k \to M_{2} (k)$ by $η (λ) = λ I_{2}$ . Entrywise operations and [F15], [F17] give $η (1) = I_{2}, η (λ + μ) = η (λ) + η (μ), η (λ μ) = η (λ) η (μ),$ and for every $X \in M_{2} (k)$ , $η (λ) X = λ X = X η (λ) .$ Thus [F16] makes $A$ a unital associative $k$ -algebra. The regular left and right actions in [F23] commute by associativity, and the displayed centrality shows they agree on $k$ , so $M$ is $k$ -central.

1.4F7F21F22given

For any $X, Y \in M_{2} (k)$ , [F21] and [F22] imply $tr (X Y - Y X) = 0$ . Since every element of $D (A, A)$ is a finite $k$ -linear combination of commutators by [F7], [F22] implies $D (A, A) \subseteq \ker (tr)$ .

1.5F1F2F3F7given

If $M = 0$ , then $C_{0} = 0$ , $b_{1} = 0$ , $D (A, M) = 0$ , and both sides of the isomorphism are zero. If $A = k$ , $k$ -centrality gives $a m = m a$ for every $a \in k, m \in M$ , so $D (k, M) = 0$ and [F3] gives $H H_{0} (k, M) = M$ . In both cases the canonical quotient map is the asserted isomorphism. No basis, projectivity, or choice is used; in particular, AC is neither assumed nor invoked.

2.1F2F6F7step 1.2given

Conversely, by [F7] an arbitrary $d \in D (A, M)$ has the form $d = \sum_{i < r} λ_{i} (a_{i} m_{i} - m_{i} a_{i})$ . The tensor $w = - \sum_{i < r} λ_{i} m_{i} \otimes a_{i}$ satisfies $b_{1} (w) = - \sum_{i < r} λ_{i} (m_{i} a_{i} - a_{i} m_{i}) = d$ by linearity of $b_{1}$ and [F2]. This also covers the empty sum $r = 0$ , for which $d = w = 0$ . Hence $D (A, M) \subseteq im b_{1}$ , so $im b_{1} = D (A, M)$ .

2.2F18F19F20step 1.4given

By [F18] and [F19], $E_{01} E_{11} = E_{01}, E_{11} E_{01} = 0,$ so $E_{01} = E_{01} E_{11} - E_{11} E_{01} \in D (A, A)$ . But $E_{01} E_{10} = E_{00}, tr (E_{00}) = 1_{k} \neq 0_{k}$ by [F19], [F20], and the field axiom $1_{k} \neq 0_{k}$ . Therefore $E_{00} \notin D (A, A)$ by step 1.4.

3.1F7F8F9F10F11step 1.1step 2.1given

By [F7], $D (A, M)$ is a $k$ -subspace. Steps 1.1 and 2.1 therefore identify the homology cokernel with the quotient module $M / D (A, M)$ by [F8], [F9], [F10], and [F11]. The isomorphism is induced by the identity on $M$ , so it is canonical.

3.2step 2.2F23

Since $E_{01} \in D (A, A)$ while $E_{01} E_{10} \notin D (A, A)$ , this subspace is not closed under the right regular action. Hence it is neither a two-sided ideal nor an $A$ -sub-bimodule, proving the stated qualification.

∎

DefinitionDefinition: Literature-sourcedProof: Not applicableprecheck passjudge pass (gpt-6-sol)audited 2026-09-30Open item page →

The polynomial diagonal Koszul bimodule complex

Definition

Let $k$ be a field and let $R = k [x_{1}, \dots, x_{n}]$ for $n \geq 0$ . Since $R$ is commutative, identify its enveloping algebra $R^{e} = R \otimes_{k} R^{o p}$ (Enveloping algebra and the bimodule–module dictionary) with $R \otimes_{k} R$ . Write $x_{i}^{L} = x_{i} \otimes 1$ and $x_{i}^{R} = 1 \otimes x_{i}$ for the two copies of each polynomial generator, and put

$u_{i} : = x_{i}^{L} - x_{i}^{R} .$

The diagonal Koszul bimodule complex is the Koszul complex $K (u_{1}, \dots, u_{n}; R^{e})$ defined in Koszul Complex Of A Sequence With Coefficients, augmented by the multiplication map $μ : R^{e} \to R$ . Its degree- $p$ term is free over $R^{e}$ on symbols

$θ_{i_{1}} \land \dots \land θ_{i_{p}} (1 \leq i_{1} < \dots < i_{p} \leq n),$

and its differential is

$d (θ_{i_{1}} \land \dots \land θ_{i_{p}}) = \sum_{r = 1}^{p} (- 1)^{r - 1} u_{i_{r}} θ_{i_{1}} \land \dots \land \hat{θ_{i_{r}}} \land \dots \land θ_{i_{p}} .$

The augmentation is a chain map because $μ (u_{i}) = 0$ for every $i$ ; the Koszul differential squares to zero by the defining alternating deletion formula. There are $(\binom{n}{p})$ displayed basis symbols in degree $p$ , and no terms above degree $n$ .

When $\deg_{i n t} x_{i} = 2$ , assign each $θ_{i}$ homological degree $1$ and internal degree $2$ . Then the differential lowers homological degree by $1$ and preserves internal degree; degree $p$ is a direct sum of $(\binom{n}{p})$ copies of $R^{e} {2 p}$ under the shift convention $M {r}_{d} = M_{d - r}$ from Associative graded algebras, bimodules, and internal shifts. Internal grading adds no super sign.

For $n = 0$ , the sequence and exterior generators are empty, $R = k$ and $R^{e} = k$ ; the complex is $k$ in degree zero and $μ : k \to k$ is the identity.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30Open item page →

Polynomial diagonal differences form a regular sequence

Statement

For $R = k [x_{1}, \dots, x_{n}]$ over a field $k$ , write $R^{e} ≅ k [x_{1}, \dots, x_{n}, y_{1}, \dots, y_{n}]$ and $u_{i} = x_{i} - y_{i}$ . The ordered sequence $(u_{1}, \dots, u_{n})$ is regular on the $R^{e}$ -module $R^{e}$ , and multiplication induces $R^{e} / (u_{1}, \dots, u_{n}) ≅ R$ . This includes $n = 0$ .

Facts & Assumptions

Given: A field $k$ , $R = k [x_{1}, \dots, x_{n}]$ , the two canonical copies of $R$ in $R^{e}$ , and the ordered differences $u_{i}$ .

[F1]

The diagonal construction sets $u_{i} = x_{i} \otimes 1 - 1 \otimes x_{i}$ and identifies $R^{e} = R \otimes_{k} R$ (The polynomial diagonal Koszul bimodule complex).

[F2]

A sequence is regular on a module when every successive quotient is nonzero and the next multiplication map is injective, and the final quotient is nonzero (Regular Sequence On A Module).

[F3]

Tensor products of commutative $k$ -algebras satisfy the coproduct mapping property (Universal mapping property of the tensor product of commutative algebras).

[F4]

A polynomial ring has the unique evaluation homomorphism for any assigned family of generator values (Universal property of a polynomial ring on an arbitrary family of indeterminates).

Proof

technique · direct

1.1F1F3F4givenalgebra

Put $S = k [X_{1}, \dots, X_{n}, Y_{1}, \dots, Y_{n}]$ . By [F4] the assignments $X_{i} \mapsto x_{i} \otimes 1$ and $Y_{i} \mapsto 1 \otimes x_{i}$ define a $k$ -algebra map $S \to R \otimes_{k} R$ . By [F3] the two polynomial maps from the copies of $R$ into $S$ induce a map $R \otimes_{k} R \to S$ . The composites fix every polynomial generator, so these maps are inverse; [F1] identifies $u_{i}$ with $X_{i} - Y_{i}$ .

2.1step 1.1F4givenalgebra

For $1 \leq i \leq n$ , substitution $Y_{r} \mapsto X_{r}$ for $r < i$ gives $S / (X_{1} - Y_{1}, \dots, X_{i - 1} - Y_{i - 1}) ≅ k [X_{1}, \dots, X_{n}, Y_{i}, \dots, Y_{n}]$ : the inverse includes the displayed remaining variables, and both composites fix their generators. This quotient is a nonzero polynomial ring over $k$ .

2.2step 1.1F1F4givenalgebra

Substituting every $Y_{r} \mapsto X_{r}$ gives $S / (X_{1} - Y_{1}, \dots, X_{n} - Y_{n}) ≅ k [X_{1}, \dots, X_{n}]$ by the same inverse-on-generators check. Under [F1] this is the multiplication quotient $R^{e} / (u_{1}, \dots, u_{n}) ≅ R$ , which is nonzero; for $n = 0$ both rings are $k$ and the map is the identity.

3.1step 2.1givenalgebra

In the quotient of step 2.1, write $C_{i} = k [X_{1}, \dots, X_{i - 1}, X_{i + 1}, \dots, X_{n}, Y_{i}, \dots, Y_{n}]$ , so that it is $C_{i} [X_{i}]$ and $Y_{i} \in C_{i}$ . If $0 \neq f \in C_{i} [X_{i}]$ has degree $d$ and leading coefficient $c \neq 0$ , then $(X_{i} - Y_{i}) f$ has degree $d + 1$ with the same nonzero leading coefficient $c$ . Thus multiplication by $u_{i}$ is injective on each successive quotient.

4.1step 2.1step 2.2step 3.1F2given∎

Steps 2.1 and 2.2 give every required nonzero successive quotient and the nonzero final quotient; step 3.1 gives injectivity at each position. By [F2] this is precisely regularity on $R^{e}$ . When $n = 0$ , there are no injectivity conditions and the final quotient $k$ is nonzero, so the empty sequence is regular as well.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30Open item page →

The diagonal Koszul complex is a finite free resolution of R

Statement

Let $k$ be a field, $R = k [x_{1}, \dots, x_{n}]$ for $n \geq 0$ , and $S = R^{e} = R \otimes_{k} R^{o p}$ . Regard $R$ as the regular $S$ -module through the enveloping-algebra dictionary, and let $u_{i} = x_{i} \otimes 1 - 1 \otimes x_{i}$ . The augmented Koszul complex

$K (u_{1}, \dots, u_{n}; S) \overset{μ}{\to} R$

is a finite free, hence projective, resolution of $R$ over $S$ . Its degree- $j$ term is free of rank $(\binom{n}{j})$ for $0 \leq j \leq n$ , and is zero for $j > n$ . When $n = 0$ , this is the identity resolution $k \overset{i d}{\to} k$ .

Facts & Assumptions

Given: A field $k$ , the polynomial algebra $R = k [x_{1}, \dots, x_{n}]$ , $S = R^{e}$ , the diagonal differences $u_{i}$ , the Koszul complex $K (u_{1}, \dots, u_{n}; S)$ , and its multiplication augmentation $μ$ .

[F1]

The regular bimodule $R$ is the left $R^{e}$ -module with action $(a \otimes b^{o p}) r = a r b$ (Enveloping algebra and the bimodule–module dictionary).

[F2]

The ordered sequence $(u_{1}, \dots, u_{n})$ is regular on $S$ , and multiplication induces $S / (u_{1}, \dots, u_{n}) S ≅ R$ (Polynomial diagonal differences form a regular sequence).

[F3]

Every finite $M$ -regular sequence has zero positive-degree Koszul homology (Regular Sequences Give Acyclic Koszul Complexes).

[F4]

The zeroth Koszul homology is the quotient by the sequence (Basic Koszul Homology).

[F5]

The diagonal Koszul degree- $j$ term is free on its increasing wedge symbols and there are no terms above degree $n$ (The polynomial diagonal Koszul bimodule complex).

[F6]

The increasing wedges indexed by $j$ -element subsets form a basis of the $j$ th exterior power, and that power is zero for $j > n$ (Exterior Algebra Basis Monomials).

[F7]

A free module with a finite basis is projective using only finite choice; the empty basis gives the zero projective module (Free modules are projective, with the exact choice boundary).

[F8]

The category of left $S$ -modules is abelian (Modules over a ring form an abelian category).

[F9]

A projective resolution in an abelian category is an exact augmented complex whose terms are projective (Projective resolutions in an abelian category).

Proof

technique · direct

1.1F1F2givenalgebra

By [F1], $R$ has the regular left $S$ -module structure $(a \otimes b^{o p}) r = a r b$ . The multiplication map $μ : S \to R$ is $S$ -linear: on $s = a \otimes b^{o p}$ and $t = c \otimes d^{o p}$ , $μ (s t) = a c d b = a (c d) b = s \cdot μ (t)$ by associativity. Pure tensors span $S$ , so this equality extends to arbitrary $s, t$ . By [F2], the induced quotient map $S / (u_{1}, \dots, u_{n}) S \to R$ is an isomorphism and is exactly $μ$ . The Koszul differential and $μ$ are module-linear maps, so both send zero to zero.

1.2F5F6givenalgebra

By [F5] and [F6], for $0 \leq j \leq n$ the increasing wedges indexed by $j$ -element subsets form a finite $S$ -basis of $K_{j}$ , with $(\binom{n}{j})$ basis elements, and $K_{j} = 0$ for $j > n$ . If $n = 0$ , the only basis symbol is the empty wedge, $S = R = k$ , and the augmentation is the identity.

2.1F2F3F4step 1.1given

Since $S$ is a commutative ring and $(u_{1}, \dots, u_{n})$ is $S$ -regular by [F2], apply [F3] with coefficient module $S$ to get $H_{j} (K (u_{1}, \dots, u_{n}; S)) = 0$ for every $j > 0$ . By [F4], $H_{0} (K (u_{1}, \dots, u_{n}; S)) = S / (u_{1}, \dots, u_{n}) S$ , which [F2] identifies with $R$ by the augmentation from 1.1. Thus the augmented complex is exact in positive degrees and at $K_{0}$ and $R$ .

2.2F7step 1.2givenalgebra

The basis in each nonzero degree is finite and explicitly enumerated by the lexicographic order on increasing subsets of ${1, \dots, n}$ . By [F7] each $K_{j}$ is projective as an $S$ -module; this uses only finite choice, not AC. The zero terms above degree $n$ are projective as well.

3.1step 2.1step 1.2step 2.2F8F9∎

The augmentation is an exact augmented chain complex by 2.1, all its terms are projective by 2.2, and $S - M o d$ is abelian by [F8]. Therefore [F9] makes it a projective resolution. Step 1.2 gives the claimed finite free ranks and length, including the $n = 0$ identity case. No form of AC is used.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30Open item page →

Polynomial Hochschild homology from the diagonal Koszul complex

Statement

Assume the Axiom of Choice (AC). Let $k$ be a field, $R = k [x_{1}, \dots, x_{n}]$ for $n \geq 0$ , and $M$ a $k$ -central $R$ -bimodule. Put $S = R^{e}$ . For $0 \leq j \leq n$ set

$K_{j} (M) : = M \otimes_{k} Λ_{k}^{j} (θ_{1}, \dots, θ_{n}) = ⨁_{1 \leq i_{1} < \dots < i_{j} \leq n} M θ_{i_{1}} \land \dots \land θ_{i_{j}},$

and put $K_{j} (M) = 0$ for $j > n$ . The differential is

$d (m θ_{i_{1}} \land \dots \land θ_{i_{j}}) = \sum_{r = 1}^{j} (- 1)^{r - 1} (x_{i_{r}} m - m x_{i_{r}}) θ_{i_{1}} \land \dots \land \hat{θ_{i_{r}}} \land \dots \land θ_{i_{j}} .$

There is an isomorphism $H H_{j} (R, M) ≅ H_{j} (K_{∙} (M)) (j \geq 0),$ natural in $M$ . For the grading fixed by the diagonal Koszul definition, when $\deg_{i n t} x_{i} = 2$ and $M$ is a graded $k$ -central $R$ -bimodule, assign $\deg_{i n t} θ_{i} = 2$ ; the isomorphism preserves internal degree.

In top degree, $H H_{n} (R, M) ≅ M^{R} θ_{1} \land \dots \land θ_{n}, M^{R} : = {m \in M : r m = m r for every r \in R} .$ When $n = 0$ , the centralizer condition is vacuous because $M$ is $k$ -central, and the displayed top wedge is the empty wedge.

Facts & Assumptions

Given: AC, a field $k$ , $R = k [x_{1}, \dots, x_{n}]$ , $S = R^{e}$ , and a $k$ -central $R$ -bimodule $M$ . For the graded clause, $M$ is graded and each $x_{i}$ has internal degree $2$ .

[F1]

Under AC, Hochschild homology with coefficients is naturally isomorphic to ${Tor}_{j}^{R^{e}} (R, M)$ (Hochschild homology is Tor over the enveloping algebra).

[F2]

The $k$ -central bimodule $M$ is a left $S$ -module by $(a \otimes b^{o p}) m = a m b$ (Enveloping algebra and the bimodule–module dictionary).

[F3]

The diagonal Koszul complex $K (u_{1}, \dots, u_{n}; S)$ , augmented to $R$ , is a finite free projective resolution of $R$ over $S$ ; its degree- $p$ basis is the increasing $p$ -fold wedge basis (The diagonal Koszul complex is a finite free resolution of R).

[F4]

The two-sided bar term is ${Bar}_{q} (R) = R \otimes_{k} R^{\otimes_{k} q} \otimes_{k} R$ , with its specified right $S$ -action and adjacent-multiplication differential (The augmented two-sided bar complex).

[F5]

Under AC, ${Bar}_{∙} (R) \to R$ is a projective resolution of the regular right $S$ -module $R$ (The two-sided bar complex is a projective $A^{e}$ -resolution).

[F6]

The maps $(a_{0} \otimes \dots \otimes a_{q + 1}) \otimes m \mapsto (a_{q + 1} m a_{0}) \otimes a_{1} \otimes \dots \otimes a_{q}$ give a chain isomorphism ${Bar}_{∙} (R) \otimes_{S} M ≅ C_{∙} (R, M)$ , natural in $M$ (Hochschild chains are bar tensor chains).

[F7]

Any two projective resolutions of the same object are homotopy equivalent over that object under DC (Projective resolutions of the same object are homotopy equivalent over that object).

[F8]

Chain-homotopic chain maps induce the same homology map (Chain-homotopic maps induce the same map on homology).

[F9]

AC means every family of nonempty sets has a choice function (The Axiom of Choice).

[F10]

In ZF, $A C \Rightarrow D C$ (AC implies DC implies countable choice).

[F11]

The diagonal Koszul differential deletes an increasing wedge factor with sign $(- 1)^{r - 1}$ and coefficient $u_{i_{r}}$ ; when each $x_{i}$ has internal degree $2$ , each $θ_{i}$ has internal degree $2$ and the differential has internal degree zero (The polynomial diagonal Koszul bimodule complex).

[F12]

A graded bimodule has homogeneous left and right actions, and a graded $k$ -central bimodule has agreeing scalar actions (Associative graded algebras, bimodules, and internal shifts).

[F13]

The graded balanced tensor product uses total internal degree and adds no sign to its balancing relation (Graded balanced tensor product and homogeneous Hom).

[F14]

Under DC, balanced Tor may be computed from a specified projective resolution of the right module by $H_{j} (Q_{∙} \otimes_{S} M)$ (The balanced Tor bifunctor).

[F15]

Hochschild homology is the homology of the chain complex $C_{∙} (R, M)$ for a $k$ -central $R$ -bimodule $M$ (Hochschild chains and Hochschild homology with coefficients).

Proof

technique · direct

1.1F1F5F6F9F10F14F15given

Identify HH with bar-tensor homology through Tor. [F1, F5, F6, F9, F10, F14, F15, given] AC implies DC by [F9]--[F10]. By [F1], it suffices to compute the balanced enveloping-algebra Tor group. The right-resolution definition [F14] and bar resolution [F5] compute it using $H_{j} ({Bar}_{∙} (R) \otimes_{S} M)$ , and the natural chain isomorphism [F6] and the definition [F15] identify this homology with $H H_{j} (R, M)$ .

1.2F3F4F5givenalgebra

The bar and diagonal Koszul complexes resolve the same right $S$ -module. [F3, F4, F5, given, algebra] Let $K_{∙} = K (u_{1}, \dots, u_{n}; S)$ . By [F3], $K_{∙} \to R$ is a projective resolution as a left $S$ -module. The ring $S = R \otimes_{k} R$ is commutative, so the same modules and maps are a projective resolution of $R$ as a right $S$ -module. Thus [F5] and $K_{∙}$ are projective resolutions of the same right $S$ -module.

1.3F2F3F11givenalgebra

Tensor the diagonal Koszul resolution with $M$ and compute its differential. [F2, F3, F11, given, algebra] Write $u_{i} = x_{i} \otimes 1 - 1 \otimes x_{i}$ . The degree- $j$ Koszul term is free over $S$ on the symbols $θ_{I}$ with $∣ I ∣ = j$ . Tensoring over $S$ with $M$ identifies each basis copy $S \otimes_{S} M$ with $M$ , so $K_{j} \otimes_{S} M ≅ ⨁_{∣ I ∣ = j} M θ_{I} ≅ M \otimes_{k} Λ_{k}^{j} (θ_{1}, \dots, θ_{n}) .$ By [F2], $u_{i}$ acts on $m \in M$ as $(x_{i} \otimes 1) m - (1 \otimes x_{i}^{o p}) m = x_{i} m - m x_{i}$ . Applying the Koszul differential [F11] therefore gives exactly the displayed formula. The operators $m \mapsto x_{i} m - m x_{i}$ commute: expand their composites and use commutativity of the $x_{i}$ on each side and commutation of the two bimodule actions. Hence terms deleting a fixed pair of wedge factors cancel in opposite orders, so $d^{2} = 0$ , also directly confirming that these are chain groups.

2.1F2F7F8step 1.1step 1.2

Compare the projective resolutions and tensor their homotopies with $M$ . [F2, F7, F8, step 1.1, step 1.2] Apply [F7] to obtain comparison maps in both directions over $R$ , with composites homotopic to the respective identity maps. Tensoring those maps and homotopies over $S$ with the left $S$ -module $M$ preserves the chain-map and homotopy identities. By [F8], the induced homology maps are inverse. Consequently $H_{j} ({Bar}_{∙} (R) \otimes_{S} M) ≅ H_{j} (K_{∙} \otimes_{S} M) .$ The maps are independent of $M$ , so this comparison is natural in coefficient bimodule maps.

2.2F3F11step 1.3algebra

Identify top homology with the centralizer of $R$ in $M$ . [F3, F11, step 1.3, algebra] For $n \geq 1$ , the degree- $n$ term has the single basis wedge $θ_{1} \land \dots \land θ_{n}$ and there is no degree- $(n + 1)$ term. Thus $H_{n} (K_{∙} (M)) = \ker d_{n}$ . Its differential is $d_{n} (m θ_{1} \land \dots \land θ_{n}) = \sum_{i = 1}^{n} (- 1)^{i - 1} (x_{i} m - m x_{i}) θ_{1} \land \dots \land \hat{θ_{i}} \land \dots \land θ_{n} .$ The target has the distinct displayed basis wedges, so this is zero exactly when $x_{i} m = m x_{i}$ for each generator $x_{i}$ . Since these generators generate the polynomial algebra, that is equivalent to $r m = m r$ for every $r \in R$ : the equality extends from generators to their products by induction and then to polynomial linear combinations by additivity. If $n = 0$ , the condition is vacuous because $R = k$ and $M$ is $k$ -central; the degree-zero complex is $M$ with zero differential, so $H_{0} = M$ .

3.1F2F3F4F9F11step 1.2

Construct degree-zero comparison maps by homogeneous lifts. [F2, F3, F4, F9, F11, step 1.2, step 2.1] The comparison in step 2.1 can be chosen to preserve internal degree. Indeed, $R^{\otimes_{k} q}$ has its homogeneous monomial basis. The map $R^{\otimes_{k} q} \otimes_{k} S ⟶ {Bar}_{q} (R), (a_{1} \otimes \dots \otimes a_{q}) \otimes (a \otimes b^{o p}) ⟼ b \otimes a_{1} \otimes \dots \otimes a_{q} \otimes a$ is an isomorphism of right $S$ -modules. Its inverse sends $b \otimes a_{1} \otimes \dots \otimes a_{q} \otimes a$ to $(a_{1} \otimes \dots \otimes a_{q}) \otimes (a \otimes b^{o p})$ ; it is right $S$ -linear because $(a \otimes b^{o p}) (c \otimes d^{o p}) = a c \otimes (d b)^{o p}$ and the right bar action sends the outer slots $(b, a)$ to $(d b, a c)$ . Thus these monomials give a homogeneous free basis for every bar term; for $q = 0$ the middle tensor is $k$ and the basis is ${1}$ . The Koszul terms are free on homogeneous wedge symbols by [F3], and [F11] makes their differentials and augmentations degree-zero maps. Recursively construct a comparison map $f : P_{∙} \to Q_{∙}$ in either direction. At degree zero, for each homogeneous free generator $g \in P_{0}$ , choose a homogeneous lift in $Q_{0}$ of its image in $R$ under the augmentation of $P_{0}$ . At degree $q > 0$ , after $f_{q - 1}$ is defined, the element $z = f_{q - 1} (d_{P} g)$ is a cycle because $d_{P}^{2} = 0$ and the already constructed maps commute with the differentials. Exactness of $Q_{∙} \to R$ gives a preimage $y \in Q_{q}$ with $d_{Q} y = z$ . Since $z$ is homogeneous and $d_{Q}$ has internal degree zero, the component of $y$ in the degree of $z$ is also a preimage. Choose such homogeneous lifts using AC [F9], and extend $S$ -linearly on the free basis. This constructs degree-zero comparison maps in both directions.

4.1F7F8F9F11F12F13step 1.1step 3.1

Construct degree-zero homotopies between comparison composites and identities. [F7, F8, F9, F11, F12, F13, step 1.1, step 3.1] The homotopies between the comparison composites and the identity maps can also be chosen degree-zero. For either composite $u$ and identity $v$ on a resolution $P_{∙}$ , set $s_{- 1} = 0$ . At degree zero, $u_{0} - v_{0}$ has zero augmentation because both maps lift $1_{R}$ , so on each homogeneous generator choose a homogeneous preimage under $d_{P}$ . At degree $q > 0$ , after $s_{q - 1}$ is defined, put $r_{q} = (u_{q} - v_{q}) - s_{q - 1} d_{P}$ . The chain-map identities and the homotopy equation in degree $q - 1$ give $d_{P} r_{q} = 0$ . Exactness makes $r_{q} (g)$ a boundary for each homogeneous generator $g$ ; taking the required-degree component of a preimage gives a degree-zero lift $s_{q} (g)$ . AC chooses the lifts on all homogeneous free generators and $S$ -linear extension gives degree-zero homotopies. [F7, F9, step 1.1, step 3.1] After tensoring with graded $M$ , [F12]--[F13] give the total internal grading on $K_{∙} \otimes_{S} M$ ; the degree-zero comparison maps and homotopies therefore induce an internal-degree-preserving isomorphism on homology.

5.1F6F11F12step 1.1step 1.1step 2.1step 3.1step 4.1step 1.3step 2.2

Combine the comparisons to obtain the natural graded homology isomorphism. [F6, F11, F12, step 1.1, step 2.1, step 4.1, step 1.3, step 2.2, step 3.1] Combining steps 1.1, 1.3, and 2.1 gives the asserted homology isomorphism. For a bimodule map $f : M \to N$ , the tensor maps $1 \otimes f$ commute with the fixed comparison maps and with the bar-to-Hochschild chain isomorphism [F6], so the isomorphism is natural. In the graded case, each $θ_{i}$ has degree two; therefore a summand $M θ_{I}$ has the internal degree of $M$ shifted by $2 ∣ I ∣$ , and the isomorphism built in steps 3.1 and 4.1 preserves this degree. Step 2.2 establishes the top-degree centralizer description.

5.2

Check zero, one, empty and endpoint cases, and record AC use. [F3, F5, F7, F9, F10, F11, F14, step 2.1, step 3.1, step 4.1, step 1.3, step 2.2] If $M = 0$ , all terms and homology groups vanish. If $n = 0$ , then $R = S = k$ , the exterior algebra has only its empty wedge in degree zero, and the complex is $M$ in degree zero with zero differential; the diagonal resolution is the identity resolution and the comparison above gives $H H_{0} (k, M) = M$ and $H H_{j} (k, M) = 0$ for $j > 0$ . If $n = 1$ , the only nonzero differential is $m θ_{1} \mapsto x_{1} m - m x_{1}$ , with positive sign, and the formula yields its kernel and cokernel in degrees one and zero. For every $n$ , there are no terms above degree $n$ , so $H H_{j} (R, M) = 0$ for $j > n$ . The empty wedge gives $K_{0} (M) = M$ and the differential out of degree zero is zero. The only choice principle used is AC: it supplies the bar projectivity through [F5], implies DC for [F7], and selects homogeneous lifts in steps 3.1 and 4.1; the monomial basis and the finite Koszul wedge basis are explicit. [F3, F5, F9, F10, F11, step 2.1, step 1.3] $□$

Source comparison

Weibel, An Introduction to Homological Algebra, §9.1.3 and Exercise 9.1.3, printed pp.302–304 (PDF pp.2–4), gives the enveloping-algebra/bar setup and poses the polynomial diagonal Koszul computation as an exercise; that exercise is a prompt, not a proof. Khovanov, “Hochschild homology,” PDF p.1, describes the polynomial algebra's shorter Koszul resolution and the coefficient contractions $x_{i} m - m x_{i}$ with the exterior deletion signs. The local proof above supplies the resolution comparison, its homotopy inverse, and the degree-preserving lift argument.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30Open item page →

Diagonal Hochschild homology of a polynomial ring

Statement

Assume AC. Let $k$ be a field and $R = k [x_{1}, \dots, x_{n}]$ for $n \geq 0$ , regarded as its regular bimodule. Give $H H_{j} (R, R)$ the $R$ -module structure induced by multiplication on the coefficient factor in Hochschild chains. For $0 \leq j \leq n$ there is an isomorphism of $R$ -modules $H H_{j} (R, R) ≅ R \otimes_{k} Λ_{k}^{j} (k^{n})$ , and $H H_{j} (R, R) = 0$ for $j > n$ . For the graded assertion, place $k$ in internal degree $0$ and grade $R$ by $\deg_{i n t} x_{i} = 2$ ; give each standard exterior generator internal degree $2$ . Then the isomorphism is one of graded $R$ -modules and, for $0 \leq j \leq n$ , $H H_{j} (R, R) ≅ R^{(\binom{n}{j})} {2 j} .$ In particular, for $n = 0$ one has $H H_{0} (k, k) = k$ and $H H_{j} (k, k) = 0$ for $j > 0$ .

Facts & Assumptions

Given: AC, a field $k$ , $R = k [x_{1}, \dots, x_{n}]$ , and the regular $R$ -bimodule. For the graded assertion, place $k$ in internal degree $0$ and give each $x_{i}$ internal degree $2$ .

[F1]

Under AC, the polynomial Hochschild theorem identifies $H H_{j} (R, M)$ with the homology of the coefficient diagonal Koszul complex, naturally in $M$ ; in its graded clause each $θ_{i}$ has internal degree $2$ (Polynomial Hochschild homology from the diagonal Koszul complex).

[F2]

For every $p$ , the increasing wedges indexed by $p$ -subsets form a basis of the $p$ th exterior power of a finite free module, and that exterior power vanishes for $p > n$ (Exterior Algebra Basis Monomials).

[F3]

AC means every family of nonempty sets has a choice function (The Axiom of Choice).

[F4]

Hochschild chains have terms $C_{n} (A, M) = M \otimes_{k} A^{\otimes_{k} n}$ , and their boundary is the alternating sum of the first action, internal multiplication, and last action faces (Hochschild chains and Hochschild homology with coefficients).

[F5]

The graded shift is defined by $(N {r})_{d} = N_{d - r}$ (Associative graded algebras, bimodules, and internal shifts).

Proof

technique · direct

1.1F1F3given

Apply the polynomial Hochschild theorem with the regular coefficient bimodule. [F1, F3, given] Since $R$ is $k$ -central, the theorem applies to $M = R$ . Under AC it gives $H H_{j} (R, R) ≅ H_{j} (K_{∙} (R))$ , where the coefficient differential deletes $θ_{i}$ with coefficient $x_{i} m - m x_{i}$ . Its naturality in $M$ will also identify the coefficient $R$ -module structure below.

2.1F1F4step 1.1algebra

The Hochschild-chain boundary is $R$ -linear on the regular coefficient chains. [F4, step 1.1, algebra] For $r \in R$ , let $r$ act on $C_{q} (R, R) = R \otimes_{k} R^{\otimes_{k} q}$ by multiplication on the first factor. This action commutes with every face: the first face uses $(r m) a_{1} = r (m a_{1})$ , internal faces leave the coefficient unchanged, and the last face uses $a_{q} (r m) = r (a_{q} m)$ because $R$ is commutative. Thus each boundary is $R$ -linear and the homology inherits this $R$ -action. For every $r \in R$ , coefficient multiplication $μ_{r} (m) = r m$ is an $R$ -bimodule endomorphism of $R$ . Naturality in [F1] shows the comparison with $H_{j} (K_{∙} (R))$ commutes with $μ_{r}$ , so it is $R$ -linear.

2.2F1step 1.1algebra

Every differential in the coefficient Koszul complex for $M = R$ is zero. [F1, step 1.1, algebra] For every $m \in R$ and generator $x_{i}$ , commutativity gives $x_{i} m - m x_{i} = 0$ . Hence each coefficient of every Koszul deletion map is zero, so $d_{q} = 0$ for all $q$ .

3.1F1F2step 2.1step 2.2algebra

Read homology from the exterior basis. [F1, F2, step 2.2, algebra] Since all differentials vanish, $H_{j} (K_{∙} (R)) = K_{j} (R)$ . For $0 \leq j \leq n$ , the increasing wedges $θ_{I}$ with $∣ I ∣ = j$ form a basis, so $K_{j} (R) = ⨁_{∣ I ∣ = j} R θ_{I} ≅ R \otimes_{k} Λ_{k}^{j} (k^{n})$ with rank $(\binom{n}{j})$ . For $j > n$ the exterior power and Koszul term are zero by [F2], giving $H H_{j} (R, R) = 0$ . The isomorphism is $R$ -linear by step 2.1.

4.1F1F5step 3.1algebra

Determine the internal grading and shift. [F1, F5, step 3.1, algebra] In the graded clause of [F1], each $θ_{i}$ has degree $2$ , so every $θ_{I}$ with $∣ I ∣ = j$ has degree $2 j$ . Thus each summand $R θ_{I}$ is the internal shift $R {2 j}$ under [F5], and the isomorphism in step 3.1 preserves internal degree.

5.1

Check the empty, one-variable, top, and degree-zero cases. [F1, F2, step 3.1, step 4.1, given] If $n = 0$ , there is only the empty wedge in degree zero and the complex is $k$ with zero differential, so $H H_{0} (k, k) = k$ and higher homology vanishes. If $n = 1$ , the two terms are $R$ and $R θ_{1}$ ; the differential is zero, so $H H_{0} = R$ , $H H_{1} = R {2}$ , and higher groups vanish. In general, degree zero uses $Λ^{0} (k^{n}) = k$ and has no outgoing differential; top degree $j = n$ has one basis wedge of degree $2 n$ (the empty wedge when $n = 0$ ); above top degree all terms vanish. The stated AC is used only through [F1]; the zero-differential calculation and exterior basis read-off make no further choices. There is no zero coefficient case because the coefficient is fixed to the nonzero regular module $R$ over a field. This claim is not an equivalence. [F1, F2, F3, step 1.1, step 2.2, step 3.1, step 4.1, algebra] $□$

Source comparison

Weibel, An Introduction to Homological Algebra, Exercise 9.1.3, printed p.304/PDF p.4, asks for the polynomial Hochschild calculation with the Koszul resolution; it is an exercise prompt, not a proof. Weibel's Exercise 9.1.1, printed p.300/PDF p.1, asks for the commutative-algebra action on Hochschild chains. Khovanov, “Hochschild homology,” PDF p.1, lines 33–62, states the polynomial diagonal Koszul complex and its coefficient contractions. These passages corroborate the conventions; the zero differential, $R$ -linearity, exterior-basis calculation, and boundary cases are proved above.

5 · Examples, counterexamples and false statements

None yet.

Hochschild Homology and Diagonal Koszul Resolutions

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Enveloping algebra and the bimodule–module dictionary

Definition

The augmented two-sided bar complex

Definition

The bar boundary squares to zero and is augmented

Statement

Facts & Assumptions

Proof

Remark

The two-sided bar complex is a projective AeAe-resolution

Statement

Facts & Assumptions

Proof

Hochschild chains and Hochschild homology with coefficients

Definition

Facts & Assumptions

Proof

Hochschild chains are bar tensor chains

Statement

Facts & Assumptions

Proof

Hochschild homology is Tor over the enveloping algebra

Statement

Facts & Assumptions

Proof

Source comparison

Functoriality and coefficient long exact sequences for Hochschild homology

Statement

Facts & Assumptions

Proof

Source notes

Degree-zero Hochschild homology is bimodule coinvariants

Statement

Facts & Assumptions

Source notes

Proof

The polynomial diagonal Koszul bimodule complex

Definition

Polynomial diagonal differences form a regular sequence

Statement

Facts & Assumptions

Proof

The diagonal Koszul complex is a finite free resolution of R

Statement

Facts & Assumptions

Proof

Polynomial Hochschild homology from the diagonal Koszul complex

Statement

Facts & Assumptions

Proof

Source comparison

Diagonal Hochschild homology of a polynomial ring

Statement

Facts & Assumptions

Proof

Source comparison

5 · Examples, counterexamples and false statements

Sources

The two-sided bar complex is a projective $A^{e}$ -resolution