TITLE: Is every unramified extension of DVRs simple?
QUESTION [6 upvotes]: Let $A$ be a discrete valuation ring with maximal ideal $\mathfrak{m}$, fraction field $K$, and $L$ a finite separable extension of $K$ degree $n$, unramified w.r.t. $A$. Let $B$ be the integral closure of $A$ in $L$. Is it true that $B$ has the form $A[x]/(f)$ for some $f\in A[x]$?
Here's what I have so far:
Certainly $B$ is finite etale over $A$ of degree $n$, and if $\alpha\in B$ is a generator of $L$ over $K$ with monic minimal polynomial $f\in A[x]$, then $A[x]/(f)$ is finite flat over $A$ of degree $n$, so if $A[x]/(f)$ is etale over $A$, then the natural injection $A[x]/(f)\hookrightarrow B$ would have to be etale of rank 1, hence an isomorphism. Thus, it suffices to show that $A[x]/(f)$ is etale over $A$, or equivalently, that the image $\overline{f}$ of $f$ in $(A/\mathfrak{m})[x]$ is a separable polynomial.
Another question: is the unramified condition necessary? What's an example of a finite non-simple extension of DVR's?

REPLY [2 votes]: The answer to your question is no. You can try to do the following exercise (taken from Serre, Local Fields, Ch. III, section 6):
Define $A,B,K,L$ in the way you did above, and assume $B$ is 'completely decomposed', i.e., there are $n = [L:K]$ primes of $B$ above the prime $\mathfrak m$ of $A$. Then $B$ is of the form $A[x]$ (for some $x\in B$) if and only if $n\leq \text{Card }\overline K$, with $\overline K$ the residue field of $K$.
This also shows that the second question is irrelevant.