Before we consider the prime decomposition

Question

Let $L/K$ be a number field extension. Let $I$ be a prime ideal of $O_K$. How to prove that $IO_L\neq O_L$?

It looks there should be a very fast way to see this, but I don't know how.

Answer 1

One method to prove $IO_L\neq O_L$: Consider the localization $O_{L,I}=\{a/b:a\in L,b\in O_K-I\}$, $O_{K,I}=\{a/b:a\in K,b\in O_K-I\}$. First one proves that $O_{K,I}$ is a PID. Then by an argument similar to the proof that $O_K$ is a free $\mathbb{Z}$ module, we can prove that $O_{L,I}$ is a free $O_{K,I}$ module: $O_{L,I}=O_{K,I}\alpha_1\oplus \dotsm \oplus O_{K,I}\alpha_n$. Now $IO_L=O_L$ is equivalent to $IO_{L,I}=O_{L,I}$ which implies $IO_{K,I}=O_{K,I}$, because $IO_{L,I}=IO_{K,I}\alpha_1\oplus \dotsm \oplus IO_{K,I}\alpha_n$. But $IO_{K,I}=O_{K,I}$ is the same as $IO_K=O_K$. Done.