Finite-type field extensions with zero Ω

Converse, setup. Assume that $L\mathrm{/}k$ is finite and separable. By [F18] there is $\alpha \in L$ with $L=k(\alpha )$ (if $L=k$ take any $\alpha \in k$). Let $f\in k[x]$ be the minimal polynomial of $\alpha$; it is monic, irreducible, of degree $n\ge 1$, and separable because $\alpha$ is separable over $k$. If $n=1$ then $f^{\mathrm{\prime }}=1$ and $f^{\mathrm{\prime }}(\alpha )=1\mathrm{\ne }0$. If $n\ge 2$, then $f$ is irreducible and separable, so $f^{\mathrm{\prime }}\mathrm{\ne }0$ by [F18]; as $\mathrm{deg}{f}^{\mathrm{\prime }}<n=\mathrm{deg}f$ and $f$ is irreducible, $f\nmid f^{\mathrm{\prime }}$, so $f^{\mathrm{\prime }}(\alpha )\mathrm{\ne }0$, since $f^{\mathrm{\prime }}(\alpha )=0$ would give $f\mid f^{\mathrm{\prime }}$ by [F18]. In both cases $f^{\mathrm{\prime }}(\alpha )\mathrm{\ne }0$.

Forward, the finite-type model. Put $B:=k[y_1,\dots ,y_m]\subseteq L$. By [F1] the ring $B$ is a finitely generated $k$-algebra, a domain, and $\mathrm{Frac}(B)=L$. Writing $B=k[x_1,\dots ,x_m]\mathrm{/}I$ for the kernel $I$ of the evaluation $x_i\mapsto y_i$, [F2] shows that $I$ is finitely generated and that ${\mathrm{\Omega }}_{B\mathrm{/}k}$ is a quotient of the free module $B^m$; hence ${\mathrm{\Omega }}_{B\mathrm{/}k}$ is a finitely generated $B$-module.

Choice and the algebraic closure. Assume the Axiom of Choice (The Axiom of Choice). By [F5] there is an algebraic closure $K$ of $k$, so $K$ is algebraically closed with $\mathrm{char}K=\mathrm{char}k$; by [F8] every vector space over a field has a basis and every module over a field is flat; and by [F14] every proper ideal of a nonzero ring lies in a maximal ideal.

Converse, conclusion. The evaluation homomorphism $k[x]\to L$, $x\mapsto \alpha$, has kernel $(f)$ by [F18], so $k[x]\mathrm{/}(f)\cong k[\alpha ]=L$; the one-relation form of the Jacobian presentation [F2] gives ${\mathrm{\Omega }}_{L\mathrm{/}k}\cong L\mathrm{/}(f^{\mathrm{\prime }}(\alpha ))$. Since $f^{\mathrm{\prime }}(\alpha )\mathrm{\ne }0$ in the field $L$, the ideal $(f^{\mathrm{\prime }}(\alpha ))$ is all of $L$ and ${\mathrm{\Omega }}_{L\mathrm{/}k}=0$. This proves claim 2 for every finite separable extension.

Forward, localising at the zero prime. The set $S:=B\setminus \{0\}$ is a multiplicative subset of the domain $B$ with $S^{-1}B=\mathrm{Frac}(B)=L$ [step 1.2], so [F3] gives $S^{-1}{\mathrm{\Omega }}_{B\mathrm{/}k}\cong {\mathrm{\Omega }}_{L\mathrm{/}k}$; under the hypothesis of claim 1 this is $0$.

Clearing denominators. The $B$-module ${\mathrm{\Omega }}_{B\mathrm{/}k}$ is finitely generated [step 1.2] and vanishes at the prime ideal $(0)$ of the domain $B$ [step 2.2], so [F4] provides $s\in B\setminus \{0\}$ with $({\mathrm{\Omega }}_{B\mathrm{/}k}{)}_s=0$. Fix such an $s$ and put $C:=B_s{\otimes }_{k}K$, the principal localisation $B_s$ being as in [F7].

The differentials of $C$ vanish. By [F3] applied to the multiplicative set $\{1,s,s^2,\dots \}$ we have ${\mathrm{\Omega }}_{B_s\mathrm{/}k}\cong ({\mathrm{\Omega }}_{B\mathrm{/}k}{)}_s=0$, and [F6] gives ${\mathrm{\Omega }}_{C\mathrm{/}K}={\mathrm{\Omega }}_{B_s{\otimes }_{k}K\mathrm{/}K}\cong {\mathrm{\Omega }}_{B_s\mathrm{/}k}{\otimes }_{B_s}C=0$.

$C$ is nonzero. The localisation map $B\to B_s$ is injective, since $B$ is a domain with $s\mathrm{\ne }0$. The canonical map $B_s\to C$, $b\mapsto b\otimes 1$, is obtained by tensoring the injection $k\hookrightarrow K$ with the $k$-module $B_s$, which is flat by [F8]; hence it is injective by [F8], and $C\mathrm{\ne }0$ because $B_s\mathrm{\ne }0$.

$C$ is a finitely generated $K$-algebra. The $k$-algebra $B=k[y_1,\dots ,y_m]$ is generated by $y_1,\dots ,y_m$, so $B_s$ is generated by the images of $y_1,\dots ,y_m$ and of $s^{-1}$ [F7]; hence $C=B_s{\otimes }_{k}K$ is generated as a $K$-algebra by the images of these same elements, using the presentation $B_s\cong k[x_1,\dots ,x_m,z]\mathrm{/}(I,z\sigma -1)$, where $\sigma$ represents $s$ in $k[x_1,\dots ,x_m]$ and $z$ represents $s^{-1}$. Its base change is $K[x_1,\dots ,x_m,z]\mathrm{/}(I,z\sigma -1)K[x_1,\dots ,x_m,z]$ [F7]. So $C$ is of finite type over $K$.

$C$ is Noetherian. The field $K$ is a Noetherian ring [F2], and $C$ is a finitely generated $K$-algebra [step 4.3], so $C$ is Noetherian by [F15]; in particular every ideal of $C$ is a finitely generated $C$-module.

Maximal ideals are rational and have vanishing cotangent space. Let $\mathfrak{m}\subseteq C$ be a maximal ideal; one exists because $C\mathrm{\ne }0$ [step 4.2] and every proper ideal lies in a maximal ideal [F14]. By [F9] the field $C\mathrm{/}\mathfrak{m}$ is a finite extension of $K$, and since $K$ is algebraically closed [step 1.3] it has no nontrivial finite extension [F9], so $C\mathrm{/}\mathfrak{m}=K$: thus $\mathfrak{m}$ is a $K$-rational point of $\mathrm{Spec}C$. By [F10], together with ${\mathrm{\Omega }}_{C\mathrm{/}K}=0$ [step 4.1], $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2\cong {\mathrm{\Omega }}_{C\mathrm{/}K}{\otimes }_C(C\mathrm{/}\mathfrak{m})=0.$

The local ring at each maximal ideal is a field. Let $\mathfrak{m}$ be a maximal ideal and $\mathfrak{n}:=\mathfrak{m}{C}_{\mathfrak{m}}$, the maximal ideal of the local ring $C_{\mathfrak{m}}$ [F11]. Localising the short exact sequence $0\to {\mathfrak{m}}^2\to \mathfrak{m}\to \mathfrak{m}\mathrm{/}{\mathfrak{m}}^2\to 0$ at $C\setminus \mathfrak{m}$ is exact [F11] and gives $\mathfrak{n}\mathrm{/}{\mathfrak{n}}^2\cong (\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2{)}_{\mathfrak{m}}=0$ [step 5.2], so $\mathfrak{n}={\mathfrak{n}}^2$. The ideal $\mathfrak{m}$ is finitely generated [step 5.1], hence so is the $C_{\mathfrak{m}}$-module $\mathfrak{n}$; since $\mathfrak{n}=J(C_{\mathfrak{m}})$ is the Jacobson radical of the local ring $C_{\mathfrak{m}}$ [F12], Nakayama's lemma [F12] with $I=\mathfrak{n}$ and $M=\mathfrak{n}$ gives $\mathfrak{n}=0$. Therefore the maximal ideal of the nonzero ring $C_{\mathfrak{m}}$ is zero, so $C_{\mathfrak{m}}$ is a field, and its residue field is, by [F11], $C_{\mathfrak{m}}\mathrm{/}\mathfrak{n}=C_{\mathfrak{m}}\cong \mathrm{Frac}(C\mathrm{/}\mathfrak{m})=C\mathrm{/}\mathfrak{m}=K$ [step 5.2]; in particular $C_{\mathfrak{m}}\cong K$.

Primes inside a maximal ideal. Let $\mathfrak{p}\subseteq \mathfrak{m}$ be a prime ideal of $C$ with $\mathfrak{m}$ maximal. Taking $S=C\setminus \mathfrak{m}$ in [F13], the primes of $C_{\mathfrak{m}}$ correspond bijectively to the primes of $C$ contained in $\mathfrak{m}$; the field $C_{\mathfrak{m}}$ [step 6.1] has only the prime ideal $(0)$, so exactly one prime of $C$ is contained in $\mathfrak{m}$. Since $\mathfrak{m}$ itself is a prime ideal contained in $\mathfrak{m}$ and $\mathfrak{p}$ is another, $\mathfrak{p}=\mathfrak{m}$.

Every prime of $C$ is maximal, and the minimal primes are the maximal ideals. Let $\mathfrak{p}$ be a prime ideal of $C$. Since $C\mathrm{\ne }0$ [step 4.2] and $\mathfrak{p}\mathrm{\ne }C$, the quotient $C\mathrm{/}\mathfrak{p}$ is a nonzero ring, so it has a maximal ideal; by [F14] its preimage $\mathfrak{m}$ in $C$ is a maximal ideal with $\mathfrak{p}\subseteq \mathfrak{m}$, and step 7.1 gives $\mathfrak{p}=\mathfrak{m}$. So every prime is maximal, and conversely every maximal ideal is prime [F14]. Hence the primes of $C$ are exactly the maximal ideals; no prime is strictly contained in another, so each prime is a minimal prime ideal.

$C$ is reduced. By [F15] the nilradical of $C$ is the intersection of the prime ideals, which by step 8.1 is the intersection of all maximal ideals. Let $x\in \mathrm{Nil}(C)$ and let $\mathfrak{m}$ be any maximal ideal. The image $x\mathrm{/}1\in C_{\mathfrak{m}}$ is nilpotent and $C_{\mathfrak{m}}$ is a field [step 6.1], so $x\mathrm{/}1=0$; by [F11] there is $t\mathrm{\notin }\mathfrak{m}$ with $tx=0$, so the annihilator of $x$ is not contained in $\mathfrak{m}$. As this holds for every maximal ideal and every proper ideal lies in a maximal ideal [F14], the annihilator of $x$ is $C$ and $x=0$. Hence $\mathrm{Nil}(C)=0$ and $C$ is reduced [F15].

$C$ is a finite product of copies of $K$. By [F15] the Noetherian ring $C$ [step 5.1] has only finitely many minimal primes, which by step 8.1 are exactly its maximal ideals ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_r$; here $r\ge 1$ because $C\mathrm{\ne }0$ has a maximal ideal [step 4.2, F14]. Distinct maximal ideals are comaximal, and the intersection ${\bigcap }_{i=1}^r{\mathfrak{m}}_i$ of all primes is the nilradical of $C$ [F15], which is zero [step 9.1]. The Chinese remainder theorem [F16] therefore gives $C\cong C/{\bigcap }_{i=1}^r{\mathfrak{m}}_i\cong {\prod }_{i=1}^{r}C\mathrm{/}{\mathfrak{m}}_i=K^r,$ using $C\mathrm{/}{\mathfrak{m}}_i=K$ from step 5.2.

$B_s$ is finite-dimensional over $k$. Choose a $k$-basis $(v_i{)}_{i\in I}$ of $B_s$ [F8]. For any finitely many basis elements $v_{i_1},\dots ,v_{i_N}$, put $V:={⨁}_{j=1}^{N}k{v}_{i_j}$, so that $V\hookrightarrow B_s$ is injective; tensoring with the flat $k$-module $K$ [F8] gives an injection $V{\otimes }_{k}K\hookrightarrow B_s{\otimes }_{k}K=C$ [step 3.1], and [F8] gives $V{\otimes }_{k}K\cong {⨁}_{j=1}^{N}K(v_{i_j}\otimes 1),$ using $k{\otimes }_{k}K\cong K$. Hence the elements $v_{i_j}\otimes 1$ are $K$-linearly independent in $C$. Therefore $\{v_i\otimes 1:i\in I\}$ is a $K$-linearly independent subset of $C$, and since $C\cong K^r$ has a spanning set with $r$ elements [step 10.1], [F21] shows that it is finite with at most $r$ elements. The map $B_s\to C$ is injective [step 4.2], so the image of the basis also has $\mathrm{\mid }I\mathrm{\mid }$ elements and $\mathrm{\mid }I\mathrm{\mid }\le r$: the $k$-vector space $B_s$ is finite-dimensional.

$L=B_s$ is finite over $k$. The ring $B_s$ is a domain, being a subring of the field $L$, and finite-dimensional over $k$ [step 11.1]. For $0\mathrm{\ne }b\in B_s$ the multiplication map $b\cdot \text{\hspace{0.17em}⁣}:B_s\to B_s$ is $k$-linear with kernel zero; by rank-nullity [F17] it is surjective, so some $b^{\mathrm{\prime }}$ satisfies $b{b}^{\mathrm{\prime }}=1$ and $b$ is a unit. Hence $B_s$ is a field. Since $B\subseteq B_s\subseteq L$ and $\mathrm{Frac}(B)=L$ [step 1.2], we get $L=\mathrm{Frac}(B)\subseteq \mathrm{Frac}(B_s)=B_s\subseteq L$, so $B_s=L$; in particular $L$ is finite-dimensional over $k$, that is, $L\mathrm{/}k$ is finite.

An element with non-separable minimal polynomial. Suppose now that $L\mathrm{/}k$ is not separable. Since it is finite [step 12.1], it is algebraic [F18], and by [F18] some $\alpha \in L$ fails to be separable over $k$, which for an algebraic element means that its minimal polynomial $f\in k[x]$ is not separable. If $\mathrm{char}k=0$ then $k$ would be perfect [F5], so every irreducible polynomial over $k$ would be separable, a contradiction; hence $\mathrm{char}k=p>0$. By [F19] there are a unique $e\ge 0$ and an irreducible separable $g\in k[x]$ with $f(x)=g(x^{p^e})$, and since $f$ is not separable the case $e=0$ does not occur, so $e\ge 1$. Then $g$ is nonconstant of some degree $d\ge 1$, and $\mathrm{deg}f=p^{e}d$.

A nonzero nilpotent in $B^{\mathrm{\prime }}{\otimes }_{k}K$. The field $K$ has characteristic $p$ and is perfect, so by [F5] its $p^e$-th power map is surjective. Write $g={\sum }_i{a}_i{x}^i$ and choose $b_i\in K$ with $b_i^{p^e}=a_i$, and set $g_1:={\sum }_i{b}_i{x}^i\in K[x]$. Since $K[x]$, like $K$, has characteristic $p$, the binomial theorem gives $(u+v{)}^{p^e}=u^{p^e}+v^{p^e}$ in $K[x]$ [F5], whence $g_1(x{)}^{p^e}={\sum }_i(b_i{x}^i{)}^{p^e}={\sum }_i{b}_i^{p^e}{x}^{i{p}^e}={\sum }_i{a}_i{x}^{i{p}^e}=g(x^{p^e})=f(x)$ in $K[x]$; also $1\le d=\mathrm{deg}{g}_1<\mathrm{deg}f=p^{e}d$. Let $B^{\mathrm{\prime }}:=k[\alpha ]\subseteq L$, which by [F18] satisfies $B^{\mathrm{\prime }}\cong k[x]\mathrm{/}(f)$ and is a field; by [F7] the $K$-algebra $M:=B^{\mathrm{\prime }}{\otimes }_{k}K\cong K[x]\mathrm{/}(f)=K[x]\mathrm{/}(g_1^{p^e})$ contains the class $z$ of $g_1$, which is nonzero because $\mathrm{deg}{g}_1<\mathrm{deg}{g}_1^{p^e}=\mathrm{deg}f$, while $z^{p^e}=0$ because $g_1^{p^e}=f\equiv 0$.

The canonical map $M\to C$ is injective. The field $B^{\mathrm{\prime }}=k[\alpha ]$ is a subfield of $L$, so $B^{\mathrm{\prime }}\hookrightarrow L$ is injective, and $M=B^{\mathrm{\prime }}{\otimes }_{k}K$ is flat over the field $B^{\mathrm{\prime }}$ because it is the extension of scalars of the flat $k$-module $K$ [F8, F20]. Hence $M\cong M{\otimes }_{B^{\mathrm{\prime }}}{B}^{\mathrm{\prime }}\hookrightarrow M{\otimes }_{B^{\mathrm{\prime }}}L$ is injective [F8], and by the canonical identification $(B^{\mathrm{\prime }}{\otimes }_{k}K){\otimes }_{B^{\mathrm{\prime }}}L\cong L{\otimes }_{k}K=B_s{\otimes }_{k}K=C$ [F20, step 12.1] this map is the canonical map $M\to C$, $b\otimes c\mapsto b\otimes c$.

Conclusion. The image of the nonzero nilpotent $z$ of step 14.1 under the injective map of step 15.1 is a nonzero element of $C$ whose $p^e$-th power is $0$; this contradicts step 10.1, since in the product of fields $C\cong K^r$ the only nilpotent element is $0$. Hence $L\mathrm{/}k$ is separable, and together with step 12.1 it is finite and separable, so claim 1 holds; claim 2 is step 2.1.