Norm and Algebraic Integers I am reading a book about algebraic number theory and I'm wondering whether the following statement is true:
Let $K$ be a number field and $\mathcal{O}_K$ its algebraic integers. Let $\beta \in \mathcal{O}_K$, is it true that $N(\beta)/\beta \in \mathcal{O}_K$? ($N$ is the norm of K over $\mathbb{Q}$)
If $K$ is a Galois extension of $\mathbb{Q}$ then we have that the norm is just the product of all Galois conjugates of a given element. Thus $N(\beta)/\beta$ would be the product of all Galois conjugates of $\beta$ except the one corresponding to the identity. Since Galois conjugates of an algebraic integer are also algebraic integers, the above follows. 
Now I'm struggling with what happens when $K$ is only a separable extension. (separable always holds since we are in characteristic 0)
 A: This still works. Let $L$ be the Galois closure of $K$ over $\Bbb Q$.
Then $N(\beta)/\beta$ is a product of conjugates of $\beta$, all of which lie
in $L$ and all of which are algebraic integers. Therefore $N(\beta)/\beta\in\mathcal{O}_L$. But $N(\beta)/\beta\in K$ and $\mathcal{O}_L\cap K=\mathcal{O}_K$
so that $N(\beta)/\beta\in\mathcal{O}_K$.
A: You can see this without appealing to Galois theory. Each conjugate of an algebraic integer is an algebraic integer, because they satisfy the same polynomial. Thus, $\mathrm{N}(\beta)/\beta$ is an algebraic integer, being the product of $n-1$ algebraic integers, taking this multiplication to occur in the ring of all algebraic integers.
This will establish that the number we're talking about is in $\mathcal{O}_K$, as long as we also know that it belongs to $K$, because $\mathcal{O}_K$ is the intersection of $K$ with the set of all algebraic integers.
That last step is no problem, though, because $N(\beta)\in K$, and $K$ is a field.
This is the same argument as the one given by Angina Seng, but instead of using the ring of integers in a Galois closure of $K$, we simply use the ring of all algebraic integers.
