TITLE: Rings of algebraic integers
QUESTION [5 upvotes]: A basic question on algebraic numbers. 

If $L/K$ is a finite extension of number fields with respective rings of integers $\mathcal O_L$ and $\mathcal O_K$ then is it true that $\mathcal O_L$ is also the integral closure of $\mathcal O_K$ in $L$?  

A proof (or a counterexample) or any references will be gratefully received.

REPLY [4 votes]: The following are equivalent:

An element $x\in L$ is integral over ${\cal O}_K$
${\cal O}_K[x]$ is f.g. as an ${\cal O}_K$-module
${\cal O}_K[x]$ is f.g. as a $\Bbb Z$-module
$\Bbb Z[x]$ is f.g. as a $\Bbb Z$-module

We know $(1)\Leftrightarrow(2)$ already (or should). Second, $(2)\Rightarrow(3)$ because if $\{x^i:0\le i\le k\}$ spans ${\cal O}_K[x]$ as an ${\cal O}_K$-module and $\{c_j\}$ spans ${\cal O}_K$ as a $\Bbb Z$-module, then $\{c_jx^i\}$ spans ${\cal O}_K[x]$ as a $\Bbb Z$-module. Third, any submodule of a f.g. (torsionfree$\,\Rightarrow$) free $\Bbb Z$-module is also f.g.  so $(3)\Rightarrow(4)$, and finally, $(4)\Rightarrow(1)$ because if $x$ satisfies a monic $\Bbb Z$-coefficient polynomial then it also satisfies a monic ${\cal O}_K$-coefficient polynomial, as $\Bbb Z\subseteq{\cal O}_K$. Note $(4)\Leftrightarrow x$ is integral over $\Bbb Z$.