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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).

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I don't have answer, but here is an example of an extreme situation. Let $K$ be an number field of class number 2. Then its Hilbert class field is a quadratic extension of $K$ where the relative discriminant $D_{H/K}$ is the unit ideal. A search on Wikipedia about Hilbert class field gives the example $K=\mathbb{Q}(\sqrt{-15})$ and $H=\mathbb{Q}(\sqrt{-3}, \sqrt{5})$. –  Jiangwei Xue May 9 '11 at 10:41
    
@Arturo : Thank you for the fix. @Jiangwei : so it might be hopeless to find an answer depending on $K$ and $d$... Can we say something if we suppose that $O_K$ is factorial ? –  user10676 May 9 '11 at 15:08
    
@user10676: Second pings don't usually work; I would suggest moving the second part of your comment to a new comment if you want to catch Jianwei's attention. –  Arturo Magidin May 9 '11 at 15:09
    
@user01676, assuming $\mathcal{O}_K$ will certainly make things easier. However, please know that algorithms that computes the ring of integers and discriminants are well established. You may check out chapter 6 of "A course in computational number theory" by H. Cohen. (I must admit that I am not familiar with the algorithm). Such algorithms are implemented in Pari/Gp, which you can download for free. –  Jiangwei Xue May 9 '11 at 16:57
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The title is "A Course in Computational Algebraic Number Theory" –  Plop May 9 '11 at 17:56

1 Answer 1

up vote 7 down vote accepted

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.

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So what are the more details? I am rather intrigued. Thanks for any information. –  awllower Feb 2 '13 at 4:34

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