The Stacks project

Proof. This follows from Lemma 15.114.3. Details omitted. $\square$

Proof. A discrete valuation ring is Nagata if and only if it is N-2. Let $K_1/K$ be a finite purely inseparable field extension. We have to show that the integral closure $A_1$ of $A$ in $K_1$ is finite over $A$, see Algebra, Lemma 10.161.12. Since $L/K$ is separable and $K_1/K$ is purely inseparable, the algebra $L \otimes _ K K_1$ is a field (by Algebra, Lemmas 10.43.6 and 10.46.10). Let $B_1$ be the integral closure of $B$ in $L \otimes _ K K_1$. Since $B$ is Nagata, $B_1$ is finite over $B$. Since $B \otimes _ A A_1 \subset B_1$ and $B$ is Noetherian, we see that $B \otimes _ A A_1$ is finite over $B$. As $A \to B$ is faithfully flat, this implies $A_1$ is finite over $A$, see Algebra, Lemma 10.83.2. $\square$

Proof. Since $A$ is finite over $A'$ and since $A'_ f = A_ f$ we can Cchoose $t > 0$ such that $f^ t A \subset A'$. Set $n_0 = 2t$. Given $n, B', g, \varphi '$ as in the statement of the lemma, denote $N \subset B'$ the set of elements $b \in B'$ such that $b \bmod g^ nB' \in \varphi '(f^ tA)$. Set $B = g^{-t}N$. As $f^ tA' \subset f^ tA$ and $\varphi '$ sends $f$ to $g$ we have $g^ tB' \subset N$, hence $B' \subset B$. Since $f^ tA \cdot f^ tA \subset f^ t \cdot f^ tA$ and $\varphi '$ sends $f$ to $g$, we see that $N \cdot N \subset g^ t N$. Hence we obtain a multiplication on $B$ extending the multiplication of $B'$. We have an isomorphism of $A'/f^ nA'$-modules

where the module structures on the right are defined using $\varphi '$. Since $A/f^ tA'$ is a finite $A'$-module, we conclude that $B/g^ tB'$ is a finite $B'$-module and hence we see that $B' \to B$ is finite. Finally, we leave it to the reader to see that the displayed isomorphism of modules sends $fA$ into $gB$ and induces an isomorphism of rings $\varphi : A/fA \to B/gB$ compatible with $\varphi '$ (it even induces an isomorphism $A/f^ tA \to B/g^ tB$ but we don't need this). $\square$

Proof. Let us use notation as in Remark 15.114.1; we will use all the observations made there. Since $L/K$ is separable, the algebra $L_1 = L \otimes _ K K_1$ is reduced (Algebra, Lemma 10.43.6). Since $B$ is Nagata, the ring extension $B \subset B_1$ is finite where $B_1$ is the integral closure of $B$ in $L_1$ and $B_1$ is a Nagata ring. Similarly, the ring $A$ is Nagata by Lemma 15.116.2 hence $A \subset A_1$ is finite and $A_1$ is a Nagata ring too. Moreover, the same assertions are true for $K_2$, i.e., $L_2 = L \otimes _ K K_2$ is reduced, the ring extensions $A_1 \subset A_2$ and $B_1 \subset B_2$ are finite where $A_2$, resp. $B_2$ is the integral closure of $A$, resp. $B$ in $K_2$, resp. $L_2$.

Let $\pi \in A$ be a uniformizer. Observe that $\pi $ is a nonzerodivisor on $K_1$, $K_2$, $A_1$, $A_2$, $L_1$, $L_2$, $B_1$, and $B_2$ and we have $K_1 = (A_1)_\pi $, $K_2 = (A_2)_\pi $, $L_1 = (B_1)_\pi $, and $L_2 = (B_2)_\pi $. We may write $K_2 = K_1(\alpha )$ where $\alpha ^ p = a_1 \in K_1$, see Fields, Lemma 9.14.5. After multiplying $\alpha $ by a power of $\pi $ we may and do assume $a_1 \in A_1$. For the rest of the proof it is convenient to write $K_2 = K_1[x]/(x^ p - a_1)$ and $L_2 = L_1[x]/(x^ p - a_1)$. Consider the extensions of rings

We may apply Lemma 15.116.3 to $A'_2 \subset A_2$ and $f = \pi ^2$ and to $B'_2 \subset B_2$ and $f = \pi ^2$. Choose an integer $n$ large enough which works for both of these.

Consider the algebras

Observe that $K_3/K_1$ and $L_3/L_1$ are finite étale algebra extensions of degree $p$. Consider the subrings

of $K_3 = (A'_2)_\pi $ and $L_3 = (B'_3)_\pi $. We are going to construct a commutative diagram

Namely, $\varphi '$ is the unique $A_1$-algebra isomorphism sending the class of $x$ to the class of $x$. Similarly, $\psi '$ is the unique $B_1$-algebra isomorphism sending the class of $x$ to the class of $x$. By our choice of $n$ we obtain, via Lemma 15.116.3 and Remark 15.116.4 finite ring extensions $A'_3 \subset A_3$ and $B'_3 \subset B_3$ such that $A'_3 \to B'_3$ extends to a ring map $A_3 \to B_3$ and a commutative diagram

with all the properties asserted in the references mentioned above (in particular $\varphi $ and $\psi $ are isomorphisms).

With all of this data in hand, we can finish the proof. Namely, we first observe that $A_3$ and $B_3$ are finite products of Dedekind domains with $\pi $ contained in all of the maximal ideals. Namely, if $\mathfrak p \subset A_3$ is a maximal ideal, then $\pi \in \mathfrak p$ as $A \to A_3$ is finite. Then $\mathfrak p/\pi ^2 A_3$ corresponds via $\varphi $ to a maximal ideal in $A_2 / \pi ^2A_2$ which is principal as $A_2$ is a finite product of Dedekind domains. We conclude that $\mathfrak p/\pi ^2 A_3$ is principal and hence by Nakayama we see that $\mathfrak p (A_3)_\mathfrak p$ is principal. The same argument works for $B_3$. We conclude that $A_3$ is the integral closure of $A$ in $K_3$ and that $B_3$ is the integral closure of $B$ in $L_3$. Let $\mathfrak q \subset B_3$ be a maximal ideal lying over $\mathfrak p \subset A_3$. To finish the proof we have to show that $(A_3)_\mathfrak p \to (B_3)_\mathfrak q$ is formally smooth in the $\mathfrak q$-adic topology. By the criterion of Lemma 15.111.5 it suffices to show that $\mathfrak p (B_3)_\mathfrak q = \mathfrak q (B_3)_\mathfrak q$ and that the field extension $\kappa (\mathfrak q)/\kappa (\mathfrak p)$ is separable. This is true because we may check both assertions by looking at the ring map $A_3/\pi ^2 A_3 \to B_3/\pi ^2 B_3$ and this is isomorphic to the ring map $A_2/\pi ^2 A_2 \to B_2/\pi ^2 B_2$ where the corresponding statement holds by our assumption that $K_2$ is a solution for $A \subset B$. Some details omitted. $\square$

Proof. The lemma is trivial if the characteristic of $K$ is zero; thus we may and do assume that the characteristic of $K$ is $p > 0$.

Let $K_2/K$ be a solution for $A \to B$. We will use induction on the inseparable degree $[K_2 : K]_ i$ (Fields, Definition 9.14.7) of $K_2/K$. If $[K_2 : K]_ i = 1$, then $K_2$ is separable over $K$ and we are done. If not, then there exists a subfield $K_2/K_1/K$ such that $K_2/K_1$ is purely inseparable of degree $p$ (Fields, Lemmas 9.14.6 and 9.14.5). By Lemma 15.116.5 there exists a separable extension $K_3/K_1$ which is a solution for $A \subset B$. Then $[K_3 : K]_ i = [K_1 : K]_ i = [K_2 : K]_ i/p$ (Fields, Lemma 9.14.9) is smaller and we conclude by induction. $\square$

Proof. Let $A \to C$ be a finite type ring map such that $B$ is a localization of $C$ at a prime $\mathfrak p$. Then $C' = C \otimes _ A A'$ is a finite type $A'$-algebra, in particular Noetherian. Since $A \to A'$ is integral, so is $C \to C'$. Thus $B = C_\mathfrak p \subset C'_\mathfrak p$ is integral too. It follows that the dimension of $C'_\mathfrak p$ is $1$ (Algebra, Lemma 10.112.4). Of course $C'_\mathfrak p$ is Noetherian. Let $\mathfrak q_1, \ldots , \mathfrak q_ n$ be the minimal primes of $C'_\mathfrak p$. Let $B'_ i$ be the integral closure of $B = C_\mathfrak p$, or equivalently by the above of $C'_\mathfrak p$ in the field of fractions of $C'_{\mathfrak p'}/\mathfrak q_ i$. It follows from Krull-Akizuki (Algebra, Lemma 10.119.12 applied to the finitely many localizations of $C'_\mathfrak p$ at its maximal ideals) that each $B'_ i$ is Noetherian. Moreover the residue field extensions in $C'_\mathfrak p \to B'_ i$ are finite by Algebra, Lemma 10.119.10. Finally, we observe that $B' = \prod B'_ i$ is the integral closure of $B$ in $L' = (L \otimes _ K K')_{red}$. $\square$

Proof. Observe that a weak solution is a solution if the residue field of $A$ is perfect, see Lemma 15.111.5. Thus the proposition follows immediately from Theorem 15.115.18 if the residue characteristic of $A$ is $0$ (and in fact we do not need the assumption that $A \to B$ is essentially of finite type). If the residue characteristic of $A$ is $p > 0$ we will also deduce it from Epp's theorem.

Let $x_ i \in A$, $i \in I$ be a set of elements mapping to a $p$-base of the residue field $\kappa $ of $A$. Set

where the transition maps send $t_{i, n + 1}$ to $t_{i, n}^ p$. Observe that $A'$ is a filtered colimit of weakly unramified finite extensions of discrete valuation rings over $A$. Thus $A'$ is a discrete valuation ring and $A \to A'$ is weakly unramified. By construction the residue field $\kappa ' = A'/\mathfrak m_ A A'$ is the perfection of $\kappa $.

Let $K'$ be the fraction field of $A'$. We may apply Lemma 15.116.7 to the extension $K'/K$. Thus $B'$ is a finite product of Dedekind domains. Let $\mathfrak m_1, \ldots , \mathfrak m_ n$ be the maximal ideals of $B'$. Using Epp's theorem (Theorem 15.115.18) we find a weak solution $K'_ i/K'$ for each of the extensions $A' \subset B'_{\mathfrak m_ i}$. Since the residue field of $A'$ is perfect, these are actually solutions. Let $K'_1/K'$ be a finite extension which contains each $K'_ i$. Then $K'_1/K'$ is still a solution for each $A' \subset B'_{\mathfrak m_ i}$ by Lemma 15.116.1.

Let $A'_1$ be the integral closure of $A$ in $K'_1$. Note that $A'_1$ is a Dedekind domain by the discussion in Remark 15.114.1 applied to $K' \subset K'_1$. Thus Lemma 15.116.7 applies to $K'_1/K$. Therefore the integral closure $B'_1$ of $B$ in $L'_1 = (L \otimes _ K K'_1)_{red}$ is a Dedekind domain and because $K'_1/K'$ is a solution for each $A' \subset B'_{\mathfrak m_ i}$ we see that $(A'_1)_{A'_1 \cap \mathfrak m} \to (B'_1)_{\mathfrak m}$ is formally smooth in the $\mathfrak m$-adic topology for each maximal ideal $\mathfrak m \subset B'_1$.

By construction, the field $K'_1$ is a filtered colimit of finite extensions of $K$. Say $K'_1 = \mathop{\mathrm{colim}}\nolimits _{i \in I} K_ i$. For each $i$ let $A_ i$, resp. $B_ i$ be the integral closure of $A$, resp. $B$ in $K_ i$, resp. $L_ i = (L \otimes _ K K_ i)_{red}$. Then it is clear that

Since the ring maps $A_ i \to A'_1$ and $B_ i \to B'_1$ are injective integral ring maps and since $A'_1$ and $B'_1$ have finite spectra, we see that for all $i$ large enough the ring maps $A_ i \to A'_1$ and $B_ i \to B'_1$ are bijective on spectra. Once this is true, for all $i$ large enough the maps $A_ i \to A'_1$ and $B_ i \to B'_1$ will be weakly unramified (once the uniformizer is in the image). It follows from multiplicativity of ramification indices that $A_ i \to B_ i$ induces weakly unramified maps on all localizations at maximal ideals of $B_ i$ for such $i$. Increasing $i$ a bit more we see that

induces surjective maps on residue fields (because the residue fields of $B'_1$ are finitely generated over those of $A'_1$ by Lemma 15.116.7). Picture of residue fields at maximal ideals lying under a chosen maximal ideal of $B'_1$:

Thus $\kappa _{B_ i}$ is a finitely generated extension of $\kappa _{A_ i}$ such that the compositum of $\kappa _{B_ i}$ and $\kappa _{A'_1}$ in $\kappa _{B'_1}$ is separable over $\kappa _{A'_1}$. Then that happens already at a finite stage: for example, say $\kappa _{B'_1}$ is finite separable over $\kappa _{A'_1}(x_1, \ldots , x_ n)$, then just increase $i$ such that $x_1, \ldots , x_ n$ are in $\kappa _{B_ i}$ and such that all generators satisfy separable polynomial equations over $\kappa _{A_ i}(x_1, \ldots , x_ n)$. This means that $A_ i \to (B_ i)_\mathfrak m$ is formally smooth in the $\mathfrak m$-adic topology for all maximal ideals $\mathfrak m$ of $B_ i$ and the proof is complete. $\square$

Proof. Observe that if $A$ is Nagata, then so is $B$ (Algebra, Lemma 10.162.6 and Proposition 10.162.15). Thus the lemma follows on combining Proposition 15.116.8 and Lemma 15.116.6. $\square$