The Stacks project

Proof. This lemma proves itself. $\square$

Proof. We will use the equivalence of Lemma 15.111.5 without further mention. Let $\pi \in A$ and $\pi _ i \in (A_1)_{\mathfrak m_ i}$ be uniformizers. As $\kappa _ A \subset \kappa _ B$ is separable, the ring

is a product of fields each separable over $\kappa _{\mathfrak m_ i}$. Hence the element $\pi _ i$ in $B \otimes _ A (A_1)_{\mathfrak m_ i}$ is a nonzerodivisor and the quotient by this element is a product of fields. It follows that $B \otimes _ A A_1$ is a Dedekind domain in particular reduced. Thus $B \otimes _ A A_1 \subset B_1$ is an equality. $\square$

Proof. Let $\pi \in A$ be a uniformizer. Let $\pi _1$ be a uniformizer of $(A_1)_{\mathfrak m_ i}$. Write $\pi = u \pi _1^{e_1}$ with $u$ a unit of $(A_1)_{\mathfrak m_ i}$ and $e_1$ the ramification index of $A \subset (A_1)_{\mathfrak m_ i}$.

Claim: we may assume that $u$ is an $e$th power in $K_1$. Namely, let $K_2$ be an extension of $K_1$ obtained by adjoining a root of $x^ e = u$; thus $K_2$ is a factor of $K_1[x]/(x^ e - u)$. Then $K_2/K_1$ is a finite separable extension (by our assumption on $e$) and hence $A_1 \subset A_2$ is finite. Since $(A_1)_{\mathfrak m_ i} \to (A_1)_{\mathfrak m_ i}[x]/(x^ e - u)$ is finite étale (as $e$ is prime to the residue characteristic and $u$ a unit) we conclude that $(A_2)_{\mathfrak m_ i}$ is a factor of a finite étale extension of $(A_1)_{\mathfrak m_ i}$ hence finite étale over $(A_1)_{\mathfrak m_ i}$ itself. The same reasoning shows that $B_1 \subset B_2$ induces finite étale extensions $(B_1)_{\mathfrak m_{ij}} \subset (B_2)_{\mathfrak m_{ij}}$. Pick a maximal ideal $\mathfrak m'_{ij} \subset B_2$ lying over $\mathfrak m_{ij} \subset B_1$ (of course there may be more than one) and consider

where $\mathfrak m'_ i \subset A_2$ is the image. Now the horizontal arrows have ramification index $1$ and induce finite separable residue field extensions. Thus, using the equivalence of Lemma 15.111.5, we see that it suffices to show that the right vertical arrow is formally smooth in the $\mathfrak m'_{ij}$-adic topology. Since $u$ has a $e$th root in $K_2$ we obtain the claim.

Assume $u$ has an $e$th root in $K_1$. Since $e | e_1$ and since $u$ has a $e$th root in $K_1$ we see that $\pi = \theta ^ e$ for some $\theta \in K_1$. Let $K'_1 = K[\theta ] \subset K_1$ be the subfield generated by $\theta$. By Lemma 15.114.2 the integral closure $A'_1$ of $A$ in $K[\theta ]$ is the discrete valuation ring $A'_1 = A[\theta ]$ which has ramification index $e$ over $A$. If we can prove the lemma for the extension $K'_1/K$, then we conclude by Lemma 15.114.3 applied to the diagram

for all $j = 1, \ldots , m_ i$. This reduces us to the case discussed in the next paragraph.

Assume $K_1 = K[\pi ^{1/e}]$ and set $\theta = \pi ^{1/e}$. Let $\pi _ B$ be a uniformizer for $B$ and write $\pi = w \pi _ B^ e$ for some unit $w$ of $B$. Then we see that $L_1 = L \otimes _ K K_1$ is obtained by adjoining $\theta / \pi _ B$ which is an $e$th root of the unit $w$. Thus $B \subset B_1$ is finite étale. Thus for any maximal ideal $\mathfrak m \subset B_1$ consider the commutative diagram

Here the numbers along the arrows are the ramification indices. By multiplicativity of ramification indices (Lemma 15.111.3) we conclude $e_\mathfrak m = 1$. Looking at the residue field extensions we find that $\kappa (\mathfrak m)$ is a finite separable extension of $\kappa _ B$ which is separable over $\kappa _ A$. Therefore $\kappa (\mathfrak m)$ is separable over $\kappa _ A$ which is equal to the residue field of $A_1$ and we win by Lemma 15.111.5. $\square$

Proof. Let $C$ be the integral closure of $A$ in $M$. Every maximal ideal $\mathfrak m'$ of $C$ lies over a maximal ideal $\mathfrak m$ of $B$. Then the lemma follows from the multiplicativity of ramification indices (Lemma 15.111.3) and the fact that we have the tower $\kappa (\mathfrak m')/\kappa (\mathfrak m)/\kappa _ A$ of finite extensions of fields. $\square$

Proof. We will use the results of the discussion in Remark 15.111.6 without further mention. Let $C/B/A$ be the integral closures of $A$ in $M/L/K$. Since $C$ is a finite ring extension of $B$, we see that $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(B)$ is surjective. Hence for ever maximal ideal $\mathfrak m \subset B$ there is a maximal ideal $\mathfrak m' \subset C$ lying over $\mathfrak m$. By the multiplicativity of ramification indices (Lemma 15.111.3) and the assumption, we conclude that the ramification index of $B_\mathfrak m$ over $A$ is prime to the residue characteristic. Since $\kappa (\mathfrak m')/\kappa _ A$ is finite separable, the same is true for $\kappa (\mathfrak m)/\kappa _ A$. $\square$

Proof. Observe that $A'$ is a discrete valuation ring with fraction field $K'$, see Lemma 15.114.2. Of course the ramification index of $A'$ over $A$ is $e$. Thus if (2) holds, then $L'$ is tamely ramified with respect to $A$ by Lemma 15.114.5. Hence $L$ is tamely ramified with respect to $A$ by Lemma 15.114.6.

The implication (3) $\Rightarrow$ (2) is immediate.

Assume that (1) holds. Let $B$ be the integral closure of $A$ in $L$ and let $\mathfrak m_1, \ldots , \mathfrak m_ n$ be its maximal ideals. Denote $e_ i$ the ramification index of $A \to B_{\mathfrak m_ i}$. Let $e_0$ be the least common multiple of $e_1, \ldots , e_ r$. This is invertible in $\kappa _ A$ by our assumption (1). Let $e = de_0$ as in (3). Set $A' = A[\pi ^{1/e}]$. Then $A \to A'$ is an extension of discrete valuation rings with fraction field $K' = K[\pi ^{1/e}]$, see Lemma 15.114.2. Choose a product decomposition

where $L'_ j$ are fields. Let $B'_ j$ be the integral closure of $A$ in $L'_ j$. Let $\mathfrak m_{ijk}$ be the maximal ideals of $B'_ j$ lying over $\mathfrak m_ i$. Observe that $(B'_ j)_{\mathfrak m_ i}$ is the integral closure of $B_{\mathfrak m_ i}$ in $L'_ j$. By Abhyankar's lemma (Lemma 15.114.4) applied to $A \subset B_{\mathfrak m_ i}$ and the extension $K'/K$ we see that $A' \to (B'_ j)_{\mathfrak m_{ijk}}$ is formally smooth in the $\mathfrak m_{ijk}$-adic topology. This implies that the ramification index is $1$ and that the residue field extension is separable (Lemma 15.111.5). In this way we see that $L'_ j$ is unramified with respect to $A'$. This finishes the proof: we take $L' = L'_ j$ for some $j$. $\square$

Proof. Proof of (2). Choose a uniformizer $\pi \in A$. We can choose an integer $e$ invertible in $\kappa _ A$ and extensions $L_ i'/K' = K[\pi ^{1/e}]$ unramified with respect to $A' = A[\pi ^{1/e}]$ with $L'_ i/L_ i$ as extensions of $K$, see Lemma 15.114.7. By Lemma 15.111.8 we can find an extension $L'/K'$ which is unramified with respect to $A'$ such that $L'_ i/K$ is isomorphic to a subextension of $L'/K'$ for $i = 1, 2$. This finishes the proof of (3) as $L'/K$ is tamely ramified (use same lemma as above).

Proof of (1). We may first replace $L$ by a larger extension and assume that $L$ is an extension of $K' = K[\pi ^{1/e}]$ unramified with respect to $A' = A[\pi ^{1/e}]$ where $e$ is invertible in $\kappa _ A$, see Lemma 15.114.7. Let $M$ be the normal closure of $L$ over $K$, see Fields, Definition 9.16.4. Then $M/K$ is Galois by Fields, Lemma 9.21.5. On the other hand, there is a surjection

of $K$-algebras, see Fields, Lemma 9.16.6. Let $B$ be the integral closure of $A$ in $L$ as in Remark 15.111.6. The condition that $L$ is unramified with respect to $A' = A[\pi ^{1/e}]$ exactly means that $A' \to B$ is an étale ring map, see Algebra, Lemma 10.143.7. Claim:

is a product of field extensions $K'_ i/K$ tamely ramified with respect to $A$. Then if $A'_ i$ is the integral closure of $A$ in $K'_ i$ we see that

is finite étale over $\prod A'_ i$ and hence a product of Dedekind domains (Lemma 15.44.4). We conclude that $M$ is the fraction field of one of these Dedekind domains which is finite étale over $A'_ i$ for some $i$. It follows that $M/K'_ i$ is unramified with respect to every maximal ideal of $A'_ i$ and hence $M/K$ is tamely ramified by Lemma 15.114.5.

It remains the prove the claim. For this we write $A' = A[x]/(x^ e - \pi )$ and we see that

The normalization of this ring certainly contains the elements $y_ i = x_ i/x_1$ for $i = 2, \ldots , r$ subject to the relations $y_ i^ e - 1 = 0$ and we obtain

This ring is finite étale over $A'$ because $e$ is invertible in $A'$. Hence it is a product of Dedekind domains each unramified over $A'$ as desired (see references given above in case of confusion). $\square$

Proof. The algebra $K' \otimes _ K L$ is a finite product of fields as it is a finite étale algebra over $L$. Let $A'$ be the integral closure of $A$ in $K'$.

In case (1) the ring map $A \to A'$ is finite étale. Hence $B' = B \otimes _ A A'$ is finite étale over $B$ and is a finite product of Dedekind domains (Lemma 15.44.4). Hence $B'$ is the integral closure of $B$ in $K' \otimes _ K L$. It follows immediately that each $L'_ i$ is unramified with respect to $B$.

Choose a uniformizer $\pi \in A$. To prove (2) we may replace $K'$ by a larger extension tame ramified with respect to $A$ (details omitted; hint: use Lemma 15.114.6). Thus by Lemma 15.114.7 we may assume there exists some $e \geq 1$ invertible in $\kappa _ A$ such that $K'$ contains $K[\pi ^{1/e}]$ and such that $K'$ is unramified with respect to $A[\pi ^{1/e}]$. Choose a product decomposition

For every $i$ there exists a $j_ i$ such that $L'_ i/L_{e, j_ i}$ is a finite separable extension. Let $B_{e, j}$ be the integral closure of $B$ in $L_{e, j}$. By (1) applied to $K'/K[\pi ^{1/e}]$ and $A[\pi ^{1/e}] \subset (B_{e, j_ i})_\mathfrak m$ we see that $L'_ i$ is unramified with respect to $(B_{e, j_ i})_\mathfrak m$ for every maximal ideal $\mathfrak m \subset B_{e, j_ i}$. Hence the proof will be complete if we can show that $L_{e, j}$ is tamely ramified with respect to $B$, see Lemma 15.114.5.

Choose a uniformizer $\theta$ in $B$. Write $\pi = u \theta ^ t$ where $u$ is a unit of $B$ and $t \geq 1$. Then we have

where $e' = e/\gcd (e, t)$. The map sends $x$ to $z y^{t/\gcd (e, t)}$. Since the right hand side is a product of Dedekind domains each tamely ramified over $B$ the proof is complete (details omitted). $\square$