Separable residue and the cotangent sequence of a local algebra

A conormal computation. Let $S$ be any commutative $k$-algebra, let $I\subseteq S$ be an ideal and $B=S\mathrm{/}I$. Then $I\mathrm{/}{I}^2\to B{\otimes }_S{\mathrm{\Omega }}_{S\mathrm{/}k}\to {\mathrm{\Omega }}_{B\mathrm{/}k}\to 0$ is exact, the first map sending the class of $x\in I$ to $1\otimes \mathrm{d}x$. The second map exists by [F1] applied to $S\to B$, and it is surjective because the elements $\mathrm{d}b$ generate ${\mathrm{\Omega }}_{B\mathrm{/}k}$; the composite is zero because $x\in I$ maps to $\mathrm{d}(x+I)=0$. The first map is well defined because for $x,y\in I$ one has $\mathrm{d}(xy)=x\mathrm{d}y+y\mathrm{d}x$, which lies in the image of $I{\otimes }_S{\mathrm{\Omega }}_{S\mathrm{/}k}$ in ${\mathrm{\Omega }}_{S\mathrm{/}k}$, so the classes of $xy$ and of $0$ in $I\mathrm{/}{I}^2$ have the same image. Let $Q$ be the cokernel of the first map. A $B$-linear map $Q\to M$ is the same thing as a $k$-derivation $D\text{\hspace{0.17em}⁣}:S\to M$ with $D(x)=0$ for all $x\in I$, because $B$-linear maps out of $B{\otimes }_S{\mathrm{\Omega }}_{S\mathrm{/}k}$ correspond by [F1] to $k$-derivations of $S$, and the quotient imposes exactly the vanishing on the classes $\mathrm{d}x$, $x\in I$. Such a $D$ factors through a $k$-derivation $B\to M$: it is constant on cosets, since $D(x+b)=D(b)$ for $x\in I$, and it satisfies Leibniz on classes, since for $x\in I$ the identity $D((x+b)b^{\mathrm{\prime }})=xD(b^{\mathrm{\prime }})+b^{\mathrm{\prime }}D(x)+bD(b^{\mathrm{\prime }})=bD(b^{\mathrm{\prime }})+b^{\mathrm{\prime }}D(b)$ holds because $xM=0$ for the $B$-module $M$. Conversely every $k$-derivation of $B$ pulled back along $S\to B$ is such a $D$. By [F1] the functor $M\mapsto {\mathrm{Hom}}_B(Q,M)$ is therefore isomorphic to $M\mapsto {\mathrm{Hom}}_B({\mathrm{\Omega }}_{B\mathrm{/}k},M)$, so the canonical map $Q\to {\mathrm{\Omega }}_{B\mathrm{/}k}$ is an isomorphism.

Preparation of the section. Write $\kappa =k(T)(\alpha )$ with $T=(t_1,\dots ,t_r)$ as in [F2] and [F3], and let $P\in k(T)[X]$ be the minimal polynomial of $\alpha$ over $k(T)$; by [F4] and [F5] the polynomial $P$ is monic, irreducible and separable, so $P^{\mathrm{\prime }}(\alpha )\mathrm{\ne }0$ by [F6], since $P^{\mathrm{\prime }}$ is nonzero and $\gcd (P,P^{\mathrm{\prime }})=1$ makes its class a unit of $\kappa =k(T)[X]\mathrm{/}(P)$. Put $\bar{R}:=R\mathrm{/}{\mathfrak{m}}^2$, a local ring with maximal ideal $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ and residue field $\kappa$, and write $\pi \text{\hspace{0.17em}⁣}:\bar{R}\to \kappa$ for the quotient map. Choose $y_1,\dots ,y_r,a\in \bar{R}$ with $\pi (y_i)=t_i$ and $\pi (a)=\alpha$, and let ${\phi }_0\text{\hspace{0.17em}⁣}:k[T]\to \bar{R}$ be the $k$-algebra map with $T_i\mapsto y_i$. It is injective because the $t_i$ are algebraically independent over $k$, so $k[y_1,\dots ,y_r]$ is a polynomial ring and every nonzero element $q(y)$ of it is a unit of the local ring $\bar{R}$, since $\pi (q(y))=q(t)\mathrm{\ne }0$ places it outside the maximal ideal. By the universal property of localisation, ${\phi }_0$ extends to a $k$-algebra map $\phi \text{\hspace{0.17em}⁣}:k(T)\to \bar{R}$, and $\pi \circ \phi$ is the inclusion $k(T)\hookrightarrow \kappa$ because they agree on the generators $T_i$.

Finite separable residue. If $\kappa \mathrm{/}k$ is finite separable, then by [F3] there is $\alpha \in \kappa$ with $\kappa =k(\alpha )$; its minimal polynomial $P\in k[X]$ is separable by [F5], so $P^{\mathrm{\prime }}(\alpha )\mathrm{\ne }0$ by [F4] and [F6]. Every $k$-derivation $D\text{\hspace{0.17em}⁣}:\kappa \to M$ into a $\kappa$-module satisfies $0=D(P(\alpha ))={\sum }_{i}D(c_i){\alpha }^i+P^{\mathrm{\prime }}(\alpha )D(\alpha )=P^{\mathrm{\prime }}(\alpha )D(\alpha )$ with $c_i\in k$ and $D(c_i)=0$, so $D(\alpha )=0$ because $\kappa$ is a field, and then $D=0$ on $\kappa =k[\alpha ]$. By [F1] this forces ${\mathrm{\Omega }}_{\kappa \mathrm{/}k}=0$.

Applying step 1.1 with $S=R$, $I=\mathfrak{m}$ and $B=\kappa$ gives exactness of $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2\to {\mathrm{\Omega }}_{R\mathrm{/}k}{\otimes }_R\kappa \to {\mathrm{\Omega }}_{\kappa \mathrm{/}k}\to 0$; it remains to prove that the first map is injective.

Correction of the lift. With the notation of step 1.2 form $\delta :=\phi (P)(a)\in \bar{R}$, where $\phi (P)\in \bar{R}[X]$ is $P$ with coefficients transported by $\phi$. Then $\pi (\delta )=P(\alpha )=0$, so $\delta \in \mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$, and $(\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2{)}^2=0$ in $\bar{R}$ because $\mathfrak{m}\cdot \mathfrak{m}\subseteq {\mathfrak{m}}^2$. Also $\pi (\phi (P^{\mathrm{\prime }})(a))=P^{\mathrm{\prime }}(\alpha )\mathrm{\ne }0$, so $\phi (P^{\mathrm{\prime }})(a)$ is a unit of $\bar{R}$. Set $a^{\mathrm{\prime }}:=a-\phi (P^{\mathrm{\prime }})(a{)}^{-1}\delta$, an element of $\bar{R}$ with the same residue $\pi (a^{\mathrm{\prime }})=\pi (a)=\alpha$. Taylor expansion in the commutative ring $\bar{R}$ terminates after the linear term because $a^{\mathrm{\prime }}-a\in \mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ has square zero, so $\phi (P)(a^{\mathrm{\prime }})=\phi (P)(a)+\phi (P^{\mathrm{\prime }})(a)(a^{\mathrm{\prime }}-a)=\delta -\delta =0$. Hence the $k$-algebra map $k(T)[X]\to \bar{R}$ with $T_i\mapsto y_i$ and $X\mapsto a^{\mathrm{\prime }}$ kills $P$, and by [F4] it descends along $\kappa \cong k(T)[X]\mathrm{/}(P)$ to a $k$-algebra map $s\text{\hspace{0.17em}⁣}:\kappa \to \bar{R}$ with $\pi \circ s={\mathrm{id}}_{\kappa }$.

A derivation inverse to $\mathrm{d}$. Define $D:={\mathrm{id}}_{\bar{R}}-s\circ \pi \text{\hspace{0.17em}⁣}:\bar{R}\to \bar{R}$; its image lies in $\ker \pi =\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$, and $D(x)=x$ for $x\in \mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ because $\pi (x)=0$. For $u,v\in \bar{R}$ write $u=s(\pi (u))+D(u)$ and $v=s(\pi (v))+D(v)$; since $\pi$ is a ring map, $D(u)$ and $D(v)$ have square zero and their product with any element of $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ vanishes, so expanding gives $uv=s(\pi (uv))+uD(v)+vD(u)$ and hence $D(uv)=uD(v)+vD(u)$. Thus $D$ is a $k$-derivation of $\bar{R}$ into the $\bar{R}$-module $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$, and composing with $R\to \bar{R}$ gives a $k$-derivation $R\to \mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ whose value at $x\in \mathfrak{m}$ is the class of $x$ modulo ${\mathfrak{m}}^2$.

Conclusion. By [F1] the derivation of step 3.1 corresponds to an $R$-linear map ${\mathrm{\Omega }}_{R\mathrm{/}k}\to \mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$, which factors through ${\mathrm{\Omega }}_{R\mathrm{/}k}{\otimes }_R\kappa \to \mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ because the target is annihilated by $\mathfrak{m}$, giving a $\kappa$-linear $\theta$ with $\theta (1\otimes \mathrm{d}x)=x<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>{\mathfrak{m}}^2$ for $x\in R$. The first map of step 2.1 sends the class of $x\in \mathfrak{m}$ to $1\otimes \mathrm{d}x$, so $\theta$ is a left inverse of it and that map is injective; with step 2.1, the displayed sequence is exact.

Finite separable residue concluded. By step 1.3 the hypothesis of the final clause gives ${\mathrm{\Omega }}_{\kappa \mathrm{/}k}=0$, so the exact sequence of step 4.1 reads $0\to \mathfrak{m}\mathrm{/}{\mathfrak{m}}^2\to {\mathrm{\Omega }}_{R\mathrm{/}k}{\otimes }_R\kappa \to 0$, that is, $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2\cong {\mathrm{\Omega }}_{R\mathrm{/}k}{\otimes }_R\kappa$ via the first map.