Proof about Number Fields

Question

It is a known result that if $\alpha$ is an algebraic integer in a number field $K$, i.e. $\alpha \in \mathcal{O}_K$, then its trace and norm are integers.

I am looking over a proof of this, which proceeds as follows:

1. Assume $\alpha \in \mathcal{O}_K$, and let its $K$-conjugates be $\alpha_1,...,\alpha_n$
2. Let $L$ be the splitting field of the minimum polynomial of $\alpha$, say $m_\alpha$
3. Then clearly the trace and norm of $\alpha$ are in $\mathcal{O}_L$
4. The trace and norm of $\alpha$ are in $\mathbb{Q}$
5. $\mathbb{Q} \cap \mathcal{O}_L=\mathbb{Z}$
6. The trace and norm of $\alpha$ are in $\mathbb{Z}$

My problem is with claim 5. Consider the following:

Define $K=\mathbb{Q}(\sqrt2)$, $\alpha=\sqrt3$. Then, $L=\mathbb{Q}(\sqrt2,\sqrt3)$. Now, $\frac{1}{2} \in \mathbb{Q}$ and $\frac{1}{2} \in \mathcal{O}_L$ (consider the monic $2x-1$), but $\frac{1}{2} \notin \mathbb{Z}$, so 5 doesn't hold.

Can anyone point out where I'm going wrong here...

EDIT

It has been pointed out to me (and is very obvious) that since $2x-1$ is not in fact monic, my question is a little senseless. Instead then, can anyone answer the following:

Why is $\mathbb{Q} \cap \mathcal{O}_L=\mathbb{Z}$?

Answer 1

Let $f(x)=x^n + a_{n-1}x^{n-1}+\dots +a_0\in\mathbb Z[x]$ be an integer monic polynomial with $\frac{p}{q}$ as a rational root, with $p,q$ relatively prime.

Then $$q^nf(p/q)=p^n + a_{n-1}p^{n-1}q\dots + a_0q^n=0$$

So $p^n$ must be divisible by $q$. Since $p,q$ relatively prime, this means that $q=\pm 1$, and therefore $p/q$ is an integer.

Answer 2

$$\Bbb Q\cap\mathcal O(L)=\Bbb Z\iff$$ every rational number integral (over the integers, of course) is an integer...but we know $\;\Bbb Z\;$ is integrally closed, so...