CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-6-sol)audited 2026-09-27

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A conormal left map with nonzero kernel

Statement refuted

“For an ideal $I$ of a ring $P$ , the conormal map $I / I^{2} \to B \otimes_{P} Ω_{P / A}$ is injective.”

Counterexample

Let $k$ be a field, $P = k [x]$ , and let $I = (x^{2}) \subseteq P$ with quotient $B = P / I = k [x] / (x^{2})$ . In the conormal sequence $I / I^{2} ⟶ B \otimes_{P} Ω_{P / k} ⟶ Ω_{B / k} ⟶ 0$ the class $[x^{3}] \in I / I^{2} = (x^{2}) / (x^{4})$ is nonzero, but its image $1 \otimes d (x^{3}) = 3 x^{2} d x$ is $0$ in $B \otimes_{P} Ω_{P / k} ≅ B d x$ , because $x^{2} = 0$ in $B$ . Hence the left map has a nonzero kernel, in every characteristic: for $p = 3$ the element $3 x^{2}$ is already $0$ in $P$ , and for $p \neq 3$ it becomes $0$ after the identification. In particular the sequence is only right exact.

Facts & Assumptions

Given: A field $k$ , the polynomial algebra $P = k [x]$ , the principal ideal $I = (x^{2}) = x^{2} P$ , the quotient $B = P / I$ and the class $x^{3} \in I$ .

[F1]

Conormal exact sequence for an algebra quotient: for a ring map $A \to P$ , an ideal $I \subseteq P$ and $B = P / I$ , the sequence $I / I^{2} \overset{α}{\to} B \otimes_{P} Ω_{P / A} \to Ω_{B / A} \to 0$ is exact with $α$ sending the class of $i \in I$ to $1 \otimes d i$ ; no injectivity of $α$ is asserted.

[F2]

Polynomial differentials are free with $n = 1$ : $Ω_{P / k}$ is free with basis $d x$ , and $d g = (\partial g / \partial x) d x$ for every $g \in P$ , where $\partial / \partial x$ is the $k$ -derivation with $\partial x / \partial x = 1$ ; for the powers this gives $\partial (x^{n}) / \partial x = n x^{n - 1}$ .

[F3]

Derivation of an algebra: the map $d$ is a $k$ -derivation, so it is additive, kills $k$ , and satisfies the Leibniz rule.

[F4]

Over an integral domain, degrees add under multiplication of nonzero polynomials: in an integral domain, nonzero polynomials satisfy $\deg (f g) = \deg f + \deg g$ and $f g \neq 0$ .

Verification

1.1

Apply [F1] with $A = k$ : the sequence $I / I^{2} \overset{α}{\to} B \otimes_{P} Ω_{P / k} \to Ω_{B / k} \to 0$ is exact, $α$ sends the class of $i \in I$ to $1 \otimes d i$ , and the statement of [F1] explicitly leaves injectivity of $α$ open.

F1given

1.2

The powers of the ideal: $I = x^{2} P$ and $I^{2} = x^{4} P$ , since $I^{2}$ is generated by the products of two elements of $I$ , and $x^{2} f \cdot x^{2} g = x^{4} f g$ . Hence $I / I^{2} = (x^{2}) / (x^{4})$ . The class $[x^{3}]$ is nonzero: if $x^{3} \in (x^{4})$ , there would be $g \in P$ with $x^{3} = x^{4} g$ , so $\deg (x^{3}) = 3$ equals $\deg (x^{4} g) = 4 + \deg g$ for $g \neq 0$ and is impossible for $g = 0$ , by the degree rule of [F4] applied in the integral domain $P = k [x]$ .

F4given

1.3

The target: by [F2] the module $Ω_{P / k}$ is free with basis $d x$ , so $B \otimes_{P} Ω_{P / k} ≅ B d x$ as $B$ -modules, via $b \otimes d x \mapsto b d x$ ; here $B = B$ is written as usual and $x^{2} = 0$ in $B$ because $I = (x^{2})$ .

F2given

2.1

The image of the class: by step 1.1, $α ([x^{3}]) = 1 \otimes d (x^{3})$ . By [F2] the derivation $d$ satisfies $d (x^{3}) = (\partial (x^{3}) / \partial x) d x = 3 x^{2} d x$ , the coefficient $3$ being read in $k$ ; by the Leibniz rule [F3] this is the same as $\partial (x^{3}) / \partial x = 3 x^{2}$ obtained from $\partial (x^{n}) / \partial x = n x^{n - 1}$ .

F2F3

3.1

Under the identification of step 1.3 the element $1 \otimes 3 x^{2} d x$ of step 2.1 is $3 x^{2} \cdot d x \in B d x$ , and this is $0$ since $x^{2} = 0$ in $B$ . So $α ([x^{3}]) = 0$ , while $[x^{3}] \neq 0$ by step 1.2; the class $[x^{3}]$ therefore lies in the nonzero kernel of $α$ , in every characteristic: for $p = 3$ the coefficient $3 x^{2}$ is already zero in $P$ , and for $p \neq 3$ it is nonzero in $P d x$ but its image in $B d x$ vanishes.

step 1.2step 1.3step 2.1

4.1

Consequently the left map $α ⁣ : I / I^{2} \to B \otimes_{P} Ω_{P / k}$ of the conormal sequence is not injective, so the sequence is exact at $B \otimes_{P} Ω_{P / k}$ and at $Ω_{B / k}$ but not at $I / I^{2}$ ; the sequence is only right exact. The witness is the single element $[x^{3}]$ , whose image is $3 x^{2} d x = 0$ . This refutes the displayed statement and shows why injectivity of $α$ is deliberately excluded from [F1].

step 1.1step 3.1∎

Depends on

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Sources

Stacks Algebra 10.131.9 (standard reference, not scraped)
Vakil 22.2.12-13, pp.579-580 (standard reference, not scraped)