
15.100 Abhyankar's lemma and tame ramification

In this section we prove what we think is the most general version of Abhyankar's lemma for discrete valuation rings. After doing so, we apply this to prove some results about tamely ramified extensions of the fraction field of a discrete valuation ring.

Remark 15.100.1. Let $A \to B$ be an extension of discrete valuation rings with fraction fields $K \subset L$. Let $K \subset K_1$ be a finite extension of fields. Let $A_1 \subset K_1$ be the integral closure of $A$ in $K_1$. On the other hand, let $L_1 = (L \otimes _ K K_1)_{red}$. Then $L_1$ is a nonempty finite product of finite field extensions of $L$. Let $B_1$ be the integral closure of $B$ in $L_1$. We obtain compatible commutative diagrams

$\vcenter { \xymatrix{ L \ar[r] & L_1 \\ K \ar[u] \ar[r] & K_1 \ar[u] } } \quad \text{and}\quad \vcenter { \xymatrix{ B \ar[r] & B_1 \\ A \ar[u] \ar[r] & A_1 \ar[u] } }$

In this situation we have the following

1. By Algebra, Lemma 10.119.18 the ring $A_1$ is a Dedekind domain and $B_1$ is a finite product of Dedekind domains.

2. Note that $L \otimes _ K K_1 = (B \otimes _ A A_1)_\pi$ where $\pi \in A$ is a uniformizer and that $\pi$ is a nonzerodivisor on $B \otimes _ A A_1$. Thus the ring map $B \otimes _ A A_1 \to B_1$ is integral with kernel consisting of nilpotent elements. Hence $\mathop{\mathrm{Spec}}(B_1) \to \mathop{\mathrm{Spec}}(B \otimes _ A A_1)$ is surjective on spectra (Algebra, Lemma 10.35.17). The map $\mathop{\mathrm{Spec}}(B \otimes _ A A_1) \to \mathop{\mathrm{Spec}}(A_1)$ is surjective as $A_1/\mathfrak m_ A A_1 \to B/\mathfrak m_ AB \otimes _{\kappa _ A} A_1/\mathfrak m_ A A_1$ is an injective ring map with $A_1/\mathfrak m_ A A_1$ Artinian. We conclude that $\mathop{\mathrm{Spec}}(B_1) \to \mathop{\mathrm{Spec}}(A_1)$ is surjective.

3. Let $\mathfrak m_ i$, $i = 1, \ldots n$ with $n \geq 1$ be the maximal ideals of $A_1$. For each $i = 1, \ldots , n$ let $\mathfrak m_{ij}$, $j = 1, \ldots , m_ i$ with $m_ i \geq 1$ be the maximal ideals of $B_1$ lying over $\mathfrak m_ i$. We obtain diagrams

$\xymatrix{ B \ar[r] & (B_1)_{\mathfrak m_{ij}} \\ A \ar[u] \ar[r] & (A_1)_{\mathfrak m_ i} \ar[u] }$

of extensions of discrete valuation rings.

4. If $A$ is henselian (for example complete), then $A_1$ is a discrete valuation ring, i.e., $n = 1$. Namely, $A_1$ is a union of finite extensions of $A$ which are domains, hence local by Algebra, Lemma 10.148.4.

5. If $B$ is henselian (for example complete), then $B_1$ is a product of discrete valuation rings, i.e., $m_ i = 1$ for $i = 1, \ldots , n$.

6. If $K \subset K_1$ is purely inseparable, then $A_1$ and $B_1$ are both discrete valuation rings, i.e., $n = 1$ and $m_1 = 1$. This is true because for every $b \in B_1$ a $p$-power power of $b$ is in $B$, hence $B_1$ can only have one maximal ideal.

7. If $K \subset K_1$ is finite separable, then $L_1 = L \otimes _ K K_1$ and is a finite product of finite separable extensions too. Hence $A \subset A_1$ and $B \subset B_1$ are finite by Algebra, Lemma 10.155.8.

8. If $A$ is Nagata, then $A \subset A_1$ is finite.

9. If $B$ is Nagata, then $B \subset B_1$ is finite.

Lemma 15.100.2. Let $A$ be a discrete valuation ring with uniformizer $\pi$. Let $n \geq 2$. Then $K_1 = K[\pi ^{1/n}]$ is a degree $n$ extension of $K$ and the integral closure $A_1$ of $A$ in $K_1$ is the ring $A[\pi ^{1/n}]$ which is a discrete valuation ring with ramification index $n$ over $A$.

Proof. This lemma proves itself. $\square$

Lemma 15.100.3. Let $A \to B$ be an extension of discrete valuation rings with fraction fields $K \subset L$. Assume that $A \to B$ is formally smooth in the $\mathfrak m_ B$-adic topology. Then for any finite extension $K \subset K_1$ we have $L_1 = L \otimes _ K K_1$, $B_1 = B \otimes _ A A_1$, and each extension $(A_1)_{\mathfrak m_ i} \subset (B_1)_{\mathfrak m_{ij}}$ (see Remark 15.100.1) is formally smooth in the $\mathfrak m_{ij}$-adic topology.

Proof. We will use the equivalence of Lemma 15.97.5 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

$(B \otimes _ A (A_1)_{\mathfrak m_ i})/\pi _ i (B \otimes _ A (A_1)_{\mathfrak m_ i}) = B/\pi B \otimes _{A/\pi A} (A_1)_{\mathfrak m_ i}/\pi _ i (A_1)_{\mathfrak m_ i}$

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. $\square$

The following lemma is our version of Abhyankar's lemma for discrete valuation rings. Observe that $\kappa _ B/\kappa _ A$ is not assumed to be an algebraic extension of fields.

Lemma 15.100.4 (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 15.100.1 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 15.97.5, 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 15.100.2 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 15.100.3.

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 15.97.3) 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 15.97.5. $\square$

Lemma 15.100.5. Let $A$ be a discrete valuation ring with fraction field $K$. Let $M/L/K$ be finite separable extensions. Let $B$ be the integral closure of $A$ in $L$. If $L/K$ is tamely ramified with respect to $A$ and $M/L$ is tamely ramified with respect to $B_\mathfrak m$ for every maximal ideal $\mathfrak m$ of $B$, then $M/K$ is tamely ramified with respect to $A$.

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 15.97.3) and the fact that we have the tower $\kappa (\mathfrak m')/\kappa (\mathfrak m)/\kappa _ A$ of finite extensions of fields. $\square$

Lemma 15.100.6. Let $A$ be a discrete valuation ring with fraction field $K$. If $M/L/K$ are finite separable extensions and $M$ is tamely ramified with respect to $A$, then $L$ is tamely ramified with respect to $A$.

Proof. We will use the results of the discussion in Remark 15.97.6 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 15.97.3) 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 $\kappa (\mathfrak m)/\kappa _ A$. $\square$

Lemma 15.100.7. Let $A$ be a discrete valuation ring with fraction field $K$. Let $\pi \in A$ be a uniformizer. Let $L/K$ be a finite separable extension. The following are equivalent

1. $L$ is tamely ramified with respect to $A$,

2. there exists an $e \geq 1$ invertible in $\kappa _ A$ and an extension $L'/K' = K[\pi ^{1/e}]$ unramified with respect to $A' = A[\pi ^{1/e}]$ such that $L$ is contained in $L'$, and

3. there exists an $e_0 \geq 1$ invertible in $\kappa _ A$ such that for every $d \geq 1$ invertible in $\kappa _ A$ (2) holds with $e = de_0$.

Proof. Observe that $A'$ is a discrete valuation ring with fraction field $K'$, see Lemma 15.100.2. 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 15.100.5. Hence $L$ is tamely ramified with respect to $A$ by Lemma 15.100.6.

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 15.100.2. Choose a product decomposition

$L \otimes _ K K' = \prod L'_ j$

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 15.100.4) applied to $A \subset B_{\mathfrak m_ i}$ and the extension $K \subset K'$ 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 15.97.5). 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 $j$. $\square$

Lemma 15.100.8. Let $A$ be a discrete valuation ring with fraction field $K$.

1. If $L/K$ is a finite separable extension which is tamely ramified with respect to $A$, then there exists a Galois extension $M/K$ containing $L$ which is tamely ramified with respect to $A$.

2. If $L_1/K$, $L_2/K$ are finite separable extensions which are tamely ramified with respect to $A$, then there exists a a finite separable extension $L/K$ which is tamely ramified with respect to $A$ containing $L_1$ and $L_2$.

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 $L_ i \subset L'_ i$ as extensions of $K$, see Lemma 15.100.7. By Lemma 15.97.8 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 15.100.7. 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

$L \otimes _ K \ldots \otimes _ K L \longrightarrow M$

of $K$-algebras, see Fields, Lemma 9.16.6. Let $B$ be the integral closure of $A$ in $L$ as in Remark 15.97.6. 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 10.141.7. Claim:

$K' \otimes _ K \ldots \otimes _ K K' = \prod K'_ i$

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

$\prod A'_ i \otimes _{(A' \otimes _ A \ldots \otimes _ A A')} (B \otimes _ A \ldots \otimes _ A B)$

is finite étale over $\prod A'_ i$ and hence a product of Dedekind domains (Lemma 15.43.4). 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 15.100.5.

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

$A' \otimes _ A \ldots \otimes _ A A' = A'[x_1, \ldots , x_ r]/(x_1^ e - \pi , \ldots , x_ r^ e - \pi )$

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

$A[x_1, y_2, \ldots , y_ r]/(x_1^ e - \pi , y_2^ e - 1, \ldots , y_ r - 1) = A'[y_2, \ldots , y_ r]/(y_2^ e - 1, \ldots , y_ r^ e - 1)$

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). $\square$

Lemma 15.100.9. Let $A \subset B$ be an extension of discrete valuation rings. Denote $L/K$ the corresponding extension of fraction fields. Let $K'/K$ be a finite separable extension. Then

$K' \otimes _ K L = \prod L'_ i$

is a finite product of fields and the following is true

1. If $K'$ is unramified with respect to $A$, then each $L'_ i$ is unramified with respect to $B$.

2. If $K'$ is tamely ramified with respect to $A$, then each $L'_ i$ is tamely ramified with respect to $B$.

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 15.43.4). 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 15.100.6). Thus by Lemma 15.100.7 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

$K[\pi ^{1/e}] \otimes _ K L = \prod L_{e, j}$

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 $A[\pi ^{1/e}]$, see Lemma 15.100.5.

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

$A[\pi ^{1/e}] \otimes _ A B = B[x]/(x^ e - u \theta ^ t) \subset B[y, z]/(y^{e'} - \theta , z^ e - u)$

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). $\square$

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