# Proof about Number Fields

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

• Well, $2x-1$ is not a monic polynomial. It has leading coefficient $2$, not $1$. Commented Apr 18, 2014 at 17:02
• Well that's embarrassing... Ignoring my shoddy thought process initially, can you explain why 5. holds? (I've made an edit to make it clear that this is what I want to know) Commented Apr 18, 2014 at 17:05
• Note (elementarily) that if $0\neq f\in\mathbb Z[X]$ and $m/n\in\mathbb Q$ is irreducible, then $m$ divides $f(0)$ and $n$ divides the leading coefficient of $f$. Commented Apr 18, 2014 at 17:11
• Hint $\$ It is equivalent to the *monic* special-case of the [Rational Root Test,](en.wikipedia.org/wiki/Rational_root_theorem) i.e. a root $\,r\,$ of a monic polynomal $\,f\in\Bbb Z[x]\,$ is integral if rational, i.e. $\,f(r)=0,\ r\in\Bbb Q\,\Rightarrow\,r\in\Bbb Z.\ \$ Commented Apr 18, 2014 at 17:59

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.

• Thanks, are we saying that it is true for any number field $L$ then? It seems your reasoning is independent of the extension field we are working in. Commented Apr 18, 2014 at 17:20
• Can you be more specific about what you mean by your question? If you mean whether, when $L\subset K$ is $\mathcal O_{K}\cap L=\mathcal O_L$? In general, we can't always write elements of $L$ as a fraction of relatively prime elements of $\mathcal O_L$. Commented Apr 18, 2014 at 17:25
• Yes, that's exactly what I mean. Is there a condition that ensures that is the case? Commented Apr 18, 2014 at 17:31
• Certainly, the above proof still works if $\mathcal O_L$ is a unique factorization domain. Not sure more generally, although Don's answer would indicate $O_L$ being "integrally closed" might be the condition needed. (Wikipedia's definition of "integrally closed" is relative to a containing ring, presumably $L$ in this case?) Commented Apr 18, 2014 at 18:00

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

• "We know?" I mean, isn't that what he's trying to prove, really? Commented Apr 18, 2014 at 17:11
• I think that is basic, @ThomasAndrews: it can even be proved without any algebraic number theory, only with the rational root of integral polynomials lemma. Of course,I could be wrong, so let us wait until the OP addresses these concerns. Commented Apr 18, 2014 at 17:13
• I haven't actually encountered the expression "integrally closed" before. @ThomasAndrews has posted an answer which I can understand more intuitively. Thanks Commented Apr 18, 2014 at 17:19
• ...and using the lemma I mentioned above, @Mathmo. Sometimes what's elementary for one isn't for other. Commented Apr 18, 2014 at 17:21