The Stacks project

Lemma 15.104.4 (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 15.104.1 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

\[ \xymatrix{ (B_1)_{\mathfrak m_{ij}} \ar[r] & (B_2)_{\mathfrak m'_{ij}} \\ (A_1)_{\mathfrak m_ i} \ar[u] \ar[r] & (A_2)_{\mathfrak m'_ i} \ar[u] } \]

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 15.101.5, 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 $K[\theta ] \subset K_1$ be the subfield generated by $\theta $. By Lemma 15.104.2 the integral closure of $A$ in $K[\theta ]$ is the discrete valuation ring $A[\theta ]$. If we can prove the lemma for the extension $K \subset K[\theta ]$, then $K \subset K[\theta ]$ is a solution for $A \subset B$ and we conclude by Lemma 15.104.3.

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

\[ \xymatrix{ B \ar[r]_1 & (B_1)_{\mathfrak m} \\ A \ar[u]^ e \ar[r]^ e & A_1 \ar[u]_{e_\mathfrak m} } \]

Here the numbers along the arrows are the ramification indices. By multiplicativity of ramification indices (Lemma 15.101.3) 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 15.101.5. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BRM. Beware of the difference between the letter 'O' and the digit '0'.