Let $K$ be a number field and $d \in \mathcal{O}_K \setminus \mathcal{O}_K^2$. What is the discriminant of the extension $K[\sqrt{d}]/K$ ? Do we know its ring of integers and which primes are split or not ? (I know that these questions are solved when $K=\mathbb{Q}$, but I can't find any result otherwise).
[Math] the discriminant of a quadratic extension over a number field
algebraic-number-theorynumber theory
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As far as I know there is no way of finding the discriminant without finding the full ring of integers in the process.
Indeed, once you know the ring of integers, finding the discriminant is a trivial piece of linear algebra; and conversely, if you know the discriminant in advance, that makes finding the ring of integers much easier (because once you've found enough integers to generate a subring with the right discriminant, you know you can stop). So determining the discriminant cannot be all that much easier than determining the integers.
As for how to find the integers, it is a very well-studied problem; if you're interested in such things, Henri Cohen's "A course in computational algebraic number theory" is a very good reference.
The answer to the title question is no and the answer to the body question is yes.
First, Stickelberger's theorem asserts that the discriminant $\Delta_K$ of a number field $K$ is congruent to $0, 1 \bmod 4$. So that already rules out half of the possibilities.
Second, Minkowski's bound implies that if $K$ has degree $n$ then
$$\sqrt{|\Delta_K|} \ge \left( \frac{\pi}{4} \right)^{n/2} \frac{n^n}{n!}.$$
The RHS is always strictly bigger than $1$, so the discriminant can also never be equal to $1$; that is, $\mathbb{Q}$ has no unramified extensions. Moreover it follows that number fields of a given discriminant cannot have arbitrarily high degree, and in fact there are only finitely many number fields of a given discriminant, so for particular small discriminants one can rule them out via casework.
To start, let's record the value of the Minkowski bound for some small values of $n$.
- For $n = 2$ it is about $1.57$, so a number field of degree at least $2$ has discriminant of absolute value at least $3$.
- For $n = 3$ it is about $3.13$, so a number field of degree at least $3$ has discriminant of absolute value at least $10$.
- For $n = 4$ it is about $6.58$, so a number field of degree at least $4$ has discriminant of absolute value at least $44$.
Let's also recall that for a squarefree integer $d$ the discriminant of $\mathbb{Q}(\sqrt{d})$ is $d$ if $d \equiv 1 \bmod 4$ and $4d$ otherwise. Now let's go through the smallest few discriminants in order, skipping the ones that are impossible by Stickelberger to see what Minkowski has to say about them.
- $1$: impossible by Minkowski.
- $-3$: realized uniquely by $\mathbb{Q}(\sqrt{-3})$.
- $4$: impossible by Minkowski (can only be realized by a quadratic field and isn't).
- $-4$: realized uniquely by $\mathbb{Q}(i)$.
- $5$: realized uniquely by $\mathbb{Q}(\sqrt{5})$.
- $-7$: realized uniquely by $\mathbb{Q}(\sqrt{-7})$.
- $8$: realized uniquely by $\mathbb{Q}(\sqrt{2})$.
- $-8$: realized uniquely by $\mathbb{Q}(\sqrt{-2})$.
- $9$: impossible by Minkowski (can only be realized by a quadratic field and isn't).
- $-11$: realized by $\mathbb{Q}(\sqrt{-11})$.
- $12$: realized by $\mathbb{Q}(\sqrt{3})$.
$-12$ is the first discriminant whose status I can't determine from just Stickelberger and Minkowski. If it occurs as a discriminant it must be the discriminant of a cubic field. I think it's known that in fact the smallest possible discriminant (in absolute value) of a cubic field is $-23$, realized by $\mathbb{Q}(x)/(x^3 - x - 1)$, but I don't know how to prove this.
In any case, here's a discriminant I can rule out using an additional technique: $25$ is congruent to $1 \bmod 4$ and, by the Minkowski bound, could only be the discriminant of a cubic field. However, I claim it isn't the discriminant of a cubic field, and so can't be a discriminant at all. The reason is that it's a square, which means that the corresponding cubic field is Galois with Galois group $A_3 \cong C_3$. By the Kronecker-Weber theorem it must therefore be a subfield of the cyclotomic integers $\mathbb{Q}(\zeta_n)$, and it's known that we should in fact be able to take $n = 5$. However, $\mathbb{Q}(\zeta_5)$ has no cubic subfields since it has degree $4$.
Best Answer
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.