Showing that 2 does not split completely in a number field of degree greater than 2

Question

Let $K$ be a number field with $[K:\mathbb{Q}]>2$, and such that the ring of algebraic integers $O_K=\mathbb{Z}[\alpha]$.

I wish to show that 2 cannot split completely, that is, that it is not possible to write $$(2)=P_1P_2...P_n$$ with the $P_i$ distinct prime ideals, $N(P_i)=2$.

So, we have that $K$ contains $\mathbb{Q}(\alpha)$. Let $f_{\alpha}$ be the minimal polynomial of $\alpha$ over $\mathbb{Q}$. Applying Dedekind's criterion to 2 (which may be done since $[\mathbb{O_K}:\mathbb{Z}[\alpha]]=1$, 2 splits completely if and only if modulo 2, $$f_{\alpha} \equiv \prod_{i}g_i(t)^{k_i}$$ so that $$(2)=\prod(2,g_i(\alpha))^{k_i}$$ with $k_i=1$ for all $i$ and $N(2,g_i(\alpha))$=2.

This is where I get stuck. I know I can assume that $f_{\alpha}$ has at most 2 linear factors mod 2, as if it has more than this then one is repeated and we have ramification. $f_\alpha$ has degree at least 3 since $[K:\mathbb{Q}]\geq 3$. So, one of the $g_i$ can be assumed to have degree at least 2. Presumably this should contradict the condition on the norms, but I do not understand how.

EDIT: makes absolutely clear that the condition $O_K=\mathbb{Z}[\alpha]$ is a condition, not a general fact.

Answer 1

As Jyrki Lahtonen points out in the comments, we know that $N(2)=2^{[K:\mathbb{Q}]}=2^{\sum_ik_i}$ by the multiplicity of norms, as we assume each ideal in the factorisation has norm 2. However, we also have that $[K:\mathbb{Q}] =deg(f)=\sum_ik_ideg(g_i)$. So comparing the two expressions for $[K:\mathbb{Q}]$, we conclude that $deg(g_i)=1$ for all $i$, implying that modulo 2 $f$ has at least three linear factors, and hence a repeated factor.

Answer 2

This is not right. Your error is in thinking that if the minimal polynomial for $\alpha$ over $\Bbb Q$ has repeated factors modulo $p$, then $p$ must be ramified in $\Bbb Q(\alpha)$. Counterexample: the minimal polynomial for $\sqrt{-7}$ is $X^2+7\equiv(X+1)^2\pmod2$. But $2$ splits in $\Bbb Q(\sqrt{-7}\,)$. Note the following:

Lemma. Let $K$ be a normal extension of $\Bbb Q$ that has an embedding into $\Bbb Q_p$. Then $p$ splits completely in $K$.
Proof. Let $G=G^K_{\Bbb Q}$ be the Galois group, and let $\sigma:K\to\Bbb Q_p$ be the given embedding. Then for every $g\in G$, $\sigma\circ g$ is an embedding of $K$ into $\Bbb Q_p$, and these embeddings, $|G|=[K:\Bbb Q]$ in number, are all distinct. Each induces a different prime of $K$ above $p$. Alternatively, each has $e=f=1$, unramified with trivial residue-field extension.

For an example, take $\Bbb Q(\sqrt{-7},\sqrt{-15}\,)$, $p=2$.

EDIT: In view of your correction, I see that I have answered the claim of the title, modified in your first paragraph.