# When does a dyadic prime ramify in a relative quadratic extension?

In a quadratic extension $\mathbb{Q}(\sqrt{d})$of $\mathbb{Q}$ it is clear that 2 ramifies if and only if $d\equiv 2,3\mod 4$ (easy to see if you compute the discriminant). But if I take a relative quadratic extension, to make it simple let's say $L=K(\sqrt{\pi})$ where $K$ is an arbitrary number field and $\pi$ is an irreducible element of the ring of integers of $K$, when does a dyadic prime ramify from $K$ to $L$?

Certainly $\pi$ itself ramifies in this case, that's easy to see without knowing the discriminant or even ring of integers. But how do I know when primes above $2$ are ramifying, i.e. being contributed to the discriminant?

In general, I know that computing the discriminant (or ring of integers) of such a field, is computationally complex. But is there a known congruence condition on $\pi$ or some algebraic way to decide when primes above 2 ramify?

• $L=K(\sqrt{\mathfrak{p}})$ makes no sense. There is no such thing as extending by the "square-root of an ideal" (unless you define it here). You can extend by the square-root of an element or, more generally, by the root of a polynomial with coefficients lying in the base field. So please clarify. – GH from MO Mar 16 '14 at 11:26
• Thanks, right, I just want $\mathfrak{p}$ to be a prime element in the base field $K$. – user48331 Mar 16 '14 at 11:41
• @user48331 Probably you mean that $\pi$ is an irreducible element of the ring of integers of $K$, and you're looking at the ring of integers of $K(\sqrt{\pi})$. But the set of irreducible elements is not so nice in general. So I think that your attempted simplification is actually confusing the issue. Why not just ask: Is there a congruence condition on $\delta\in O_K$ that determines whether primes above 2 are ramified in $O_{K(\sqrt{\delta})}$? – Joe Silverman Mar 16 '14 at 13:09
• This is a purely local question, easily answered, but someone else will do it before I get back — gotta run. – Lubin Mar 16 '14 at 13:11

## 2 Answers

There's an appendix on Kummer theory to Milne's class field theory notes (available here) that gives an answer to this question (EDIT: in some cases). Assuming that $K$ contains a primitive $p$th root of unity $\zeta$, $\pi = (\zeta - 1)$ and $a \in K^{\times}$ is such that

$\bullet$ $a$ is not a $p$th power in $K$,

$\bullet$ $a$ is relatively prime to $p$, and

$\bullet$ $X^{p} \equiv a \pmod{p \pi}$ has a nonzero solution $X \in \mathcal{O}_{K}$,

then none of the primes $\mathfrak{p}$ above $p$ ramify in $K[a^{1/p}]/K$.(This is Remark A.6 on page 223.)

Milne doesn't give a proof, instead referring to exercises in Washington's Cyclotomic Fields text (pages 182 and 183 there), and Cassels and Frohlich.

• Many thanks, that is precisely what I was looking for. – user48331 Mar 17 '14 at 9:22
• No hypothesis on $a$ being a unit? After all, with $a=8$, $\mathbb Q(\sqrt8)$ is ramified over $\mathbb Q$ but $x^2\equiv8\pmod4$ has a perfectly good solution in $\mathbb Q$, namely zero. – Lubin Mar 17 '14 at 12:56
• The Milne notes do specify that the solution $x$ must be nonzero. But in general this is another good question. What happens when $a$ is even? – user48331 Mar 17 '14 at 14:17
• There were more hypotheses from Milne that I missed - my answer has been edited to be correct. (The version is somewhat less helpful, however.) – Jeremy Rouse Mar 17 '14 at 17:36
• Some care is needed: $\mathbb{Q}(\sqrt{20})/\mathbb{Q}$ is unramified at $2$. Given $a \in K_{\mathfrak{p}}$, one can write $a = \pi^{k} a'$, where $\pi$ is a generator of the unique maximal ideal of $\mathcal{O}_{K_{\mathfrak{p}}}$ and $a' \in \mathcal{O}_{K_{\mathfrak{p}}}^{\times}$. If $k$ is odd, then $K_{\mathfrak{p}}(\sqrt{a})$ is ramified. If $k$ is even, then $K_{\mathfrak{p}}(\sqrt{a}) = K_{\mathfrak{p}}(\sqrt{a'})$ (reducing to an "odd" value of $a$). – Jeremy Rouse Mar 18 '14 at 12:09

While I was away I realized that I didn’t have a complete answer to this question. But the key to finding whether $K(\sqrt\delta)\supset K$ is ramified over $2$ is to take a prime $\mathfrak q$ of $K$ above $2$, and consider $V=v_{\mathfrak q}$, the (additive) valuation at $\mathfrak q$ and $V(\delta-1)$. I’ll normalize $V$ so that $V(2)=1$.

Now, in view of the fact that the binomial expansion $(1+4t)^{1/2}$ has its coefficients in $\mathbb Z$, you see that if $V(\delta-1)>2$, the prime $\mathfrak q$ splits in the extension. Similarly, if $\delta=1+4u$ for a $\mathfrak q$-unit $u$, you see directly that the extension is unramified: may or may not split.

The same kind of computation shows that if $V(\delta-1)<2$, $V(\sqrt\delta-1)=\frac12V(\delta-1)$. In case $V(\delta)=m/e$ with $m$ odd, and where the local ramification is $e$, you’re home free: the extension is definitely ramified. But I don’t see offhand how you can exclude the possibility that $m$ is even, and there I’m not so sure.