The Stacks project

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

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. $\square$

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 15.114.1. 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

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 15.111.5 this proves (1). A similar argument works for (2). $\square$

Proof. By Algebra, Lemma 10.159.2 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 $K'/K$ 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 10.155.9 there exists a commutative diagram as in the statement of the lemma. Parts (1) and (2) of the lemma are clear.

Let $K'_1/K'$ 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

is a finite product of rings isomorphic to $L' \otimes _{K'} K'_1$. Thus we see that $K_1/K$ is a (weak) solution for $A \to B'$. Hence it is also a (weak) solution for $A \to B$ by Lemma 15.115.4. $\square$

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. $\square$

Proof. If the characteristic of $K$ is zero, then we can take the extension given by $\pi _ B^ q = \pi$, see Lemma 15.114.2. 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. $\square$

Proof. This lemma proves itself. $\square$

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 15.111.2 and 15.111.4. 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

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 $K_1/K$ as in Lemma 15.115.7 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 15.115.3. In $B_1$ we have

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

Since the residue field of $B_1$ equals the residue field of $B$ we see from Lemma 15.115.8 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 10.162.18. Whence the second case of the statement of the lemma holds. $\square$

Proof. Separable extensions are formally smooth by Algebra, Proposition 10.158.9. 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 $\sigma$. $\square$

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 15.112 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

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

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 15.115.11) 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 15.115.10. We lift the second of these to a map of sets $\sigma : \kappa _ B \to B$¹. Then we can write

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

and

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

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 15.115.11 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 15.115.3). By Lemma 15.115.11 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

with $b \in B$ (see above). Thus by Lemma 15.115.11 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

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

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 15.115.7). 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 15.115.3). Computing in the integral closure $B_1$ we get

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

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

to obtain

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 15.115.11. This finishes the proof. $\square$

Proof. It suffices to prove this when

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

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. $\square$

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 15.112 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

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. $\square$

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 $K_1/K$ totally ramified with respect to $A$ (for example using Lemma 15.115.7). Let $A_1$ be the integral closure of $A$ in $K_1$. By Lemma 15.115.3 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 $K_1/K$ 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 15.115.8) 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

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 $\xi = \pi b / w^ p$. $\square$

Proof. Let $\pi \in A$ be a uniformizer. Let $w \in B$ and $P \in B[t]$ be as in Lemma 15.115.13 (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 15.115.12. To explain what is going on, observe that

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

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 15.115.10. We lift the second of these to a map of sets $\sigma : \kappa _ B \to B$. Then we can write

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

and

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 15.115.13 we have

and by equations (15.115.16.1) and (15.115.16.2) we have

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 15.115.14 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 15.115.3). By Lemma 15.115.14 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

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

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

by using equations (15.115.16.1) and (15.115.16.2) as above. Writing

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 15.115.3). Computing in the integral closure $B_1$ we get

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

Similar to our choice in the previous paragraph we set

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

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 15.115.14. This finishes the proof. $\square$

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 10.162.8. Moreover, $C'$ is a discrete valuation ring, see discussion in Remark 15.114.1. Moreover $C'$ is complete as a $B$-module, hence complete as a discrete valuation ring, see Algebra, Section 10.96. 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 15.115.4 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

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 15.112.5 and on the other hand the extension $M^ I/L$ is unramified with respect to $B$ by Lemma 15.112.8. 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 $K_1/K$ 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 $K'_1/K$ is a weak solution for $A \to B^1$ as well by Lemma 15.115.3. In other words, we may and do assume that $K_1$ is a normal extension of $K$. Having done so we consider the sequence

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 15.115.6. In particular all the extensions of discrete valuation rings $A_1 \to (C_1)_\mathfrak m$ are isomorphic and a weak solution for one will be a weak solution for all of them. We can apply the induction hypothesis to the sequence

to get a weak solution $K_2/K_1$ for $A_1 \to (C_1)_\mathfrak m$. The extension $K_2/K$ will then be a weak 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 $M/L$ 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 15.114.4).

Case (b). We divide this case into the mixed characteristic case and the equicharacteristic case. In the equicharacteristic case this is Lemma 15.115.12. 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 15.115.15. Then the level is a rational number $l \in [0, p)$, see discussion preceding Lemma 15.115.16. 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 15.115.16. 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 15.115.9. 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. $\square$

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 15.114.4). 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 15.114.2 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 15.114.4 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 15.115.5 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

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

of extensions of discrete valuation rings. Let $K^\wedge$ be the fraction field of $A^\wedge$. Suppose that we can find a finite extension $M/K^\wedge$ 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 15.113.2 we can find a finite extension $K_1/K$ 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 15.111.5) we see that $A_1 \to A_1^\wedge$ is formally smooth in the $\mathfrak m_1^\wedge$-adic topology (Lemma 15.114.3 and $A_1$ and $A_1^\wedge$ are discrete valuation rings by discussion in Remark 15.114.1). We conclude from Lemma 15.115.4 part (2) that $K_1/K$ is a weak solution for $A \to B^\wedge$. Applying Lemma 15.115.4 part (1) we see that $K_1/K$ 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 10.160.8) and in the mixed characteristic case this follows as $\mathbf{Z}_ p \to \Lambda$ is formally smooth in the adic topology (Lemmas 15.111.5 and 15.37.5). Applying Lemma 15.115.4 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 15.111.5). 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 15.37.5). Since $\Lambda$ and $B$ are complete discrete valuation rings with the same residue field, $B$ is finite over $\Lambda$ (Algebra, Lemma 10.96.12). This reduces us to the special case discussed in Lemma 15.115.17. $\square$