# Primes that ramify in a field

Consider the number field $L/\mathbb{Q}$. I know that the only primes $p$ that ramify over $L$ are the ones that divide $\Delta_{L}$, the discriminant of $L$. But what if I can't compute $\Delta_{L}$? Are there other ways to determine which primes ramify over $L$?

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If $f(x)\in \mathbb Q[x]$ is the minimal polynomial of a generator of $L$, then $L/\mathbb Q$ is unramified at all $p$ such that $f(x)\in \mathbb Z[1/p][x]$ and $\bar{f}(x)$ is separable in $\mathbb F_p[x]$. However you can miss some unramified primes this way. – user18119 Nov 30 '11 at 20:48

The answer is certainly yes, with the caveat that it depends on how you specify $L$.

Let me just give one example, that of "ramification calculus," the game of figuring out ramification indices by using the fact that such indices are multiplicative in towers of extensions. The idea here is to use your knowledge of previously-determined ramification in other fields to determine the ramification in your given field.

For example, if $L\subset \mathbb{Q}(\zeta_p)$ (for $\zeta_p$ a primitive $p$-th root of unity) is of degree $n\mid p-1$, then $p$ is totally ramified (i.e., with ramification index $n$) in $L$ just by the multiplicativity $$p-1=e_p(\mathbb{Q}(\zeta_p):\mathbb{Q})=e_p(\mathbb{Q}(\zeta_p):L)e_p(L:\mathbb{Q})$$ and the observation that $e_p(K_2:K_1)\leq [K_2:K_1]$ for any extension $K_2/K_1$. (Of course, one should really be careful about taking a prime $\mathfrak{p}$ above $p$ in $L$ and writing $e_{\mathfrak{p}}(\mathbb{Q}(\zeta_p):L)$, etc., but never mind that.)

In some sense, this particular example is sneakily replacing the "ramifies if and only if divides the discriminant" property with the "ramifies if and only if it divides the conductor" property -- in fact, this conductor approach might provide a more global "yes" to your original question. While the conductor and discriminant are certainly related, they come from different places (in some vague and not particularly rigorous sense) . In any case, note that you didn't have to compute the discriminant of $L$ for this to work -- the one-time overhead of computing the discriminant of $\mathbb{Q}(\zeta_p)$ suffices to give the ramification information for all $L\subset \mathbb{Q}(\zeta_p)$. A similar analysis can be done for $\mathbb{Q}(\zeta_n)$, so if you know $L$ is abelian, you've got a pretty good headstart.

This sort of calculus can get fancier and fancier (and honestly, is pretty fun to work out). For example, I frequently encounter the following situation: Suppose you know an extension $K/\mathbb{Q}$ such that $KL/L$ and $K/\mathbb{Q}$ have easily-determined ramification. Then the two tower-multiplicativity statements

$$e_p(KL:L)e_p(L:\mathbb{Q})=e_p(KL:\mathbb{Q})=e_p(KL:K)e_p(K:\mathbb{Q})$$

might let you solve for $e_p(L:\mathbb{Q})$, i.e., determine the ramification in $L$, from known information about ramification in $K$.

One final thought is that there are a local-global relationships for ramification, coming from (for example) that the conductor is a product of local conductors. Local information can come in handy for determining seemingly-unrelated global ramification --Abhyankar's Lemma comes to mind as a first instantiation of this.

Hope that helps.

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