2
$\begingroup$

Let $K$ and $L$ be number fields, namely

$$K:=\mathbb{Q}[\sqrt{pq}]\ \ \ \textrm{and}\ \ \ L:=\mathbb{Q}[\sqrt{p},\sqrt{q}]$$ where $p,q$ are rational prime numbers, with $p\equiv 1\pmod{4}$ and $q\equiv 3\pmod{4}$. I want to show that $L|K$ is unramified.

What I did:

1) let $Q$ be a prime ideal in $\mathscr{O}_K$. Let $R$ be a prime ideal in $\mathscr{O}_L$, lying above $Q$. I need to prove that the ramification index $e(R|Q)=1$. Let $n$ be the rational prime number such that $n\mathbb{Z}=R\cap\mathbb{Z}=Q\cap\mathbb{Z}$. By multiplicativity of ramification index, i get $$e(R|n\mathbb{Z})=e(R|Q)\cdot e(Q|n\mathbb{Z})$$

2) Suppose that $n$ doesn't ramify in $L$. Then $e(R|n\mathbb{Z})=1$, thus $e(R|Q)=1$, as I wanted.

3) Suppose that $n$ ramifies in $L$. Then a priori $e(R|n\mathbb{Z})=2$ or $4$. If it is $4$, again by multiplicativity of $e$, $n$ must ramify both in $\mathbb{Q}[\sqrt{p}]$ and $\mathbb{Q}[\sqrt{q}]$, but this is impossible since the related rings of integers have coprime discriminants.

4) Hence $e(R|n\mathbb{Z})=2$ and $e(R|Q)=1$ or $2$. Suppose $e(R|Q)=2$ and $e(Q|n\mathbb{Z})=1$. Then $n$ ramifies in $L$, hence $n$ divides the discriminant of $L$, which is $16p^2q^2$, hence $n$ divides also the discriminant of $K$, which is $4pq$, hence $n$ ramifies in $K$, contraddicting $e(Q|n\mathbb{Z})$.

My question is: do you think what I did is correct? In particular, consider for example point 3). I said that if $n$ ramifies in $L$ then $e(R|n\mathbb{Z})=2,4$, but actually what I know is that at least one prime ideal above $n$ has exponent grater than $1$, how can I say that all (hence also $R$) have exponent at least $2$?

$\endgroup$
  • 1
    $\begingroup$ An alternative way is to suppose $Q$ ramifies in $L$, then the multiplicativity tells you that $n\mathbb{Z}$ would ramify in $L$...but you know exactly which $n$ will ramify in $L$. It is then a short list of checks to find that $Q$ does not ramify giving contradiction. $\endgroup$ – fretty Sep 13 '13 at 22:38
2
$\begingroup$

This looks good to me. Your concern about different ramification indices is not an issue in this case because every extension in sight is Galois. cf Corollary 2.15 of http://www.math.umass.edu/~weston/cn/notes.pdf .

$\endgroup$

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.