The Stacks project

Proof. Choose a nonzero $\pi \in \mathfrak m$. Since the dimension of $A$ is $1$ we have $\mathfrak m = \sqrt{(\pi )}$. By assumption we may write $P'(\alpha )^{-1} = \pi ^{-m} a$ for some $m \geq 0$ and $a \in A$. We may and do assume that $c \geq m + 1$. Pick $n$ such that $\mathfrak m_ A^ n \subset (\pi ^{c + m})$. Pick any $Q$ as in the statement. For later use we observe that we can write

for some $R(x, y) \in A[x, y]$. We will show by induction that we can find a sequence $\alpha _ m, \alpha _{m + 1}, \alpha _{m + 2}, \ldots $ such that

1. $\alpha _ k \equiv \alpha \bmod \pi ^ c$,

2. $\alpha _{k + 1} - \alpha _ k \in (\pi ^ k)$, and

3. $(P + Q)(\alpha _ k) \in (\pi ^{m + k})$.

Setting $\beta = \mathop{\mathrm{lim}}\nolimits \alpha _ k$ will finish the proof.

Base case. Since the coefficients of $Q$ are in $(\pi ^{c + m})$ we have $(P + Q)(\alpha ) \in (\pi ^{c + m})$. Hence $\alpha _ m = \alpha $ works. This choice guarantees that $\alpha _ k \equiv \alpha \bmod \pi ^ c$ for all $k \geq m$.

Induction step. Given $\alpha _ k$ we write $\alpha _{k + 1} = \alpha _ k + \delta $ for some $\delta \in (\pi ^ k)$. Then we have

Because the coefficients of $Q$ are in $(\pi ^{c + m})$ we see that $Q(\alpha _ k + \delta ) \equiv Q(\alpha _ k) \bmod \pi ^{c + m + k}$. On the other hand we have

Note that $P'(\alpha _ k) \equiv P'(\alpha ) \bmod (\pi ^{m + 1})$ as $\alpha _ k \equiv \alpha \bmod \pi ^{m + 1}$. Hence we obtain

Recombining the two terms we see that

Thus a solution is to take $\delta = -P'(\alpha )^{-1} (P + Q)(\alpha _ k) = - \pi ^{-m} a (P + Q)(\alpha _ k)$ which is contained in $(\pi ^ k)$ by induction assumption. $\square$

Proof. Note that $A^\wedge $ is a discrete valuation ring too (by Lemmas 15.43.4 and 15.43.1). In particular $A^\wedge $ is a domain. The proof will work more generally for Noetherian local rings $A$ such that $A^\wedge $ is a local domain of dimension $1$.

Let $\theta \in M$ be an element that generates $M$ over $K^\wedge $. (Theorem of the primitive element.) Let $P(t) \in K^\wedge [t]$ be the minimal polynomial of $\theta $ over $K^\wedge $. Let $\pi \in \mathfrak m_ A$ be a nonzero element. After replacing $\theta $ by $\pi ^ n\theta $ we may assume that the coefficients of $P(t)$ are in $A^\wedge $. Let $B = A^\wedge [\theta ] = A^\wedge [t]/(P(t))$. Note that $B$ is a complete local domain of dimension $1$ because it is finite over $A$ and contained in $M$. Since $M$ is separable over $K$ the element $\theta $ is not a root of the derivative of $P$. For any integer $n$ we can find a monic polynomial $P_1 \in A[t]$ such that $P - P_1$ has coefficients in $\pi ^ nA^\wedge [t]$. By Krasner's lemma (Lemma 15.113.1) we see that $P_1$ has a root $\beta $ in $B$ for $n$ sufficiently large. Moreover, we may assume (if $n$ is chosen large enough) that $\theta - \beta \in \pi B$. Consider the map $\Phi : A^\wedge [t]/(P_1) \to B$ of $A^\wedge $-algebras which maps $t$ to $\beta $. Since $B = \pi B + \sum _{i < \deg (P)} A^\wedge \theta ^ i$, the map $\Phi $ is surjective by Nakayama's lemma. As $\deg (P_1) = \deg (P)$ it follows that $\Phi $ is an isomorphism. We conclude that the ring extension $L = K[t]/(P_1(t))$ satisfies $K^\wedge \otimes _ K L \cong M$. This implies that $L$ is a field and the proof is complete. $\square$