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

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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.
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Differentials of dual numbers in both characteristics

Example

Let $k$ be a commutative ring and let $B = k [ϵ] / (ϵ^{2})$ be the algebra of dual numbers, with class $ϵ$ satisfying $ϵ^{2} = 0$ ; for $k$ a field, $B$ is the coordinate ring of the dual-numbers scheme $D_{k}$ (The affine scheme of dual numbers). Then the Jacobian presentation of $ϵ^{2}$ gives $Ω_{B / k} = B d ϵ / (2 ϵ d ϵ) ≅ B / (2 ϵ),$ without any assumption on the characteristic, since the derivative of $t^{2}$ is $2 t$ and the relation is the cyclic submodule generated by $2 ϵ d ϵ$ . Consequently:

if $k$ has characteristic $2$ , that is $2 = 0$ in $k$ , then $2 ϵ = 0$ and $Ω_{B / k} ≅ B d ϵ$ is free of rank one over $B$ with basis $d ϵ$ ;
if $k$ is a field of characteristic $\neq 2$ , then $2$ is invertible, so the ideal $(2 ϵ)$ is $(ϵ)$ , and $Ω_{B / k} ≅ B / (ϵ) ≅ k$ is one-dimensional over $k$ with basis $d ϵ$ ; it is not free over $B$ ;

in neither case is the answer obtained by inverting $2$ when it is not invertible, and the relation can be trivial without the ring becoming a field.

Facts & Assumptions

Given: A commutative ring $k$ , the polynomial algebra $k [t]$ , the element $f = t^{2}$ , the quotient $B = k [t] / (t^{2})$ with the class $ϵ$ of $t$ , and (in the last case) a field $k$ of characteristic $\neq 2$ .

[F1]

Jacobian presentation of Ω: for $P = A [x_{1}, \dots, x_{n}]$ and $B = P / I$ with $I = (f_{1}, \dots, f_{r})$ , the module $Ω_{B / A}$ is the cokernel of the $B$ -linear map $B^{r} \to B^{n}$ whose $j$ -th column is $(\partial f_{j} / \partial x_{1}, \dots, \partial f_{j} / \partial x_{n})$ ; in particular for $n = r = 1$ and $I = (f)$ one gets $Ω_{B / A} ≅ B / (f^{'})$ with $f^{'}$ the derivative, the cokernel of multiplication by $f^{'}$ on $B$ .

[F2]

The affine scheme of dual numbers: for a field $k$ the dual-numbers scheme is $D_{k} = Spec (k [ϵ] / (ϵ^{2}))$ , so the ring $B$ above is its coordinate ring and $Ω_{B / k}$ is the module whose associated sheaf is $Ω_{D_{k} / k}$ .

[F3]

First isomorphism theorem for rings: $R / \ker f ≅ im f$ : for a ring homomorphism $φ ⁣ : R \to S$ there is an isomorphism $R / \ker φ ≅ im φ$ ; applied to the evaluation $B \to k$ , $ϵ \mapsto 0$ , it identifies $B / (ϵ) ≅ k$ because that map is surjective with kernel the ideal $(ϵ)$ .

Verification

1.1

Apply [F1] with $A = k$ , $n = 1$ , $r = 1$ , $P = k [t]$ , $f_{1} = t^{2}$ and $B = k [t] / (t^{2})$ : the derivative is $f^{'} = 2 t$ , whose class in $B$ is $2 ϵ$ , so the cokernel of multiplication by $2 ϵ$ on $B$ is $Ω_{B / k} ≅ B / (2 ϵ)$ . Writing the image of the free generator for $B^{1}$ as $d ϵ$ , this reads $Ω_{B / k} = B d ϵ / (2 ϵ d ϵ)$ , the submodule $B (2 ϵ d ϵ)$ corresponding to the ideal $(2 ϵ) \subseteq B$ under the identification $B d ϵ ≅ B$ .

F1given

2.1

Suppose $2 = 0$ in $k$ . Then $2 ϵ = 0 \in B$ , so the ideal $(2 ϵ)$ is the zero ideal [step 1.1], and $Ω_{B / k} ≅ B / (0) = B$ : the assignment $b \mapsto b d ϵ$ is a $B$ -module isomorphism $B \to Ω_{B / k}$ with inverse induced by $d ϵ \mapsto 1$ , so $d ϵ$ is a basis of $Ω_{B / k}$ over $B$ and $Ω_{B / k}$ is free of rank one.

step 1.1given

2.2

Suppose $k$ is a field of characteristic $\neq 2$ . Then $2 \neq 0$ in $k$ , so $2$ is a unit of $k$ and hence of $B$ , and $(2 ϵ) = (ϵ)$ : the two generators differ by the unit $2$ [step 1.1]. Hence $Ω_{B / k} ≅ B / (ϵ)$ . By [F3], applied to the surjection $B \to k$ with $ϵ \mapsto 0$ whose kernel is $(ϵ)$ , one has $B / (ϵ) ≅ k$ , so $Ω_{B / k} ≅ k$ is one-dimensional over $k$ with basis image of $d ϵ$ , and $ϵ d ϵ = 0$ holds in $Ω_{B / k}$ while $d ϵ \neq 0$ .

step 1.1F3

3.1

In the case of step 2.2 the module $Ω_{B / k}$ is not free over $B$ : it has $k$ -dimension $1$ , whereas a free $B$ -module of rank one has $k$ -dimension equal to $\dim_{k} B = 2$ , since ${1, ϵ}$ is a $k$ -basis of $B$ (every class in $k [t] / (t^{2})$ is uniquely $c + d ϵ$ with $c, d \in k$ ). In the case of step 2.1 the dimension count is reversed and $Ω_{B / k}$ is free of rank one, so the two characteristics genuinely give different answers, and the presentation of step 1.1 is the common source of both. By [F2] these computations are those of the relative differentials of the dual-numbers scheme over $k$ when $k$ is a field.

step 2.1step 2.2F2given∎

Depends on

Used by

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Sources

Stacks Algebra 10.131.9 (standard reference, not scraped)
Vakil 22.2.7, pp.577-578 (standard reference, not scraped)