Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $B/A$ be a finite integral extension of discrete valuation rings of characteristic zero.

Question. Is there a uniformizer $\pi$ in $B$ such that $A[\pi]$ is a discrete valuation ring?

If not, can one give an explicit example? Note that to give such an example the extension of residue fields has to be INseparable. Is there an example with the extension of residue fields PURELY inseparable?

share|cite|improve this question

Suppose $A$ is complete with imperfect residue field $k$ of characteristic $p>0$. Let $a\in A$ whose class in $k$ is not a $p$-th power. Let $t$ be a uniformizing element of $A$ and let $K=\mathrm{Frac}(A)$. Consider the field extension $$L=K[X]/(X^{p^2}-t^pa).$$ It has degree $p^2$. Let $B$ be the integral closure of $A$ in $L$. It is a DVR and is finite over $A$ because $A$ is complete. Let $\pi$ be the class of $X$ in $L$. Then $(\pi^p/t)^p=a$. So $e_{B/A}\ge p$ and because $a$ is not a $p$-th power in $k$, the degree of the residue extension is $\ge p$. Therefore $e_{B/A}=p$ and $\pi$ is a uniformizing element of $B$.

As $A[\pi]\simeq A[X]/(X^{p^2}-t^pa)$, it is not integrally closed by considering the element $\pi^p/t$.

In this example $B=A[\pi, \pi^p/t]$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.