0
$\begingroup$

So I showed that $\mathbb{Q}_{p}[\theta]$ is an unramified extension of degree p, where $0=g(\theta)=\theta^{p}-\theta-1$.

But it also follows that $\mathbb{Q}_{p}[\phi]$ is an unramified extension of degree p, where $0=f(\phi)=\phi^{p}-\phi-a$ and $|a|_{p}=1$ .

So doesn't just follow that $\mathbb{Q}_{p}[\theta]=\mathbb{Q}_{p}[\phi]$ from the 1-1 correspondence of unramified extensions of a local field? Thanks

$\endgroup$
0
$\begingroup$

For every local field $K$ and natural number $n$ coprime to $K$'s residue characteristic, there is a unique unramified extension $L/K$ of degree $n$. (Have you proven this for $K=\Bbb Q_p$?)

Since $\Bbb Q_p[\theta]$ and $\Bbb Q_p[\phi]$ are both unramified extensions of degree $p$ they are equal, yes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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