LemmaStatement: Literature-sourcedProof: Literature-sourcedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)verified 2026-10-03 (gpt-6.1-sol)

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.

Affine blowup algebras: normal form, nonzerodivisors, reducedness, domains

Statement

Let $A$ be a commutative ring, let $I \subseteq A$ be an ideal and let $a \in I$ . Write $R (I) = ⨁_{n \geq 0} I^{n} t^{n} \subseteq A [t]$ for the Rees algebra of $I$ , graded by the degree of $t$ (The Rees algebra of an ideal and the Rees module of a filtered module, Nonnegatively graded rings and modules, homogeneous elements, and twists). The affine blowup algebra is

$A [I / a] : = (R (I))_{(a)},$

the degree-zero part of the localization of $R (I)$ at the multiplicative set generated by the degree-one element $a t \in I t$ (Multiplicative subsets and the localisation $S^{- 1} R$ as equivalence classes of fractions). Then:

the image of $a$ in $A [I / a]$ is a nonzerodivisor, $I A [I / a] = a A [I / a]$ , and $(A [I / a])_{a} = A_{a}$ ;
if $I = (a_{0}, \dots, a_{r})$ with $a = a_{0}$ (The ideal generated by a subset and principal ideals), the homomorphism $A [x_{1}, \dots, x_{r}] / (a x_{i} - a_{i}) ⟶ A [I / a]$ sending $x_{i} \mapsto a_{i} / a$ is surjective with kernel the $a$ -power torsion;
if $A$ is reduced then $A [I / a]$ is reduced; if $A$ is a domain and $a \neq 0$ then $A [I / a]$ is a domain;
the construction is independent of the generating set and of the representative used for the homogeneous localization, up to canonical $A$ -algebra isomorphism.

Facts & Assumptions

Given: A commutative ring $A$ , an ideal $I \subseteq A$ and an element $a \in I$ ; the Rees algebra $R (I) = ⨁_{n \geq 0} I^{n} t^{n} \subseteq A [t]$ of (The Rees algebra of an ideal and the Rees module of a filtered module), localized at the multiplicative set generated by the degree-one element $a t$ (Multiplicative subsets and the localisation $S^{- 1} R$ as equivalence classes of fractions), with grades read in the sense of (Nonnegatively graded rings and modules, homogeneous elements, and twists).

[F1]

The Rees algebra of an ideal and the Rees module of a filtered module: For an ideal $I$ of a commutative ring $A$ , the Rees algebra is the graded subring $R (I) = ⨁_{n \geq 0} I^{n} t^{n} \subseteq A [t]$ , whose degree- $n$ piece is $I^{n}$ (so $R (I)_{n} = I^{n} t^{n}$ ), with $I^{0} = A$ .

[F2]

Multiplicative subsets and the localisation $S^{- 1} R$ as equivalence classes of fractions: For a commutative ring $R$ and a multiplicative subset $S \subseteq R$ , the localization $S^{- 1} R$ has elements written $r / s$ with $(r, s) \sim (r^{'}, s^{'})$ iff $u (r s^{'} - r^{'} s) = 0$ for some $u \in S$ ; the operations are $r / s + r^{'} / s^{'} = (r s^{'} + r^{'} s) / (s s^{'})$ and $(r / s) (r^{'} / s^{'}) = r r^{'} / (s s^{'})$ ; every $s \in S$ maps to a unit; if $0 \in S$ the localization is the zero ring.

[F3]

Nonnegatively graded rings and modules, homogeneous elements, and twists: A nonnegatively graded ring is a commutative ring $S = ⨁_{n \geq 0} S_{n}$ with $S_{n} S_{m} \subseteq S_{n + m}$ ; an element of $S_{n}$ is homogeneous of degree $n$ . In particular the degree-zero part of a $Z$ -graded ring is closed under addition, multiplication and contains the unit class of the localization, and products of homogeneous elements add degrees.

[F4]

The ideal generated by a subset and principal ideals: If $I = (a_{0}, \dots, a_{r})$ then every element of $I$ is a finite sum $\sum_{i} c_{i} a_{i}$ with $c_{i} \in A$ , and conversely each $a_{i}$ lies in $I$ .

Proof

1.1F1F2F3

An element of the localization $R (I)_{a t}$ is a class $z / (a t)^{n}$ with $z \in R (I)$ and $n \geq 0$ ; writing $z = \sum_{i} b_{i} t^{i}$ , its degree-zero component is $b_{n} t^{n} / (a t)^{n}$ with $b_{n} \in I^{n}$ . Hence the degree-zero part $D = (R (I))_{(a)}$ consists of the classes of $b t^{n} / (a t)^{n}$ with $n \geq 0$ and $b \in I^{n}$ , so that $A [I / a] = D$ ; for $b \in I^{n}$ , $c \in I^{m}$ one has $b t^{n} / (a t)^{n} = c t^{m} / (a t)^{m}$ in $D$ if and only if $a^{k} (a^{m} b - a^{n} c) = 0$ for some $k \geq 0$ , because the defining relation is $(a t)^{k} ((a t)^{m} b t^{n} - (a t)^{n} c t^{m}) = 0$ and $t$ is a nonzerodivisor of $A [t]$ .

2.1F1F2step 1.1

The assignment $ι ⁣ : D \to A_{a}$ , $b t^{n} / (a t)^{n} \mapsto b / a^{n}$ , is a well-defined injective ring homomorphism: well-defined and injective because $b / a^{n} = c / a^{m}$ in $A_{a}$ holds exactly when $a^{k} (a^{m} b - a^{n} c) = 0$ for some $k \geq 0$ , i.e. exactly under the equality criterion of step 1.1; compatible with addition because $(b t^{n}) / (a t)^{n} + (c t^{m}) / (a t)^{m} = (a^{m} b + a^{n} c) t^{n + m} / (a t)^{n + m}$ and with multiplication because the product is $b c t^{n + m} / (a t)^{n + m}$ . Its image is the $A$ -subalgebra $B = A [b / a : b \in I] \subseteq A_{a}$ generated by the fractions $b / a$ : every $b / a$ is the image of $b t / (a t)$ , and conversely every $b \in I^{n}$ is a finite sum of products of $n$ elements of $I$ , so $b / a^{n}$ is a sum of products of these generators (with coefficients from $A$ ). Hence $A [I / a] ≅ B$ as $A$ -algebras and all computations may be performed in $B \subseteq A_{a}$ .

3.1F1step 2.1

The element $a$ is invertible in $A_{a}$ , hence a nonzerodivisor on $A_{a}$ , hence a nonzerodivisor on the subring $B$ ; in particular the image of $a$ in $A [I / a]$ is a nonzerodivisor. Moreover $a B \subseteq I B$ because $a \in I$ , and $I B \subseteq a B$ because for $b \in I$ and $c / a^{n} \in B$ with $c \in I^{n}$ one has $b \cdot c / a^{n} = a \cdot b c / a^{n + 1}$ with $b c \in I^{n + 1}$ ; hence $I A [I / a] = a A [I / a]$ .

3.2F2step 2.1

The image of $A$ lies in $B$ , since $c = a c / a$ for $c \in A$ and $a c \in I$ ; thus $A_{a} = A [1 / a] \subseteq B_{a}$ , and the inclusion $B \subseteq A_{a}$ induces $B_{a} \subseteq (A_{a})_{a} = A_{a}$ . Hence $B_{a} = A_{a}$ , that is, $(A [I / a])_{a} = A_{a}$ .

3.3F4step 2.1

Let $φ ⁣ : C = A [x_{1}, \dots, x_{r}] / (a x_{i} - a_{i}) \to B$ send $x_{i}$ to $a_{i} / a$ . Its image contains $A$ and every $a_{i} / a$ ; since $I$ is generated by the $a_{i}$ , every $b \in I$ is a finite sum $\sum_{i} c_{i} a_{i}$ , whence $b / a = \sum_{i} c_{i} (a_{i} / a)$ lies in the image; as the image is an $A$ -subalgebra of $A_{a}$ containing all $b / a$ it contains $B$ . So $φ$ is surjective.

3.4F2step 2.1

If $A$ is reduced, then so is the localization $A_{a}$ , and $B \subseteq A_{a}$ is a subring of a reduced ring, hence reduced: if $x^{N} = 0$ in $B$ then $x = 0$ already in $A_{a}$ . If $A$ is a domain and $a \neq 0$ , then $A_{a} \subseteq Frac (A)$ is a nonzero subring of a field (nonzero because $a \cdot a^{- 1} = 1$ with $a \neq 0$ ), hence a domain, and so is its subring $B$ .

4.1F2step 3.3

Localizing $C$ at the multiplicative set generated by $a$ gives $C_{a} = A_{a} [x_{1}, \dots, x_{r}] / (a x_{i} - a_{i})$ in which $a$ is a unit and hence $x_{i} = a_{i} a^{- 1}$ ; the $A_{a}$ -algebra homomorphism $C_{a} \to A_{a}$ sending $x_{i} \mapsto a_{i} / a$ is therefore surjective and has zero kernel, because evaluation at $x_{i} = a_{i} / a$ identifies the quotient $A_{a} [x_{i}] / (x_{i} - a_{i} / a)$ with $A_{a}$ . Since the image of $C$ in $A_{a}$ is $B$ by step 3.3, $\ker (C \to B) = \ker (C \to C_{a})$ , and by the defining relation of the localization an element $f \in C$ lies in this kernel exactly when $a^{N} f = 0$ for some $N \geq 0$ , i.e. exactly when $f$ is $a$ -power torsion.

5.1step 1.1step 2.1step 3.3step 4.1∎

The description $B = {b / a^{n} : n \geq 0, b \in I^{n}}$ of step 2.1 uses only the pair $(I, a)$ : no generating family is chosen, and a class in $D$ is compared with another by the equality criterion of step 1.1, so the use of a particular fraction representative is immaterial. A finite generating family $I = (a_{0}, \dots, a_{r})$ with $a = a_{0}$ only produces the presentation $C$ of step 3.3, whose kernel is described in step 4.1; the image $B$ and hence the algebra are the same for every such family, with the identity of $B$ as the canonical isomorphism. Finally, for $a^{'} \in I$ with $a^{'} \neq a$ the algebras $B \subseteq A_{a}$ and $B^{'} \subseteq A_{a^{'}}$ lie in different localizations of $A$ ; the statement asserts no identification between them, and the transition maps between the corresponding charts are supplied separately by the standard-chart theorem.

Remarks

The vanishing of $A [I / a]$ is allowed: if $a = 0$ then $0$ lies in the multiplicative set generated by $a t$ and the localization, and hence $A [I / a]$ , is the zero ring, consistently with $A_{a} = 0$ ; all four assertions then hold trivially. The description $B = {b / a^{n}}$ is the normal form used in the chart computations of the blowup.

Depends on

Dependency tree · 2 levels

Sources

The Stacks Project, Commutative Algebra, Section 10.70 (Blow up algebras)