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 $L$?
[Math] Primes that ramify in a field
algebraic-number-theorynumber theory
Related Solutions
Suppose that $K$ is ramified at $p_1,\ldots,p_n$. Then certainly $K \subset \mathbb Q(\zeta_{p_1^{m_1}},\ldots,\zeta_{p_n^{m_n}})$ for some $m_1,\ldots,m_n$.
Now you could try to pin down the field precisely by using some combination of information about the discriminant and the Galois group of $K/\mathbb Q$.
For example, suppose that our abelian extension $K$ is ramified at a single odd prime $p$. Then $Gal(\mathbb Q_{\zeta_{p^m}})/\mathbb Q) = (\mathbb Z/p^m\mathbb Z)^{\times}$ is cyclic; in fact it is a product of the cyclic groups $(1 + p\mathbb Z_p)/(1+p^m\mathbb Z_p)$ (of order $p^{m-1}$) and $(\mathbb Z/p\mathbb Z)^{\times}$ (of order $p-1$).
Any quotient is then equal to $(1+p\mathbb Z_p)/(1+p^{m'}\mathbb Z_p)$ (for $1 \leq m'\leq m$) and some quotient $H$ of $(\mathbb Z/p\mathbb Z)^{\times}$.
Thus if we choose $m$ minimally in the first place, the Galois group of $K$ over $\mathbb Q$ (which is a quotient of $Gal(\mathbb Q_{\zeta_{p^m}}/ \mathbb Q)$) is isomorphic to $(1 + p\mathbb Z_p)/(1+p^m\mathbb Z_p)\times H$, for a quotient $H$ as above.
The problem that you asked about is to compute $m$ given $K$. In this case we see that it is just a matter of computing the power of $p$ dividing the order of $Gal(K/\mathbb Q)$: the number $m$ is one more than the power of $p$ dividing this order.
As a slightly more complicated example, suppose that $K$ is ramified at two odd primes $p < q$, so that $K \subset \mathbb Q(\zeta_{p^m},\zeta_{q^n})$. Again, let's choose $m$ and $n$ minimally. Then $Gal(K/\mathbb Q)$ is a quotient of $$(\mathbb Z/p^m\mathbb Z)^{\times} \times (\mathbb Z/q^n\mathbb Z)^{\times}$$ $$ = (\mathbb Z/p\mathbb Z)^{\times} \times (1 + p\mathbb Z_p)/(1+p^m\mathbb Z_p) \times (\mathbb Z/q\mathbb Z)^{\times} \times (1+q\mathbb Z_q)/(1+q^n\mathbb Z_q)^{\times} .$$
Now we see that it makes a difference whether or not $p | q - 1$. More precisely, suppose that $p^e$ is the precise power of $p$ that divides $q-1$. Then (thinking about the possible Sylow subgroup structures of a quotient of the above product, and the fact that $m$ and $n$ were chosen minimally), we see that $n$ is one more than the power of $q$ dividing the order of $Gal(K/\mathbb Q)$. On the other hand, if $m'$ is the power of $p$ dividing this order, then what we see is that $m-1 \leq m' \leq m-1+e$, so if $e \geq 1$, then we have not pinned down $m$ precisely just by knowing $m'$.
However, if we now apply the conductor-discriminant formula, one will find that knowledge of the discriminant of $K$ should allow us to determine $m$. The point is that the contribution to $m'$ that is coming from the $(\mathbb Z/q\mathbb Z)^{\times}$ factor will lead to additional powers of $q$ in the discriminant.
As a more precise example, suppose that $e = 1$, and write $(\mathbb Z/q\mathbb Z)^{\times} = H \times H',$ where $H$ has order prime-to-$p$ and $H'$ has order $p$. Consider two possibilities for $Gal(K/Q)$. In the first case, suppose that $Gal(K/Q) = (\mathbb Z/p^2\mathbb Z)^{\times} \times H$, while in the second case suppose that $Gal(K/Q) = (\mathbb Z/p\mathbb Z)^{\times}\times (\mathbb Z/q\mathbb Z)^{\times}$. In each case $Gal(K/Q)$ is the product of a cyclic group of order $(p-1)$ and a cyclic group of order $(q-1)$.
But in the first case the discriminant (up to sign) will be $p^{2p^2 - 3p} q^{(q-p-1)/p}$, while in the second case it will be $p^{p-2}q^{q-2}$.
And in the first case we have that $K \subset \mathbb Q(\zeta_{p^2},\zeta_q)$ (but no smaller cyclotomic field), while in the second case $K = \mathbb Q(\zeta_p,\zeta_q)$.
I didn't try to work this method out more systematically, but presumably one can, and certainly it shouldn't be so bad to apply by hand to any particular small example that you have in mind.
At least when working theoretically, or by hand, it is probably easiest to solve this sort of question locally. (I don't have any feeling for how computationally implemented algorithms for this sort of question work, and so won't comment any more on them.)
In short, fix a prime $v$ of $K$, and consider the completion $K_v$ and the local extension $K_v[\sqrt{d}]/K_v$. The discriminant of $K[\sqrt{d}]/K$ will be a product of the local discriminants, and so we are reduced to computing these.
How do we analyze $K_v[\sqrt{d}]/K_v$? Well, you could first try to compute the finite index subgroup $(K_v^{\times})^2$ of $K_v^{\times}$. Note that $K_v^{\times} = \mathcal O_{K_v}^{\times} \times \pi^{\mathbb Z}$, if $\pi$ is some choice of uniformizer, and so $(K_v^{\times})^2 = (\mathcal O_{K_v}^{\times})^2\times \pi^{2\mathbb Z}$.
Computing $(\mathcal O_{K_v}^{\times})^2\subset \mathcal O_{K_v}^{\times}$ is straightforward. E.g. if $v$ has odd residue characteristic, then any element in $1 + \pi \mathcal O_{K_v}$ is a square, and so it is just a question of computing the subgroup of squares in the residue field at $v$. If $v$ has even residue characteristic, then it is still pretty easy in any particular case to compute the square units, say using the fact that any unit congruent to $1 \bmod 4\pi$ is a square (as one sees using the binomial expansion for $(1+x)^{1/2}$).
Once you've done this step, you can easily check if $d$ is a square in $K_v$, which tells you whether or not $K[\sqrt{d}]/K$ is split at $v$.
To compute the discriminant is not that much harder. If $v$ has odd residue characteristic, then we may divide $d$ through by even powers of $\pi$ so that it is either a unit, or else is exactly divisible by $\pi$. In the first case, $K_v[\sqrt{d}]/K_v$ will be unramified at $v$, and hence the contribution to the discriminant from $v$ will be $1$. In the second case, $K_v[\sqrt{d}]/K_v$ will be tamely ramified at $v$, and the contribution to the discriminant from $v$ will be one power of $v$.
If $v$ is of even residue characteristic, then the computations are more involved, because even if $d$ (perhaps after dividing through by even powers of $\pi$) is a unit, it can still happen that $K_v[\sqrt{d}]/K_v$ is ramified (consider the case $\mathbb Q_2(i)/\mathbb Q_2$), although it need not be (consider the case $\mathbb Q_2(\sqrt{5})/\mathbb Q_2$).
If you would like more details, I can provide them.
Best Answer
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 \begin{equation} p-1=e_p(\mathbb{Q}(\zeta_p):\mathbb{Q})=e_p(\mathbb{Q}(\zeta_p):L)e_p(L:\mathbb{Q}) \end{equation} 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
\begin{equation} e_p(KL:L)e_p(L:\mathbb{Q})=e_p(KL:\mathbb{Q})=e_p(KL:K)e_p(K:\mathbb{Q}) \end{equation}
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.