The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

15.101 Eliminating ramification

In this section we discuss a result of Helmut Epp, see [Epp]. We strongly encourage the reader to read the original. Our approach is slightly different as we try to handle the mixed and equicharacteristic cases by the same method. For related results, see also [Ponomarev], [Ponomarev-Abhyankar], [Kuhlmann], and [ZK].

Let $A \subset B$ be an extension of discrete valuation rings with fraction fields $K \subset L$. The goal in this section is to find a finite extension $K_1/K$ such that with

\[ \vcenter { \xymatrix{ L \ar[r] & L_1 \\ K \ar[u] \ar[r] & K_1 \ar[u] } } \quad \text{and}\quad \vcenter { \xymatrix{ B \ar[r] & B_1 \ar[r] & (B_1)_{\mathfrak m_{ij}} \\ A \ar[u] \ar[r] & A_1 \ar[r] \ar[u] & (A_1)_{\mathfrak m_ i} \ar[u] } } \]

as in Remark 15.100.1 the extensions $(A_1)_{\mathfrak m_ i} \subset (B_1)_{\mathfrak m_{ij}}$ are all weakly unramified or even formally smooth in the relevant adic topologies. The simplest (but nontrivial) example of this is Abhyankar's lemma, see Lemma 15.100.4.

Definition 15.101.1. Let $A \to B$ be an extension of discrete valuation rings with fraction fields $K \subset L$.

  1. We say a finite field extension $K \subset K_1$ is a weak solution for $A \subset B$ if all the extensions $(A_1)_{\mathfrak m_ i} \subset (B_1)_{\mathfrak m_{ij}}$ of Remark 15.100.1 are weakly unramified.

  2. We say a finite field extension $K \subset K_1$ is a solution for $A \subset B$ if each extension $(A_1)_{\mathfrak m_ i} \subset (B_1)_{\mathfrak m_{ij}}$ of Remark 15.100.1 is formally smooth in the $\mathfrak m_{ij}$-adic topology.

We say a solution $K \subset K_1$ is a separable solution if $K \subset K_1$ is separable.

In general (weak) solutions do not exist; there is an example in [Epp]. We will prove the existence of weak solutions in Theorem 15.101.19 following [Epp] in case the residue field extension satisfies a mild condition. We will then deduce the existence of solutions and sometimes separable solutions in geometrically meaningful cases in Proposition 15.101.21 and Lemma 15.101.5. The following example shows that in general one needs inseparable extensions to get a solution.

Example 15.101.2. Let $k$ be a perfect field of characteristic $p > 0$. Let $A = k[[x]]$ and $K = k((x))$. Let $B = A[x^{1/p}]$. Any weak solution $K \subset K_1$ for $A \to B$ is inseparable (and any finite inseparable extension of $K$ is a solution). We omit the proof.

Solutions are stable under further extensions (follows from Lemma 15.100.3). This may not be true for weak solutions. Weak solutions are in some sense stable under totally ramified extensions, see Lemma 15.101.3.

Lemma 15.101.3. Let $A \to B$ be an extension of discrete valuation rings with fraction fields $K \subset L$. Assume that $A \to B$ is weakly unramified. Then for any finite separable extension $K_1/K$ totally ramified with respect to $A$ we have that $L_1 = L \otimes _ K K_1$ is a field, $A_1$ and $B_1 = B \otimes _ A A_1$ are discrete valuation rings, and the extension $A_1 \subset B_1$ (see Remark 15.100.1) is weakly unramified.

Proof. Let $\pi \in A$ and $\pi _1 \in A_1$ be uniformizers. As $K_1/K$ is totally ramified with respect to $A$ we have $\pi _1^ e = u_1 \pi $ for some unit $u_1$ in $A_1$. Hence $A_1$ is generated by $\pi _1$ over $A$ and the minimal polynomial $P(t)$ of $\pi _1$ over $K$ has the form

\[ P(t) = t^ e + a_{e - 1} t^{e - 1} + \ldots + a_0 \]

with $a_ i \in (\pi )$ and $a_0 = u\pi $ for some unit $u$ of $A$. Note that $e = [K_1 : K]$ as well. Since $A \to B$ is weakly unramified we see that $\pi $ is a uniformizer of $B$ and hence $B_1 = B[t]/(P(t))$ is a discrete valuation ring with uniformizer the class of $t$. Thus the lemma is clear. $\square$

Lemma 15.101.4. Let $A \to B \to C$ be extensions of discrete valuation rings with fraction fields $K \subset L \subset M$. Let $K \subset K_1$ be a finite extension.

  1. If $K_1$ is a (weak) solution for $A \to C$, then $K_1$ is a (weak) solution for $A \to B$.

  2. If $K_1$ is a (weak) solution for $A \to B$ and $L_1 = (L \otimes _ K K_1)_{red}$ is a product of fields which are (weak) solutions for $B \to C$, then $K_1$ is a weak solution for $A \to C$.

Proof. Let $L_1 = (L \otimes _ K K_1)_{red}$ and $M_1 = (M \otimes _ K K_1)_{red}$ and let $B_1 \subset L_1$ and $C_1 \subset M_1$ be the integral closure of $B$ and $C$. Note that $M_1 = (M \otimes _ L L_1)_{red}$ and that $L_1$ is a (nonempty) finite product of finite extensions of $L$. Hence the ring map $B_1 \to C_1$ is a finite product of ring maps of the form discussed in Remark 15.100.1. In particular, the map $\mathop{\mathrm{Spec}}(C_1) \to \mathop{\mathrm{Spec}}(B_1)$ is surjective. Choose a maximal ideal $\mathfrak m \subset C_1$ and consider the extensions of discrete valuation rings

\[ (A_1)_{A_1 \cap \mathfrak m} \to (B_1)_{B_1 \cap \mathfrak m} \to (C_1)_\mathfrak m \]

If the composition is weakly unramified, so is the map $(A_1)_{A_1 \cap \mathfrak m} \to (B_1)_{B_1 \cap \mathfrak m}$. If the residue field extension $\kappa _{A_1 \cap \mathfrak m} \to \kappa _\mathfrak m$ is separable, so is the subextension $\kappa _{A_1 \cap \mathfrak m} \to \kappa _{B_1 \cap \mathfrak m}$. Taking into account Lemma 15.97.5 this proves (1). A similar argument works for (2). $\square$

Lemma 15.101.5. Let $A \subset B$ be an extension of discrete valuation rings. Assume

  1. the extension $K \subset L$ of fraction fields is separable,

  2. $B$ is Nagata, and

  3. there exists a solution for $A \subset B$.

Then there exists a separable solution for $A \subset B$.

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 $K \subset K_1$ be a finite extension. Since $L/K$ is separable, the algebra $L \otimes _ K K_1$ is reduced (Algebra, Lemma 10.42.6). Since $B$ is Nagata, the ring extension $B \subset B_1$ is finite (Remark 15.100.1) and $B_1$ is a Nagata ring. Moreover, if $K \subset K_1 \subset K_2$ is a tower of finite extensions, then the same thing is true, i.e., the ring extension $B_1 \subset B_2$ is finite too where $B_2$ is the integral closure of $B$ (or $B_1$) in $L \otimes _ K K_2$.

Let $K \subset K_2$ be a solution for $A \to B$. There exists a subfield $K \subset K_1 \subset K_2$ such that $K_1/K$ is separable and $K_2/K_1$ is purely inseparable (Fields, Lemma 9.14.6). Thus it suffices to show that if we have $K \subset K_1 \subset K_2$ with $K_2/K_1$ purely inseparable of degree $p$, then $K \subset K_1$ is a solution for $A \subset B$. Using the remarks above, we may replace $A$ by a localization $(A_1)_{\mathfrak m_ i}$ and $B$ by $(B_1)_{\mathfrak m_{ij}}$ (notation as in Remark 15.100.1) and reduce to the problem discussed in the following paragraph.

Assume that $K \subset K_1$ is a purely inseparable extension of degree $p$ which is a solution for $A \subset B$. Problem: show that $A \to B$ is formally smooth in the $\mathfrak m_ B$-adic topology. By the discussion in Remark 15.100.1 we see that $A_1$ and $B_1$ are discrete valuation rings and as discussed above $B \subset B_1$ is finite. Consider the diagrams

\[ \vcenter { \xymatrix{ B \ar[r]_{e_ u} & B_1 \\ A \ar[u]_ e \ar[r]^{e_ d} & A_1 \ar[u]^1 } } \quad \text{and}\quad \vcenter { \xymatrix{ \kappa _ B \ar[r]_{d_ u} & \kappa _{B_1} \\ \kappa _ A \ar[u]_ d \ar[r]^{d_ d} & \kappa _{A_1} \ar[u]^1 } } \]

of extensions of discrete valuation rings and residue fields. Here $e, e_ u, e_ d, 1$ denote ramification indices, so $e e_ u = e_ d$. Also $d, d_ u, d_ d, 1$ denote the inseparable degrees (Fields, Definition 9.14.7), so $dd_ u = d_ d$ (Fields, Lemma 9.14.9). By Algebra, Lemma 10.120.8 and the fact that $L \subset L \otimes _ K K_1$ is a degree $p$ field extension, we see that $e_ ud_ u = p$ (this is where we really use that $B$ is Nagata; this need not be true if the extension $B \subset B_1$ is not finite). We have $e_ d d_ d \leq p$ by Lemma 15.97.2. Thus it follows that $e = d = 1$ as desired. $\square$

Lemma 15.101.6. Let $A \to B$ be an extension of discrete valuation rings. There exists a commutative diagram

\[ \xymatrix{ B \ar[r] & B' \\ A \ar[r] \ar[u] & A' \ar[u] } \]

of extensions of discrete valuation rings such that

  1. the extensions $K \subset K'$ and $L \subset L'$ of fraction fields are separable algebraic,

  2. the residue fields of $A'$ and $B'$ are separable algebraic closures of the residue fields of $A$ and $B$, and

  3. if a solution, weak solution, or separable solution exists for $A' \to B'$, then a solution, weak solution, or separable solution exists for $A \to B$.

Proof. By Algebra, Lemma 10.153.2 there exists an extension $A \subset A'$ which is a filtered colimit of finite étale extensions such that the residue field of $A'$ is a separable algebraic closure of the residue field of $A$. Then $A \subset A'$ is an extension of discrete valuation rings such that the induced extension $K \subset K'$ of fraction fields is separable algebraic.

Let $B \subset B'$ be a strict henselization of $B$. Then $B \subset B'$ is an extension of discrete valuation rings whose fraction field extension is separable algebraic. By Algebra, Lemma 10.150.11 there exists a commutative diagram as in the statement of the lemma. Parts (1) and (2) of the lemma are clear.

Let $K' \subset K'_1$ be a (weak) solution for $A' \to B'$. Since $A'$ is a colimit, we can find a finite étale extension $A \subset A_1'$ and a finite extension $K_1$ of the fraction field $F$ of $A_1'$ such that $K'_1 = K' \otimes _ F K_1$. As $A \subset A_1'$ is finite étale and $B'$ strictly henselian, it follows that $B' \otimes _ A A_1'$ is a finite product of rings isomorphic to $B'$. Hence

\[ L' \otimes _ K K_1 = L' \otimes _ K F \otimes _ F K_1 \]

is a finite product of rings isomorphic to $L' \otimes _{K'} K'_1$. Thus we see that $K \subset K_1$ is a (weak) solution for $A \to B'$. Hence it is also a (weak) solution for $A \to B$ by Lemma 15.101.4. $\square$

Lemma 15.101.7. Let $A \to B$ be an extension of discrete valuation rings with fraction fields $K \subset L$. Let $K \subset K_1$ be a normal extension. Say $G = \text{Aut}(K_1/K)$. Then $G$ acts on the rings $K_1$, $L_1$, $A_1$ and $B_1$ of Remark 15.100.1 and acts transitively on the set of maximal ideals of $B_1$.

Proof. Everything is clear apart from the last assertion. If there are two or more orbits of the action, then we can find an element $b \in B_1$ which vanishes at all the maximal ideals of one orbit and has residue $1$ at all the maximal ideals in another orbit. Then $b' = \prod _{\sigma \in G} \sigma (b)$ is a $G$-invariant element of $B_1 \subset L_1 = (L \otimes _ K K_1)_{red}$ which is in some maximal ideals of $B_1$ but not in all maximal ideals of $B_1$. Lifting it to an element of $L \otimes _ K K_1$ and raising to a high power we obtain a $G$-invariant element $b''$ of $L \otimes _ K K_1$ mapping to $(b')^ N$ for some $N > 0$; in fact, we only need to do this in case the characteristic is $p > 0$ and in this case raising to a suitably large $p$-power $q$ defines a canonical map $(L \otimes _ K K_1)_{red} \to L \otimes _ K K_1$. Since $K = (K_1)^ G$ we conclude that $b'' \in L$. Since $b''$ maps to an element of $B_1$ we see that $b'' \in B$ (as $B$ is normal). Then on the one hand it must be true that $b'' \in \mathfrak m_ B$ as $b'$ is in some maximal ideal of $B_1$ and on the other hand it must be true that $b'' \not\in \mathfrak m_ B$ as $b'$ is not in all maximal ideals of $B_1$. This contradiction finishes the proof of the lemma. $\square$

Lemma 15.101.8. Let $A$ be a discrete valuation ring with uniformizer $\pi $. If the residue characteristic of $A$ is $p > 0$, then for every $n > 1$ and $p$-power $q$ there exists a degree $q$ separable extension $L/K$ totally ramified with respect to $A$ such that the integral closure $B$ of $A$ in $L$ has ramification index $q$ and a uniformizer $\pi _ B$ such that $\pi _ B^ q = \pi + \pi ^ n b$ and $\pi _ B^ q = \pi + (\pi _ B)^{nq}b'$ for some $b, b' \in B$.

Proof. If the characteristic of $K$ is zero, then we can take the extension given by $\pi _ B^ q = \pi $, see Lemma 15.100.2. If the characteristic of $K$ is $p > 0$, then we can take the extension of $K$ given by $z^ q - \pi ^ n z = \pi ^{1 - q}$. Namely, then we see that $y^ q - \pi ^{n + q - 1} y = \pi $ where $y = \pi z$. Taking $\pi _ B = y$ we obtain the desired result. $\square$

Lemma 15.101.9. Let $A$ be a discrete valuation ring. Assume the reside field $\kappa _ A$ has characteristic $p > 0$ and that $a \in A$ is an element whose residue class in $\kappa _ A$ is not a $p$th power. Then $a$ is not a $p$th power in $K$ and the integral closure of $A$ in $K[a^{1/p}]$ is the ring $A[a^{1/p}]$ which is a discrete valuation ring weakly unramified over $A$.

Proof. This lemma proves itself. $\square$

Lemma 15.101.10. Let $A \subset B \subset C$ be extensions of discrete valuation rings with fractions fields $K \subset L \subset M$. Let $\pi \in A$ be a uniformizer. Assume

  1. $B$ is a Nagata ring,

  2. $A \subset B$ is weakly unramified,

  3. $M$ is a degree $p$ purely inseparable extension of $L$.

Then either

  1. $A \to C$ is weakly unramified, or

  2. $C = B[\pi ^{1/p}]$, or

  3. there exists a degree $p$ separable extension $K_1/K$ totally ramified with respect to $A$ such that $L_1 = L \otimes _ K K_1$ and $M_1 = M \otimes _ K K_1$ are fields and the maps of integral closures $A_1 \to B_1 \to C_1$ are weakly unramified extensions of discrete valuation rings.

Proof. Let $e$ be the ramification index of $C$ over $B$. If $e = 1$, then we are done. If not, then $e = p$ by Lemmas 15.97.2 and 15.97.4. This in turn implies that the residue fields of $B$ and $C$ agree. Choose a uniformizer $\pi _ C$ of $C$. Write $\pi _ C^ p = u \pi $ for some unit $u$ of $C$. Since $\pi _ C^ p \in L$, we see that $u \in B^*$. Also $M = L[\pi _ C]$.

Suppose there exists an integer $m \geq 0$ such that

\[ u = \sum \nolimits _{0 \leq i < m} b_ i^ p \pi ^ i + b \pi ^ m \]

with $b_ i \in B$ and with $b \in B$ an element whose image in $\kappa _ B$ is not a $p$th power. Choose an extension $K \subset K_1$ as in Lemma 15.101.8 with $n = m + 2$ and denote $\pi '$ the uniformizer of the integral closure $A_1$ of $A$ in $K_1$ such that $\pi = (\pi ')^ p + (\pi ')^{np} a$ for some $a \in A_1$. Let $B_1$ be the integral closure of $B$ in $L \otimes _ K K_1$. Observe that $A_1 \to B_1$ is weakly unramified by Lemma 15.101.3. In $B_1$ we have

\[ u \pi = \left(\sum \nolimits _{0 \leq i < m} b_ i (\pi ')^{i + 1}\right)^ p + b (\pi ')^{(m + 1)p} + (\pi ')^{np} b_1 \]

for some $b_1 \in B_1$ (computation omitted). We conclude that $M_1$ is obtained from $L_1$ by adjoining a $p$th root of

\[ b + (\pi ')^{n - m - 1} b_1 \]

Since the residue field of $B_1$ equals the residue field of $B$ we see from Lemma 15.101.9 that $M_1/L_1$ has degree $p$ and the integral closure $C_1$ of $B_1$ is weakly unramified over $B_1$. Thus we conclude in this case.

If there does not exist an integer $m$ as in the preceding paragraph, then $u$ is a $p$th power in the $\pi $-adic completion of $B_1$. Since $B$ is Nagata, this means that $u$ is a $p$th power in $B_1$ by Algebra, Lemma 10.156.18. Whence the second case of the statement of the lemma holds. $\square$

Lemma 15.101.11. Let $A$ be a local ring annihilated by a prime $p$ whose maximal ideal is nilpotent. There exists a ring map $\sigma : \kappa _ A \to A$ which is a section to the residue map $A \to \kappa _ A$. If $A \to A'$ is a local homomorphism of local rings, then we can choose a similar ring map $\sigma ' : \kappa _{A'} \to A'$ compatible with $\sigma $ provided that the extension $\kappa _ A \subset \kappa _{A'}$ is separable.

Proof. Separable extensions are formally smooth by Algebra, Proposition 10.152.9. Thus the existence of $\sigma $ follows from the fact that $\mathbf{F}_ p \to \kappa _ A$ is separable. Similarly for the existence of $\sigma '$ compatible with $\sigma $. $\square$

Lemma 15.101.12. Let $A$ be a discrete valuation ring with fraction field $K$ of characteristic $p > 0$. Let $\xi \in K$. Let $L$ be an extension of $K$ obtained by adjoining a root of $z^ p - z = \xi $. Then $L/K$ is Galois and one of the following happens

  1. $L = K$,

  2. $L/K$ is unramified with respect to $A$ of degree $p$,

  3. $L/K$ is totally ramified with respect to $A$ with ramification index $p$, and

  4. the integral closure $B$ of $A$ in $L$ is a discrete valuation ring, $A \subset B$ is weakly unramified, and $A \to B$ induces a purely inseparable residue field extension of degree $p$.

Let $\pi $ be a uniformizer of $A$. We have the following implications:

  1. If $\xi \in A$, then we are in case (1) or (2).

  2. If $\xi = \pi ^{-n}a$ where $n > 0$ is not divisible by $p$ and $a$ is a unit in $A$, then we are in case (3)

  3. If $\xi = \pi ^{-n} a$ where $n > 0$ is divisible by $p$ and the image of $a$ in $\kappa _ A$ is not a $p$th power, then we are in case (4).

Proof. The extension is Galois of order dividing $p$ by the discussion in Fields, Section 9.25. It immediately follows from the discussion in Section 15.98 that we are in one of the cases (1) – (4) listed in the lemma.

Case (A). Here we see that $A \to A[x]/(x^ p - x - \xi )$ is a finite étale ring extension. Hence we are in cases (1) or (2).

Case (B). Write $\xi = \pi ^{-n}a$ where $p$ does not divide $n$. Let $B \subset L$ be the integral closure of $A$ in $L$. If $C = B_\mathfrak m$ for some maximal ideal $\mathfrak m$, then it is clear that $p \text{ord}_ C(z) = -n \text{ord}_ C(\pi )$. In particular $A \subset C$ has ramification index divisible by $p$. It follows that it is $p$ and that $B = C$.

Case (C). Set $k = n/p$. Then we can rewrite the equation as

\[ (\pi ^ kz)^ p - \pi ^{n - k} (\pi ^ kz) = a \]

Since $A[y]/(y^ p - \pi ^{n - k}y - a)$ is a discrete valuation ring weakly unramified over $A$, the lemma follows. $\square$

Lemma 15.101.13. Let $A \subset B \subset C$ be extensions of discrete valuation rings with fractions fields $K \subset L \subset M$. Assume

  1. $A \subset B$ weakly unramified,

  2. the characteristic of $K$ is $p$,

  3. $M$ is a degree $p$ Galois extension of $L$, and

  4. $\kappa _ A = \bigcap _{n \geq 1} \kappa _ B^{p^ n}$.

Then there exists a finite Galois extension $K_1/K$ totally ramified with respect to $A$ which is a weak solution for $A \to C$.

Proof. Since the characteristic of $L$ is $p$ we know that $M$ is an Artin-Schreier extension of $L$ (Fields, Lemma 9.25.1). Thus we may pick $z \in M$, $z \not\in L$ such that $\xi = z^ p - z \in L$. Choose $n \geq 0$ such that $\pi ^ n\xi \in B$. We pick $z$ such that $n$ is minimal. If $n = 0$, then $M/L$ is unramified with respect to $B$ (Lemma 15.101.12) and we are done. Thus we have $n > 0$.

Assumption (4) implies that $\kappa _ A$ is perfect. Thus we may choose compatible ring maps $\overline{\sigma } : \kappa _ A \to A/\pi ^ n A$ and $\overline{\sigma } : \kappa _ B \to B/\pi ^ n B$ as in Lemma 15.101.11. We lift the second of these to a map of sets $\sigma : \kappa _ B \to B$1. Then we can write

\[ \xi = \sum \nolimits _{i = n, \ldots , 1} \sigma (\lambda _ i) \pi ^{-i} + b \]

for some $\lambda _ i \in \kappa _ B$ and $b \in B$. Let

\[ I = \{ i \in \{ n, \ldots , 1\} \mid \lambda _ i \in \kappa _ A\} \]

and

\[ J = \{ j \in \{ n, \ldots , 1\} \mid \lambda _ i \not\in \kappa _ A\} \]

We will argue by induction on the size of the finite set $J$.

The case $J = \emptyset $. Here for all $i \in \{ n, \ldots , 1\} $ we have $\sigma (\lambda _ i) = a_ i + \pi ^ n b_ i$ for some $a_ i \in A$ and $b_ i \in B$ by our choice of $\sigma $. Thus $\xi = \pi ^{-n} a + b$ for some $a \in A$ and $b \in B$. If $p | n$, then we write $a = a_0^ p + \pi a_1$ for some $a_0, a_1 \in A$ (as the residue field of $A$ is perfect). We compute

\[ (z - \pi ^{-n/p}a_0)^ p - (z - \pi ^{-n/p}a_0) = \pi ^{-(n - 1)}(a_1 + \pi ^{n - 1 - n/p}a_0) + b' \]

for some $b' \in B$. This would contradict the minimality of $n$. Thus $p$ does not divide $n$. Consider the degree $p$ extension $K_1$ of $K$ given by $w^ p - w = \pi ^{-n}a$. By Lemma 15.101.12 this extension is Galois and totally ramified with respect to $A$. Thus $L_1 = L \otimes _ K K_1$ is a field and $A_1 \subset B_1$ is weakly unramified (Lemma 15.101.3). By Lemma 15.101.12 the ring $M_1 = M \otimes _ K K_1$ is either a product of $p$ copies of $L_1$ (in which case we are done) or a field extension of $L_1$ of degree $p$. Moreover, in the second case, either $C_1$ is weakly unramified over $B_1$ (in which case we are done) or $M_1/L_1$ is degree $p$, Galois, and totally ramified with respect to $B_1$. In this last case the extension $M_1/L_1$ is generated by the element $z - w$ and

\[ (z - w)^ p - (z - w) = z^ p - z - (w^ p - w) = b \]

with $b \in B$ (see above). Thus by Lemma 15.101.12 once more the extension $M_1/L_1$ is unramified with respect to $B_1$ and we conclude that $K_1$ is a weak solution for $A \to C$. From now on we assume $J \not= \emptyset $.

Suppose that $j', j \in J$ such that $j' = p^ r j$ for some $r > 0$. Then we change our choice of $z$ into

\[ z' = z - (\sigma (\lambda _ j) \pi ^{-j} + \sigma (\lambda _ j^ p) \pi ^{-pj} + \ldots + \sigma (\lambda _ j^{p^{r - 1}}) \pi ^{-p^{r - 1}j}) \]

Then $\xi $ changes into $\xi ' = (z')^ p - (z')$ as follows

\[ \xi ' = \xi - \sigma (\lambda _ j) \pi ^{-j} + \sigma (\lambda _ j^{p^ r}) \pi ^{-j'} + \text{something in }B \]

Writing $\xi ' = \sum \nolimits _{i = n, \ldots , 1} \sigma (\lambda '_ i) \pi ^{-i} + b'$ as before we find that $\lambda '_ i = \lambda _ i$ for $i \not= j, j'$ and $\lambda '_ j = 0$. Thus the set $J$ has gotten smaller. By induction on the size of $J$ we may assume no such pair $j, j'$ exists. (Please observe that in this procedure we may get thrown back into the case that $J = \emptyset $ we treated above.)

For $j \in J$ write $\lambda _ j = \mu _ j^{p^{r_ j}}$ for some $r_ j \geq 0$ and $\mu _ j \in \kappa _ B$ which is not a $p$th power. This is possible by our assumption (4). Let $j \in J$ be the unique index such that $j p^{-r_ j}$ is maximal. (The index is unique by the result of the preceding paragraph.) Choose $r > \max (r_ j + 1)$ and such that $j p^{r - r_ j} > n$ for $j \in J$. Choose a separable extension $K_1/K$ totally ramified with respect to $A$ of degree $p^ r$ such that the corresponding discrete valuation ring $A_1 \subset K_1$ has uniformizer $\pi '$ with $(\pi ')^{p^ r} = \pi + \pi ^{n + 1}a$ for some $a \in A_1$ (Lemma 15.101.8). Observe that $L_1 = L \otimes _ K K_1$ is a field and that $L_1/L$ is totally ramified with respect to $B$ (Lemma 15.101.3). Computing in the integral closure $B_1$ we get

\[ \xi = \sum \nolimits _{i \in I} \sigma (\lambda _ i) (\pi ')^{-i p^ r} + \sum \nolimits _{j \in J} \sigma (\mu _ j)^{p^{r_ j}} (\pi ')^{-j p^ r} + b_1 \]

for some $b_1 \in B_1$. Note that $\sigma (\lambda _ i)$ for $i \in I$ is a $q$th power modulo $\pi ^ n$, i.e., modulo $(\pi ')^{n p^ r}$. Hence we can rewrite the above as

\[ \xi = \sum \nolimits _{i \in I} x_ i^{p^ r} (\pi ')^{-i p^ r} + \sum \nolimits _{j \in J} \sigma (\mu _ j)^{p^{r_ j}} (\pi ')^{- j p^ r} + b_1 \]

As in the previous paragraph we change our choice of $z$ into

\begin{align*} z' & = z \\ & - \sum \nolimits _{i \in I} \left(x_ i (\pi ')^{-i} + \ldots + x_ i^{p^{r - 1}} (\pi ')^{-i p^{r - 1}}\right) \\ & - \sum \nolimits _{j \in J} \left( \sigma (\mu _ j) (\pi ')^{- j p^{r - r_ j}} + \ldots + \sigma (\mu _ j)^{p^{r_ j - 1}} (\pi ')^{- j p^{r - 1}} \right) \end{align*}

to obtain

\[ (z')^ p - z' = \sum \nolimits _{i \in I} x_ i (\pi ')^{-i} + \sum \nolimits _{j \in J} \sigma (\mu _ j) (\pi ')^{- j p^{r - r_ j}} + b_1' \]

for some $b'_1 \in B_1$. Since there is a unique $j$ such that $j p^{r - r_ j}$ is maximal and since $j p^{r - r_ j}$ is bigger than $i \in I$ and divisible by $p$, we see that $M_1 / L_1$ falls into case (C) of Lemma 15.101.12. This finishes the proof. $\square$

Lemma 15.101.14. Let $A$ be a ring which contains a primitive $p$th root of unity $\zeta $. Set $w = 1 - \zeta $. Then

\[ P(z) = \frac{(1 + wz)^ p - 1}{w^ p} = z^ p - z + \sum \nolimits _{0 < i < p} a_ i z^ i \]

is an element of $A[z]$ and in fact $a_ i \in (w)$. Moreover, we have

\[ P(z_1 + z_2 + w z_1 z_2) = P(z_1) + P(z_2) + w^ p P(z_1) P(z_2) \]

in the polynomial ring $A[z_1, z_2]$.

Proof. It suffices to prove this when

\[ A = \mathbf{Z}[\zeta ] = \mathbf{Z}[x]/(x^{p - 1} + \ldots + x + 1) \]

is the ring of integers of the cyclotomic field. The polynomial identity $t^ p - 1 = (t - 1)(t - \zeta ) \ldots (t - \zeta ^{p - 1})$ (which is proved by looking at the roots on both sides) shows that $t^{p - 1} + \ldots + t + 1 = (t - \zeta ) \ldots (t - \zeta ^{p - 1})$. Substituting $t = 1$ we obtain $p = (1 - \zeta )(1 - \zeta ^2) \ldots (1 - \zeta ^{p - 1})$. The maximal ideal $(p, w) = (w)$ is the unique prime ideal of $A$ lying over $p$ (as fields of characteristic $p$ do not have nontrivial $p$th roots of $1$). It follows that $p = u w^{p - 1}$ for some unit $u$. This implies that

\[ a_ i = \frac{1}{p} {p \choose i} u w^{i - 1} \]

for $p > i > 1$ and $- 1 + a_1 = pw/w^ p = u$. Since $P(-1) = 0$ we see that $0 = (-1)^ p - u$ modulo $(w)$. Hence $a_1 \in (w)$ and the proof if the first part is done. The second part follows from a direct computation we omit. $\square$

Lemma 15.101.15. Let $A$ be a discrete valuation ring of mixed characteristic $(0, p)$ which contains a primitive $p$th root of $1$. Let $P(t) \in A[t]$ be the polynomial of Lemma 15.101.14. Let $\xi \in K$. Let $L$ be an extension of $K$ obtained by adjoining a root of $P(z) = \xi $. Then $L/K$ is Galois and one of the following happens

  1. $L = K$,

  2. $L/K$ is unramified with respect to $A$ of degree $p$,

  3. $L/K$ is totally ramified with respect to $A$ with ramification index $p$, and

  4. the integral closure $B$ of $A$ in $L$ is a discrete valuation ring, $A \subset B$ is weakly unramified, and $A \to B$ induces a purely inseparable residue field extension of degree $p$.

Let $\pi $ be a uniformizer of $A$. We have the following implications:

  1. If $\xi \in A$, then we are in case (1) or (2).

  2. If $\xi = \pi ^{-n}a$ where $n > 0$ is not divisible by $p$ and $a$ is a unit in $A$, then we are in case (3)

  3. If $\xi = \pi ^{-n} a$ where $n > 0$ is divisible by $p$ and the image of $a$ in $\kappa _ A$ is not a $p$th power, then we are in case (4).

Proof. Adjoining a root of $P(z) = \xi $ is the same thing as adjoining a root of $y^ p = w^ p(1 + \xi )$. Since $K$ contains a primitive $p$th root of $1$ the extension is Galois of order dividing $p$ by the discussion in Fields, Section 9.24. It immediately follows from the discussion in Section 15.98 that we are in one of the cases (1) – (4) listed in the lemma.

Case (A). Here we see that $A \to A[x]/(P(x) - \xi )$ is a finite étale ring extension. Hence we are in cases (1) or (2).

Case (B). Write $\xi = \pi ^{-n}a$ where $p$ does not divide $n$. Let $B \subset L$ be the integral closure of $A$ in $L$. If $C = B_\mathfrak m$ for some maximal ideal $\mathfrak m$, then it is clear that $p \text{ord}_ C(z) = -n \text{ord}_ C(\pi )$. In particular $A \subset C$ has ramification index divisible by $p$. It follows that it is $p$ and that $B = C$.

Case (C). Set $k = n/p$. Then we can rewrite the equation as

\[ (\pi ^ kz)^ p - \pi ^{n - k} (\pi ^ kz) + \sum a_ i \pi ^{n - ik} (\pi ^ kz)^ i = a \]

Since $A[y]/(y^ p - \pi ^{n - k}y - \sum a_ i \pi ^{n - ik} y^ i - a)$ is a discrete valuation ring weakly unramified over $A$, the lemma follows. $\square$

Let $A$ be a discrete valuation ring of mixed characteristic $(0, p)$ containing a primitive $p$th root of $1$. Let $w \in A$ and $P(t) \in A[t]$ be as in Lemma 15.101.14. Let $L$ be a finite extension of $K$. We say $L/K$ is a degree $p$ extension of finite level if $L$ is a degree $p$ extension of $K$ obtained by adjoining a root of the equation $P(z) = \xi $ where $\xi \in K$ is an element with $w^ p \xi \in \mathfrak m_ A$.

This definition is relevant to the discussion in this section due to the following straightforward lemma.

Lemma 15.101.16. Let $A \subset B \subset C$ be extensions of discrete valuation rings with fractions fields $K \subset L \subset M$. Assume that

  1. $A$ has mixed characteristic $(0, p)$,

  2. $A \subset B$ is weakly unramified,

  3. $B$ contains a primitive $p$th root of $1$, and

  4. $M/L$ is Galois of degree $p$.

Then there exists a finite Galois extension $K_1/K$ totally ramified with respect to $A$ which is either a weak solution for $A \to C$ or is such that $M_1/L_1$ is a degree $p$ extension of finite level.

Proof. Let $\pi \in A$ be a uniformizer. By Kummer theory (Fields, Lemma 9.24.1) $M$ is obtained from $L$ by adjoining the root of $y^ p = b$ for some $b \in L$.

If $\text{ord}_ B(b)$ is prime to $p$, then we choose a degree $p$ separable extension $K \subset K_1$ totally ramified with respect to $A$ (for example using Lemma 15.101.8). Let $A_1$ be the integral closure of $A$ in $K_1$. By Lemma 15.101.3 the integral closure $B_1$ of $B$ in $L_1 = L \otimes _ K K_1$ is a discrete valuation ring weakly unramified over $A_1$. If $K \subset K_1$ is not a weak solution for $A \to C$, then the integral closure $C_1$ of $C$ in $M_1 = M \otimes _ K K_1$ is a discrete valuation ring and $B_1 \to C_1$ has ramification index $p$. In this case, the field $M_1$ is obtained from $L_1$ by adjoining the $p$th root of $b$ with $\text{ord}_{B_1}(b)$ divisible by $p$. Replacing $A$ by $A_1$, etc we may assume that $b = \pi ^ n u$ where $u \in B$ is a unit and $n$ is divisible by $p$. Of course, in this case the extension $M$ is obtained from $L$ by adjoining the $p$th root of a unit.

Suppose $M$ is obtained from $L$ by adjoining the root of $y^ p = u$ for some unit $u$ of $B$. If the residue class of $u$ in $\kappa _ B$ is not a $p$th power, then $B \subset C$ is weakly unramified (Lemma 15.101.9) and we are done. Otherwise, we can replace our choice of $y$ by $y/v$ where $v^ p$ and $u$ have the same image in $\kappa _ B$. After such a replacement we have

\[ y^ p = 1 + \pi b \]

for some $b \in B$. Then we see that $P(z) = \pi b/ w^ p$ where $z = (y - 1)/w$. Thus we see that the extension is a degree $p$ extension of finite level with $\xi = \pi b / w^ p$. $\square$

Let $A$ be a discrete valuation ring of mixed characteristic $(0, p)$ containing a primitive $p$th root of $1$. Let $w \in A$ and $P(t) \in A[t]$ be as in Lemma 15.101.14. Let $L$ be a degree $p$ extension of $K$ of finite level. Choose $z \in L$ generating $L$ over $K$ with $\xi = P(z) \in K$. Choose a uniformizer $\pi $ for $A$ and write $w = u \pi ^{e_1}$ for some integer $e_1 = \text{ord}_ A(w)$ and unit $u \in A$. Finally, pick $n \geq 0$ such that

\[ \pi ^ n \xi \in A \]

The level of $L/K$ is the smallest value of the quantity $n/e_1$ taking over all $z$ generating $L/K$ with $\xi = P(z) \in K$.

We make a couple of remarks. Since the extension is of finite level we know that we can choose $z$ such that $n < pe_1$. Thus the level is a rational number contained in $[0, p)$. If the level is zero then $L/K$ is unramified with respect to $A$ by Lemma 15.101.15. Our next goal is to lower the level.

Lemma 15.101.17. Let $A \subset B \subset C$ be extensions of discrete valuation rings with fractions fields $K \subset L \subset M$. Assume

  1. $A$ has mixed characteristic $(0, p)$,

  2. $A \subset B$ weakly unramified,

  3. $B$ contains a primitive $p$th root of $1$,

  4. $M/L$ is a degree $p$ extension of finite level $l > 0$,

  5. $\kappa _ A = \bigcap _{n \geq 1} \kappa _ B^{p^ n}$.

Then there exists a finite separable extension $K_1$ of $K$ totally ramified with respect to $A$ such that either $K_1$ is a weak solution for $A \to C$, or the extension $M_1/L_1$ is a degree $p$ extension of finite level $\leq \max (0, l - 1, 2l - p)$.

Proof. Let $\pi \in A$ be a uniformizer. Let $w \in B$ and $P \in B[t]$ be as in Lemma 15.101.14 (for $B$). Set $e_1 = \text{ord}_ B(w)$, so that $w$ and $\pi ^{e_1}$ are associates in $B$. Pick $z \in M$ generating $M$ over $L$ with $\xi = P(z) \in K$ and $n$ such that $\pi ^ n\xi \in B$ as in the definition of the level of $M$ over $L$, i.e., $l = n/e_1$.

The proof of this lemma is completely similar to the proof of Lemma 15.101.13. To explain what is going on, observe that

15.101.17.1
\begin{equation} \label{more-algebra-equation-first-congruence} P(z) \equiv z^ p - z \bmod \pi ^{-n + e_1}B \end{equation}

for any $z \in L$ such that $\pi ^{-n} P(z) \in B$ (use that $z$ has valuation at worst $-n/p$ and the shape of the polynomial $P$). Moreover, we have

15.101.17.2
\begin{equation} \label{more-algebra-equation-second-congruence} \xi _1 + \xi _2 + w^ p \xi _1 \xi _2 \equiv \xi _1 + \xi _2 \bmod \pi ^{-2n + pe_1}B \end{equation}

for $\xi _1, \xi _2 \in \pi ^{-n}B$. Finally, observe that $n - e_1 = (l - 1)/e_1$ and $-2n + pe_1 = -(2l - p)e_1$. Write $m = n - e_1 \max (0, l - 1, 2l - p)$. The above shows that doing calculations in $\pi ^{-n}B / \pi ^{-n + m}B$ the polynomial $P$ behaves exactly as the polynomial $z^ p - z$. This explains why the lemma is true but we also give the details below.

Assumption (4) implies that $\kappa _ A$ is perfect. Observe that $m \leq e_1$ and hence $A/\pi ^ m$ is annihilated by $w$ and hence $p$. Thus we may choose compatible ring maps $\overline{\sigma } : \kappa _ A \to A/\pi ^ mA$ and $\overline{\sigma } : \kappa _ B \to B/\pi ^ mB$ as in Lemma 15.101.11. We lift the second of these to a map of sets $\sigma : \kappa _ B \to B$. Then we can write

\[ \xi = \sum \nolimits _{i = n, \ldots , n - m + 1} \sigma (\lambda _ i) \pi ^{-i} + \pi ^{-n + m)} b \]

for some $\lambda _ i \in \kappa _ B$ and $b \in B$. Let

\[ I = \{ i \in \{ n, \ldots , n - m + 1\} \mid \lambda _ i \in \kappa _ A\} \]

and

\[ J = \{ j \in \{ n, \ldots , n - m + 1\} \mid \lambda _ i \not\in \kappa _ A\} \]

We will argue by induction on the size of the finite set $J$.

The case $J = \emptyset $. Here for all $i \in \{ n, \ldots , n - m + 1\} $ we have $\sigma (\lambda _ i) = a_ i + \pi ^{n - m}b_ i$ for some $a_ i \in A$ and $b_ i \in B$ by our choice of $\overline{\sigma }$. Thus $\xi = \pi ^{-n} a + \pi ^{-n + m} b$ for some $a \in A$ and $b \in B$. If $p | n$, then we write $a = a_0^ p + \pi a_1$ for some $a_0, a_1 \in A$ (as the residue field of $A$ is perfect). Set $z_1 = - \pi ^{-n/p} a_0$. Note that $P(z_1) \in \pi ^{-n}B$ and that $z + z_1 + w z z_1$ is an element generating $M$ over $L$ (note that $wz_1 \not= -1$ as $n < pe_1$). Moreover, by Lemma 15.101.14 we have

\[ P(z + z_1 + w z z_1) = P(z) + P(z_1) + w^ p P(z) P(z_1) \in K \]

and by equations (15.101.17.1) and (15.101.17.2) we have

\[ P(z) + P(z_1) + w^ p P(z) P(z_1) \equiv \xi + z_1^ p - z_1 \bmod \pi ^{-n + m}B \]

for some $b' \in B$. This contradict the minimality of $n$! Thus $p$ does not divide $n$. Consider the degree $p$ extension $K_1$ of $K$ given by $P(y) = -\pi ^{-n}a$. By Lemma 15.101.15 this extension is separable and totally ramified with respect to $A$. Thus $L_1 = L \otimes _ K K_1$ is a field and $A_1 \subset B_1$ is weakly unramified (Lemma 15.101.3). By Lemma 15.101.15 the ring $M_1 = M \otimes _ K K_1$ is either a product of $p$ copies of $L_1$ (in which case we are done) or a field extension of $L_1$ of degree $p$. Moreover, in the second case, either $C_1$ is weakly unramified over $B_1$ (in which case we are done) or $M_1/L_1$ is degree $p$, Galois, totally ramified with respect to $B_1$. In this last case the extension $M_1/L_1$ is generated by the element $z + y + wzy$ and we see that $P(z + y + wzy) \in L_1$ and

\begin{align*} P(z + y + wzy) & = P(z) + P(y) + w^ p P(z) P(y) \\ & \equiv \xi - \pi ^{-n}a \bmod \pi ^{-n + m}B_1 \\ & \equiv 0 \bmod \pi ^{-n + m}B_1 \end{align*}

in exactly the same manner as above. By our choice of $m$ this means exactly that $M_1/L_1$ has level at most $\max (0, l - 1, 2l - p)$. From now on we assume that $J \not= \emptyset $.

Suppose that $j', j \in J$ such that $j' = p^ r j$ for some $r > 0$. Then we set

\[ z_1 = - \sigma (\lambda _ j) \pi ^{-j} - \sigma (\lambda _ j^ p) \pi ^{-pj} - \ldots - \sigma (\lambda _ j^{p^{r - 1}}) \pi ^{-p^{r - 1}j} \]

and we change $z$ into $z' = z + z_1 + wzz_1$. Observe that $z' \in M$ generates $M$ over $L$ and that we have $\xi ' = P(z') = P(z) + P(z_1) + wP(z)P(z_1) \in L$ with

\[ \xi ' \equiv \xi - \sigma (\lambda _ j) \pi ^{-j} + \sigma (\lambda _ j^{p^ r}) \pi ^{-j'} \bmod \pi ^{-n + m}B \]

by using equations (15.101.17.1) and (15.101.17.2) as above. Writing

\[ \xi ' = \sum \nolimits _{i = n, \ldots , n - m + 1} \sigma (\lambda '_ i) \pi ^{-i} + \pi ^{-n + m}b' \]

as before we find that $\lambda '_ i = \lambda _ i$ for $i \not= j, j'$ and $\lambda '_ j = 0$. Thus the set $J$ has gotten smaller. By induction on the size of $J$ we may assume there is no pair $j, j'$ of $J$ such that $j'/j$ is a power of $p$. (Please observe that in this procedure we may get thrown back into the case that $J = \emptyset $ we treated above.)

For $j \in J$ write $\lambda _ j = \mu _ j^{p^{r_ j}}$ for some $r_ j \geq 0$ and $\mu _ j \in \kappa _ B$ which is not a $p$th power. This is possible by our assumption (4). Let $j \in J$ be the unique index such that $j p^{-r_ j}$ is maximal. (The index is unique by the result of the preceding paragraph.) Choose $r > \max (r_ j + 1)$ and such that $j p^{r - r_ j} > n$ for $j \in J$. Let $K_1/K$ be the extension of degree $p^ r$, totally ramified with respect to $A$, defined by $(\pi ')^{p^ r} = \pi $. Observe that $\pi '$ is the uniformizer of the corresponding discrete valuation ring $A_1 \subset K_1$. Observe that $L_1 = L \otimes _ K K_1$ is a field and $L_1/L$ is totally ramified with respect to $B$ (Lemma 15.101.3). Computing in the integral closure $B_1$ we get

\[ \xi = \sum \nolimits _{i \in I} \sigma (\lambda _ i) (\pi ')^{-i p^ r} + \sum \nolimits _{j \in J} \sigma (\mu _ j)^{p^{r_ j}} (\pi ')^{-j p^ r} + \pi ^{-n + m} b_1 \]

for some $b_1 \in B_1$. Note that $\sigma (\lambda _ i)$ for $i \in I$ is a $q$th power modulo $\pi ^ m$, i.e., modulo $(\pi ')^{m p^ r}$. Hence we can rewrite the above as

\[ \xi = \sum \nolimits _{i \in I} x_ i^{p^ r} (\pi ')^{-i p^ r} + \sum \nolimits _{j \in J} \sigma (\mu _ j)^{p^{r_ j}} (\pi ')^{- j p^ r} + \pi ^{-n + m}b_1 \]

Similar to our choice in the previous paragraph we set

\begin{align*} z_1 & - \sum \nolimits _{i \in I} \left(x_ i (\pi ')^{-i} + \ldots + x_ i^{p^{r - 1}} (\pi ')^{-i p^{r - 1}}\right) \\ & - \sum \nolimits _{j \in J} \left( \sigma (\mu _ j) (\pi ')^{- j p^{r - r_ j}} + \ldots + \sigma (\mu _ j)^{p^{r_ j - 1}} (\pi ')^{- j p^{r - 1}} \right) \end{align*}

and we change our choice of $z$ into $z' = z + z_1 + wzz_1$. Then $z'$ generates $M_1$ over $L_1$ and $\xi ' = P(z') = P(z) + P(z_1) + w^ p P(z) P(z_1) \in L_1$ and a calculation shows that

\[ \xi ' \equiv \sum \nolimits _{i \in I} x_ i (\pi ')^{-i} + \sum \nolimits _{j \in J} \sigma (\mu _ j) (\pi ')^{- j p^{r - r_ j}} + (\pi ')^{(-n + m)p^ r}b'_1 \]

for some $b'_1 \in B_1$. There is a unique $j$ such that $j p^{r - r_ j}$ is maximal and $j p^{r - r_ j}$ is bigger than $i \in I$. If $j p^{r - r_ j} \leq (n - m)p^ r$ then the level of the extension $M_1/L_1$ is less than $\max (0, l - 1, 2l - p)$. If not, then, as $p$ divides $j p^{r - r_ j}$, we see that $M_1 / L_1$ falls into case (C) of Lemma 15.101.15. This finishes the proof. $\square$

Lemma 15.101.18. Let $A \subset B \subset C$ be extensions of discrete valuation rings with fraction fields $K \subset L \subset M$. Assume

  1. the residue field $k$ of $A$ is algebraically closed of characteristic $p > 0$,

  2. $A$ and $B$ are complete,

  3. $A \to B$ is weakly unramified,

  4. $M$ is a finite extension of $L$,

  5. $k = \bigcap \nolimits _{n \geq 1} \kappa _ B^{p^ n}$

Then there exists a finite extension $K \subset K_1$ which is a weak solution for $A \to C$.

Proof. Let $M'$ be any finite extension of $L$ and consider the integral closure $C'$ of $B$ in $M'$. Then $C'$ is finite over $B$ as $B$ is Nagata by Algebra, Lemma 10.156.8. Moreover, $C'$ is a discrete valuation ring, see discussion in Remark 15.100.1. Moreover $C'$ is complete as a $B$-module, hence complete as a discrete valuation ring, see Algebra, Section 10.95. It follows in particular that $C$ is the integral closure of $B$ in $M$ (by definition of valuation rings as maximal for the relation of domination).

Let $M \subset M'$ be a finite extension and let $C' \subset M'$ be the integral closure of $B$ as above. By Lemma 15.101.4 it suffices to prove the result for $A \to B \to C'$. Hence we may assume that $M/L$ is normal, see Fields, Lemma 9.16.3.

If $M / L$ is normal, we can find a chain of finite extensions

\[ L = L^0 \subset L^1 \subset L^2 \subset \ldots \subset L^ r = M \]

such that each extension $L^{j + 1}/L^ j$ is either:

  1. purely inseparable of degree $p$,

  2. totally ramified with respect to $B^ j$ and Galois of degree $p$,

  3. totally ramified with respect to $B^ j$ and Galois cyclic of order prime to $p$,

  4. Galois and unramified with respect to $B^ j$.

Here $B^ j$ is the integral closure of $B$ in $L^ j$. Namely, since $M/L$ is normal we can write it as a compositum of a Galois extension and a purely inseparable extension (Fields, Lemma 9.27.3). For the purely inseparable extension the existence of the filtration is clear. In the Galois case, note that $G$ is “the” decomposition group and let $I \subset G$ be the inertia group. Then on the one hand $I$ is solvable by Lemma 15.98.5 and on the other hand the extension $M^ I/L$ is unramified with respect to $B$ by Lemma 15.98.8. This proves we have a filtration as stated.

We are going to argue by induction on the integer $r$. Suppose that we can find a finite extension $K \subset K_1$ which is a weak solution for $A \to B^1$ where $B^1$ is the integral closure of $B$ in $L^1$. Let $K'_1$ be the normal closure of $K_1/K$ (Fields, Lemma 9.16.3). Since $A$ is complete and the residue field of $A$ is algebraically closed we see that $K'_1/K_1$ is separable and totally ramified with respect to $A_1$ (some details omitted). Hence $K \subset K'_1$ is a weak solution for $A \to B^1$ as well by Lemma 15.101.3. In other words, we may and do assume that $K_1$ is a normal extension of $K$. Having done so we consider the sequence

\[ L^0_1 = (L^0 \otimes _ K K_1)_{red} \subset L^1_1 = (L^1 \otimes _ K K_1)_{red} \subset \ldots \subset L^ r_1 = (L^ r \otimes _ K K_1)_{red} \]

and the corresponding integral closures $B^ i_1$. Note that $C_1 = B^ r_1$ is a product of discrete valuation rings which are transitively permuted by $G = \text{Aut}(K_1/K)$ by Lemma 15.101.7. In particular all the extensions of discrete valuation rings $A_1 \to (C_1)_\mathfrak m$ are isomorphic and a solution for one will be a solution for all of them. We can apply the induction hypothesis to the sequence

\[ A_1 \to (B^1_1)_{B^1_1 \cap \mathfrak m} \to (B^2_1)_{B^2_1 \cap \mathfrak m} \to \ldots \to (B^ r_1)_{B^ r_1 \cap \mathfrak m} = (C_1)_\mathfrak m \]

to get a solution $K_1 \subset K_2$ for $A_1 \to (C_1)_\mathfrak m$. The extension $K \subset K_2$ will then be a solution for $A \to C$ by what we said before. Note that the induction hypothesis applies: the ring map $A_1 \to (B^1_1)_{B^1_1 \cap \mathfrak m}$ is weakly unramified by our choice of $K_1$ and the sequence of fraction field extensions each still have one of the properties (a), (b), (c), or (d) listed above. Moreover, observe that for any finite extension $\kappa _ B \subset \kappa $ we still have $k = \bigcap \kappa ^{p^ n}$.

Thus everything boils down to finding a weak solution for $A \subset C$ when the field extension $L \subset M$ satisfies one of the properties (a), (b), (c), or (d).

Case (d). This case is trivial as here $B \to C$ is unramified already.

Case (c). Say $M/L$ is cyclic of order $n$ prime to $p$. Because $M/L$ is totally ramified with respect to $B$ we see that the ramification index of $B \subset C$ is $n$ and hence the ramification index of $A \subset C$ is $n$ as well. Choose a uniformizer $\pi \in A$ and set $K_1 = K[\pi ^{1/n}]$. Then $K_1/K$ is a solution for $A \subset C$ by Abhyankar's lemma (Lemma 15.100.4).

Case (b). We divide this case into the mixed characteristic case and the equicharacteristic case. In the equicharacteristic case this is Lemma 15.101.13. In the mixed characteristic case, we first replace $K$ by a finite extension to get to the situation where $M/L$ is a degree $p$ extension of finite level using Lemma 15.101.16. Then the level is a rational number $l \in [0, p)$, see discussion preceding Lemma 15.101.17. If the level is $0$, then $B \to C$ is weakly unramified and we're done. If not, then we can replacing the field $K$ by a finite extension to obtain a new situation with level $l' \leq \max (0, l - 1, 2l - p)$ by Lemma 15.101.17. If $l = p - \epsilon $ for $\epsilon < 1$ then we see that $l' \leq p - 2\epsilon $. Hence after a finite number of replacements we obtain a case with level $\leq p - 1$. Then after at most $p - 1$ more such replacements we reach the situation where the level is zero.

Case (a) is Lemma 15.101.10. This is the only case where we possibly need a purely inseparable extension of $K$, namely, in case (2) of the statement of the lemma we win by adjoining a $p$th power of the element $\pi $. This finishes the proof of the lemma. $\square$

At this point we have collected all the lemmas we need to prove the main result of this section.

Theorem 15.101.19 (Epp). Let $A \subset B$ be an extension of discrete valuation rings with fraction fields $K \subset L$. If the characteristic of $\kappa _ A$ is $p > 0$, assume that every element of

\[ \bigcap \nolimits _{n \geq 1} \kappa _ B^{p^ n} \]

is separable algebraic over $\kappa _ A$. Then there exists a finite extension $K \subset K_1$ which is a weak solution for $A \to B$ as defined in Definition 15.101.1.

Proof. If the characteristic of $\kappa _ A$ is zero or if the residue characteristic is $p$, the ramification index is prime to $p$, and the residue field extension is separable, then this follows from Abhyankar's lemma (Lemma 15.100.4). Namely, suppose the ramification index is $e$. Choose a uniformizer $\pi \in A$. Let $K_1/K$ be the extension obtained by adjoining an $e$th root of $\pi $. By Lemma 15.100.2 we see that the integral closure $A_1$ of $A$ in $K_1$ is a discrete valuation ring with ramification index over $A$. Thus $A_1 \to (B_1)_\mathfrak m$ is formally smooth in the $\mathfrak m$-adic topology for all maximal ideals $\mathfrak m$ of $B_1$ by Lemma 15.100.4 and a fortiori these are weakly unramified extensions of discrete valuation rings.

From now on we let $p$ be a prime number and we assume that $\kappa _ A$ has characteristic $p$. We first apply Lemma 15.101.6 to reduce to the case that $A$ and $B$ have separably closed residue fields. Since $\kappa _ A$ and $\kappa _ B$ are replaced by their separable algebraic closures by this procedure we see that we obtain

\[ \kappa _ A \supset \bigcap \nolimits _{n \geq 1} \kappa _ B^{p^ n} \]

from the condition of the theorem.

Let $\pi \in A$ be a uniformizer. Let $A^\wedge $ and $B^\wedge $ be the completions of $A$ and $B$. We have a commutative diagram

\[ \xymatrix{ B \ar[r] & B^\wedge \\ A \ar[u] \ar[r] & A^\wedge \ar[u] } \]

of extensions of discrete valuation rings. Let $K^\wedge $ be the fraction field of $A^\wedge $. Suppose that we can find a finite extension $K^\wedge \subset M$ which is (a) a weak solution for $A^\wedge \to B^\wedge $ and (b) a compositum of a separable extension and an extension obtained by adjoining a $p$-power root of $\pi $. Then by Lemma 15.99.2 we can find a finite extension $K \subset K_1$ such that $K^\wedge \otimes _ K K_1 = M$. Let $A_1$, resp. $A_1^\wedge $ be the integral closure of $A$, resp. $A^\wedge $ in $K_1$, resp. $M$. Since $A \to A^\wedge $ is formally smooth in the $\mathfrak m^\wedge $-adic topology (Lemma 15.97.5) we see that $A_1 \to A_1^\wedge $ is formally smooth in the $\mathfrak m_1^\wedge $-adic topology (Lemma 15.100.3 and $A_1$ and $A_1^\wedge $ are discrete valuation rings by discussion in Remark 15.100.1). We conclude from Lemma 15.101.4 part (2) that $K \subset K_1$ is a weak solution for $A \to B^\wedge $. Applying Lemma 15.101.4 part (1) we see that $K \subset K_1$ is a weak solution for $A \to B$.

Thus we may assume $A$ and $B$ are complete discrete valuation rings with separably closed residue fields of characteristic $p$ and with $\kappa _ A \supset \bigcap \nolimits _{n \geq 1} \kappa _ B^{p^ n}$. We are also given a uniformizer $\pi \in A$ and we have to find a weak solution for $A \to B$ which is a compositum of a separable extension and a field obtained by taking $p$-power roots of $\pi $. Note that the second condition is automatic if $A$ has mixed characteristic.

Set $k = \bigcap \nolimits _{n \geq 1} \kappa _ B^{p^ n}$. Observe that $k$ is an algebraically closed field of characteristic $p$. If $A$ has mixed characteristic let $\Lambda $ be a Cohen ring for $k$ and in the equicharacteristic case set $\Lambda = k[[t]]$. We can choose a ring map $\Lambda \to A$ which maps $t$ to $\pi $ in the equicharacteristic case. In the equicharacteristic case this follows from the Cohen structure theorem (Algebra, Theorem 10.154.8) and in the mixed characteristic case this follows as $\mathbf{Z}_ p \to \Lambda $ is formally smooth in the adic topology (Lemmas 15.97.5 and 15.36.5). Applying Lemma 15.101.4 we see that it suffices to prove the existence of a weak solution for $\Lambda \to B$ which in the equicharacteristic $p$ case is a compositum of a separable extension and a field obtained by taking $p$-power roots of $t$. However, since $\Lambda = k[[t]]$ in the equicharacteristic case and any extension of $k((t))$ is such a compositum, we can now drop this requirement!

Thus we arrive at the situation where $A$ and $B$ are complete, the residue field $k$ of $A$ is algebraically closed of characteristic $p > 0$, we have $k = \bigcap \kappa _ B^{p^ n}$, and in the mixed characteristic case $p$ is a uniformizer of $A$ (i.e., $A$ is a Cohen ring for $k$). If $A$ has mixed characteristic choose a Cohen ring $\Lambda $ for $\kappa _ B$ and in the equicharacteristic case set $\Lambda = \kappa _ B[[t]]$. Arguing as above we may choose a ring map $A \to \Lambda $ lifting $k \to \kappa _ B$ and mapping a uniformizer to a uniformizer. Since $k \subset \kappa _ B$ is separable the ring map $A \to \Lambda $ is formally smooth in the adic topology (Lemma 15.97.5). Hence we can find a ring map $\Lambda \to B$ such that the composition $A \to \Lambda \to B$ is the given ring map $A \to B$ (see Lemma 15.36.5). Since $\Lambda $ and $B$ are complete discrete valuation rings with the same residue field, $B$ is finite over $\Lambda $ (Algebra, Lemma 10.95.12). This reduces us to the special case discussed in Lemma 15.101.18. $\square$

Lemma 15.101.20. Let $A \to B$ be an extension of discrete valuation rings with fraction fields $K \subset L$. Assume $B$ is essentially of finite type over $A$. Let $K \subset K'$ be an algebraic extension of fields such that the integral closure $A'$ of $A$ in $K'$ is Noetherian. Then the integral closure $B'$ of $B$ in $L' = (L \otimes _ K K')_{red}$ is Noetherian as well. Moreover, the map $\mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(A')$ is surjective and the corresponding residue field extensions are finitely generated field extensions.

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 10.111.4). 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 10.118.12 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 10.118.10. Finally, we observe that $B' = \prod B'_ i$ is the integral closure of $B$ in $L' = (L \otimes _ K K')_{red}$. $\square$

reference

Proposition 15.101.21. Let $A \to B$ be an extension of discrete valuation rings with fraction fields $K \subset L$. If $B$ is essentially of finite type over $A$, then there exists a finite extension $K \subset K_1$ which is a solution for $A \to B$ as defined in Definition 15.101.1.

Proof. Observe that a weak solution is a solution if the residue field of $A$ is perfect, see Lemma 15.97.5. Thus the proposition follows immediately from Theorem 15.101.19 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

\[ A' = \bigcup \nolimits _{n \geq 1} A[t_{i, n}]/(t_{i, n}^{p^ n} - x_ i) \]

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 15.101.20 to the extension $K \subset K'$. 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 15.101.19) we find a weak solution $K' \subset K'_ i$ for each of the extensions $A' \subset B'_{\mathfrak m_ i}$. Since the residue field of $A'$ is perfect, these are actually solutions. Let $K' \subset K'_1$ be a finite extension which contains each $K'_ i$. Then $K' \subset K'_1$ is still a solution for each $A' \subset B'_{\mathfrak m_ i}$ by Lemma 15.100.3.

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 15.100.1 applied to $K' \subset K'_1$. Thus Lemma 15.101.20 applies to $K \subset K'_1$. Therefore the integral closure $B'_1$ of $B$ in $L'_1 = (L \otimes _ K K'_1)_{red}$ is a Dedekind domain and because $K' \subset K'_1$ 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

\[ A'_1 = \mathop{\mathrm{colim}}\nolimits A_ i\quad \text{and}\quad B'_1 = \mathop{\mathrm{colim}}\nolimits B_ i \]

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

\[ B_ i \otimes _{A_ i} A'_1 \longrightarrow B'_1 \]

induces surjective maps on residue fields (because the residue fields of $B'_1$ are finitely generated over those of $A'_1$ by Lemma 15.101.20). Picture of residue fields at maximal ideals lying under a chosen maximal ideal of $B'_1$:

\[ \xymatrix{ \kappa _{B_ i} \ar[r] & \kappa _{B_{i'}} \ar[r] & & \ldots & \kappa _{B'_1} \\ \kappa _{A_ i} \ar[r] \ar[u] & \kappa _{A_{i'}} \ar[r] \ar[u] & & \ldots & \kappa _{A'_1} \ar[u] } \]

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. $\square$

Lemma 15.101.22. Let $A \to B$ be an extension of discrete valuation rings with fraction fields $K \subset L$. Assume

  1. $B$ is essentially of finite type over $A$,

  2. either $A$ or $B$ is a Nagata ring, and

  3. $L/K$ is separable.

Then there exists a separable solution for $A \to B$ (Definition 15.101.1).

Proof. Observe that if $A$ is Nagata, then so is $B$ (Algebra, Lemma 10.156.6 and Proposition 10.156.15). Thus the lemma follows on combining Proposition 15.101.21 and Lemma 15.101.5. $\square$

[1] If $B$ is complete, then we can choose $\sigma $ to be a ring map. If $A$ is also complete and $\sigma $ is a ring map, then $\sigma $ maps $\kappa _ A$ into $A$.

Comments (2)

Comment #3000 by Dino Lorenzini on

In the statement of Abhyankar's lemma and the first two lines of proof, replace 3 times by .

Line 6 of the claim: Replace by

Comment #3123 by on

Thanks! You must have very good eyes! Changes are here.


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