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

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Differentials of a plane hypersurface

Example

Let $k$ be a commutative ring, let $f \in k [x, y]$ be an arbitrary polynomial and let $B = k [x, y] / (f)$ be the quotient by the principal ideal it generates. Writing $f_{x} = \partial f / \partial x$ and $f_{y} = \partial f / \partial y$ for the partial derivatives, the module of Kähler differentials of $B$ over $k$ is presented by the single Jacobian relation of $f$ : $Ω_{B / k} ≅ (B d x \oplus B d y) / B (f_{x} d x + f_{y} d y) .$ No regularity, smoothness or non-vanishing hypothesis is imposed on $f$ , and no flatness is assumed of $B$ over $k$ : the presentation holds for every $f \in k [x, y]$ , including $f = 0$ and including polynomials whose first partial derivatives both vanish in positive characteristic. For $f = 0$ one recovers $Ω_{k [x, y] / k} ≅ B^{2}$ , and over a ring $k$ in which the integer $2 \neq 0$ the example $f = x^{2}$ has $f_{x} = 2 x \neq 0 \in B$ , so the relation is a nonzero cyclic submodule there.

Facts & Assumptions

Given: A commutative ring $k$ , the polynomial algebra $P = k [x, y]$ , a polynomial $f \in P$ , the ideal $I = (f) \subseteq P$ , the quotient $B = P / I$ , and the partial derivatives $f_{x}, f_{y} \in P$ with images in $B$ written the same way.

[F1]

Jacobian presentation of Ω: for a commutative ring $A$ , the polynomial algebra $P = A [x_{1}, \dots, x_{n}]$ , an ideal $I = (f_{1}, \dots, f_{r}) \subseteq P$ generated by finitely many elements and the quotient $B = P / I$ , the module $Ω_{B / A}$ is the cokernel of the $B$ -linear map $B^{r} \to B^{n}$ whose $j$ -th column is $(\partial f_{j} / \partial x_{1}, \dots, \partial f_{j} / \partial x_{n})$ , that is, $Ω_{B / A} ≅ B^{n} / \sum_{j} B (\partial f_{j} / \partial x_{1}, \dots, \partial f_{j} / \partial x_{n})$ ; no flatness or minimality of $r$ is assumed.

[F2]

Polynomial differentials are free with $n = 2$ : $Ω_{P / k}$ is free with basis $d x, d y$ , the derivations $\partial / \partial x, \partial / \partial y$ satisfy $\partial x / \partial x = 1$ , $\partial x / \partial y = 0$ , $\partial y / \partial x = 0$ , $\partial y / \partial y = 1$ , and $d g = (\partial g / \partial x) d x + (\partial g / \partial y) d y$ for every $g \in P$ .

[F3]

Derivation of an algebra: a $k$ -derivation satisfies the Leibniz rule $D (g h) = g D (h) + h D (g)$ and annihilates $k$ .

Verification

1.1

Apply [F1] with $A = k$ , $n = 2$ , $r = 1$ and $f_{1} = f$ : the ideal $I = (f)$ is generated by the single element $f$ , and $Ω_{B / k}$ is the cokernel of the $B$ -linear map $B^{1} \to B^{2}$ with the single column $(f_{x}, f_{y})$ . Identifying $B^{2} = B d x \oplus B d y$ by the standard basis, the image is the cyclic submodule generated by $f_{x} d x + f_{y} d y$ , so $Ω_{B / k} ≅ (B d x \oplus B d y) / B (f_{x} d x + f_{y} d y)$ .

F1given

1.2

The derivatives of the powers of $x$ : by the Leibniz rule [F3] and $\partial x / \partial x = 1$ , $\partial y / \partial x = 0$ [F2], induction on $n \geq 0$ gives $\partial (x^{n}) / \partial x = n x^{n - 1}$ and $\partial (y^{m}) / \partial x = 0$ for all $m \geq 0$ , with the case $n = 0$ read as $\partial (1) / \partial x = 0$ .

F2F3induction

2.1

The case $f = 0$ : then $B = P$ and $f_{x} = f_{y} = 0$ , so the relation submodule in step 1.1 is $B \cdot 0 = 0$ and $Ω_{B / k} ≅ B^{2}$ , which is the free module on $d x, d y$ already recorded in [F2].

step 1.1F2

2.2

Suppose $k$ has characteristic $p > 0$ and $f = x^{p}$ . Then step 1.2 gives $f_{x} = \partial (x^{p}) / \partial x = p x^{p - 1} = 0$ and $f_{y} = \partial (x^{p}) / \partial y = 0$ because $p = 0$ in $k$ , so again the relation submodule of step 1.1 vanishes and $Ω_{B / k} ≅ B^{2}$ even though $B = k [x, y] / (x^{p})$ is non-reduced: the presentation records no relation at all. If instead $2 \neq 0$ in $k$ and $f = x^{2}$ , the relation is $B \cdot (2 x d x)$ and $2 x \neq 0$ in $B$ : the ideal $(x^{2})$ contains no nonzero polynomial of degree one, so it cannot contain $2 x$ . Thus the relation is a nonzero cyclic submodule even when $2$ is a zero divisor in $k$ ; in either case the displayed presentation of step 1.1 is the one computed, and no hypothesis on $f$ beyond $f \in k [x, y]$ was used.

step 1.1step 1.2

3.1

Combining steps 1.1 with the cases 2.1 and 2.2, for every $f \in k [x, y]$ the module $Ω_{B / k}$ is the quotient displayed in the Example, the relation $f_{x} d x + f_{y} d y$ being determined by the partial derivatives of $f$ and possibly vanishing; since [F1] requires no flatness, smoothness or non-vanishing hypothesis, the presentation holds without any regularity assumption on $f$ , and the map $P \to B$ is the surjection onto the quotient by $(f)$ in all cases.

step 1.1step 2.1step 2.2∎

Depends on

Used by

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Sources

Stacks Algebra 10.131.9 (standard reference, not scraped)
Vakil 22.2.12, p.579 (standard reference, not scraped)