Is ramification of number fields first order?

Fix a prime number $p$. Is there a first order sentence $\phi_p$ in the language of fields such that $\phi_p$ holds in a number field $K$ if and only if the prime $p$ is unramified in the field extension $K/\mathbb{Q}$?

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Could you provide a bit more explanation for those of us who are not number-theory experts? – Joel David Hamkins Mar 20 '12 at 14:43
Hi Joel. There are many intuitions about what ramification means, but here is the easiest definition. Let $O_K$ be the integral closure of $\mathbb{Z}$ in $K$. Then, since $O_K$ is a Dedekind domain, you can write $pO_K$ as a product $\mathfrak{p}_1^{e_1}...\mathfrak{p_n}^{e_n}$ where the $\mathfrak{p_i}$'s are different prime ideals of $O_K$. The prime $p$ is said to be unramified if $e_1=...=e_n=1$ (i.e. if no prime ideal appears more than once in the decomposition of $pO_K$). The most intuitive kind of ramification is by taking roots of $p$ (e.g. $\mathbb{Q}(\sqrt{2})$ ramifies at $2$), but – Makhalan Duff Mar 20 '12 at 14:49
ramification can appear in more subtle ways, like this: the prime $2$ ramifies in $\mathbb{Q}(i)$. – Makhalan Duff Mar 20 '12 at 14:49
(in scheme theory, this number-theoretic notion of ramification becomes the same as the geometric notion of ramification of maps between varieties.) – Makhalan Duff Mar 20 '12 at 14:53
I am far from an expert in logic, but my gut feeling is that the algorithm on p.28 of the pdf here math.uconn.edu/~kconrad/math5230f08/schoof.pdf could yield a first order statement because $p$ ramifies in $K$ if and only if $p$ divides the discriminant of $\mathcal O_K$ as a $\mathbb Z$-algebra. – stankewicz Mar 20 '12 at 17:08