Remark 15.114.1. Let $A \to B$ be an extension of discrete valuation rings with fraction fields $K \subset L$. Let $K_1/K$ be a finite extension of fields. Let $A_1 \subset K_1$ be the integral closure of $A$ in $K_1$. On the other hand, let $L_1 = (L \otimes _ K K_1)_{red}$. Then $L_1$ is a nonempty finite product of finite field extensions of $L$. Let $B_1$ be the integral closure of $B$ in $L_1$. We obtain compatible commutative diagrams

In this situation we have the following

By Algebra, Lemma 10.120.18 the ring $A_1$ is a Dedekind domain and $B_1$ is a finite product of Dedekind domains.

Note that $L \otimes _ K K_1 = (B \otimes _ A A_1)_\pi $ where $\pi \in A$ is a uniformizer and that $\pi $ is a nonzerodivisor on $B \otimes _ A A_1$. Thus the ring map $B \otimes _ A A_1 \to B_1$ is integral with kernel consisting of nilpotent elements. Hence $\mathop{\mathrm{Spec}}(B_1) \to \mathop{\mathrm{Spec}}(B \otimes _ A A_1)$ is surjective on spectra (Algebra, Lemma 10.36.17). The map $\mathop{\mathrm{Spec}}(B \otimes _ A A_1) \to \mathop{\mathrm{Spec}}(A_1)$ is surjective as $A_1/\mathfrak m_ A A_1 \to B/\mathfrak m_ AB \otimes _{\kappa _ A} A_1/\mathfrak m_ A A_1$ is an injective ring map with $A_1/\mathfrak m_ A A_1$ Artinian. We conclude that $\mathop{\mathrm{Spec}}(B_1) \to \mathop{\mathrm{Spec}}(A_1)$ is surjective.

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

\[ \xymatrix{ B \ar[r] & (B_1)_{\mathfrak m_{ij}} \\ A \ar[u] \ar[r] & (A_1)_{\mathfrak m_ i} \ar[u] } \]of extensions of discrete valuation rings.

If $A$ is henselian (for example complete), then $A_1$ is a discrete valuation ring, i.e., $n = 1$. Namely, $A_1$ is a union of finite extensions of $A$ which are domains, hence local by Algebra, Lemma 10.153.4.

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

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

If $K \subset K_1$ is finite separable, then $L_1 = L \otimes _ K K_1$ and is a finite product of finite separable extensions too. Hence $A \subset A_1$ and $B \subset B_1$ are finite by Algebra, Lemma 10.161.8.

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

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

## Comments (2)

Comment #6628 by Junnosuke Koizumi on

Comment #6859 by Johan on