Lemma 15.114.4 (Abhyankar's lemma). Let $A \subset B$ be an extension of discrete valuation rings. Assume that either the residue characteristic of $A$ is $0$ or it is $p$, the ramification index $e$ is prime to $p$, and $\kappa _ B/\kappa _ A$ is a separable field extension. Let $K_1/K$ be a finite extension. Using the notation of Remark 15.114.1 assume $e$ divides the ramification index of $A \subset (A_1)_{\mathfrak m_ i}$ for some $i$. Then $(A_1)_{\mathfrak m_ i} \subset (B_1)_{\mathfrak m_{ij}}$ is formally smooth $\mathfrak m_{ij}$-adic topology for all $j = 1, \ldots , m_ i$.

**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 $\pi _ B/\theta $ 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. $\square$

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## Comments (1)

Comment #7778 by Mingchen on

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