# Ramification index in number fields extension

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 $2$?

• 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. – fretty Sep 13 '13 at 22:38