Suppose that the prime $p$ is totally ramified in $O_{K_1}$ and unramified in $O_{K_2}$. Prove that $K_1 \cap K_2=\Bbb Q$.

Question

Let $K_1$ and $K_2$ be algebraic number fields. Suppose that the prime $p$ is totally ramified in $O_{K_1}$ and unramified in $O_{K_2}$. Prove that $K_1 \cap K_2=\Bbb Q$.

For unramified $<p>=P_1^{e_1}$ where all $e_1=[K_1:\Bbb Q]$ in $K_1$ and $<p>=Q_1\cdots Q_s$ in $K_2$. Do we get anything from here? You can downvote me but pls give some hint!

Edit: After putting some thought and reading comments(thanks to Jyrki) I thinjk that I have to consider the tower

 $ K_1K_2
  /     \
 /       \ 
K_1       K_2
\         /
 \       /
   K_1\cap K_2
       |
       |
      \Bbb Q$

Then we have to consider some fixed field...

Answer 1

This follows from the definitions, basic properties of the splitting of primes in an extension, in particular the multiplicativity of the ramification exponent.

Denote $M=K_1\cap K_2$. All we need is the tower $\Bbb{Q}\subseteq M\subseteq K_1$.

• Because $p$ is totally ramified in $K_1/\Bbb{Q}$, there exists a unique prime ideal $\mathfrak{P}$ of $\mathcal{O}_{K_1}$ lying above $p$. Furthermore, $$p\mathcal{O}_{K_1}=\mathfrak{P}^e,$$ where $e=n=[K_1:\Bbb{Q}]$.
• Let us consider the prime ideal $\mathfrak{p}=M\cap \mathfrak{P}.$ The multiplicativity of the ramification exponent implies that $$e(\mathfrak{P}|p)=e(\mathfrak{P}|\mathfrak{p})\cdot e(\mathfrak{p}|p).$$ Here $e(\mathfrak{P}|\mathfrak{p})\le [K_1:M]$ and $e(\mathfrak{p}|p)\le [M:\Bbb{Q}]$. As $[K_1:\Bbb{Q}]=[K_1:M]\cdot [M:\Bbb{Q}]$, we must have equality in both cases, for otherwise $e(\mathfrak{P}|p)$ would be less than $n$.
• So if $[M:\Bbb{Q}]=m>1$ we would see $\mathfrak{p}^m$ in the decomposition of $p$ in $\mathcal{O}_M$. In other words, $p$ ramifies in $M/\Bbb{Q}$.
• But, $M$ is also a subfield of $K_2$. If $p$ ramified in $M/\Bbb{Q}$ it would also ramify in $K_2/\Bbb{Q}$ (again by multiplicativity of $e$). We were given that this is not the case, so we can conclude that $m=1$.
• Hence $K_1\cap K_2=M=\Bbb{Q}$.

Parts of the argument generalize to the following two maxims:

If a prime ideal $\mathfrak{p}$ is totally ramified in a finite extension of number fields $L/K$, then it is (totally) ramified also in all the intermediate extensions $L/M, M/K$ where $K\subseteq M\subseteq L$.

If a prime ideal $\mathfrak{p}$ is unramified in a finite extension of number fields $L/K$, then all primes lying above it are unramified in all the intermediate extensions $L/M, M/K$ where $K\subseteq M\subseteq L$.