This is an exercise from my lecture notes on algebraic number theory:

Let $L = \mathbb{Q}(\sqrt{2},\sqrt{5})$ and $K = \mathbb{Q}[\sqrt{10}]$. Prove that prime ideal $\mathfrak{p} \in$ Max$(\mathbb{Z}[\sqrt{10}])$ splits completely in $L$ iff $\mathfrak{p}$ is principal.

Unfortunately I am now hopelessly stuck, and I was hoping someone could maybe give me a hint!

These are my thoughts:

$Max(\mathbb{Z}[\sqrt{10}]) = \mathcal{O}_K$. I already found out that no primes in $\mathcal{O}_K$ can ramify in $L$. So I thought that for prime numbers $p = a^2 - 10b^2$ with $a$ and $b$ integers, we would have $(p) = (a + b\sqrt{10})(a - b\sqrt{10})$. These are now principal prime ideals in $\mathcal{O}_K$. I know already that the cannot ramify.. also I have the feeling that they cannot remain prime because $\sqrt{10}$ is an irreducible element in $K$ but not in $L$. But how to proceed? And how to show that non-principal primes don't split in $L$? I also found out the the class group group $L$ is cyclic or order 2, but I don't know how to use that fact. And also I'm really wondering what principality of a prime has to do with its splitting behaviour..

Many thanks!



Principality has to do with splitting behaviour in unramified abelian extensions; this is called class field theory. In special cases such as yours, this can be made explicit at a very elementary level.

In your case, prime ideal above primes $p$ with $(2/p) = (5/p) = +1$ split in $L/K$. So you want to show that these primes generate principal ideals, and that primes with $(2/p) = (5/p) = -1$ generate nonprincipal ideals in $K$. This is genus theory, but follows from the class group being generated by the prime ideal above $2$. In your case, solvability of $x^2 - 10y^2 = p$ is equivalent to the prime ideal above $p$ being principal, and solvability of $x^2 - 10y^2 = 2p$ (or $2X^2 - 5y^2 = p$) implies the prime ideal is non-principal.

[edit] So now you're dealing with these diophantine equations. The class group calculation shows that for primes with $(10/p) = +1$, exactly one of them must be solvable. Now $2x^2 - 5y^2 = p$ implies $(p/5) = (2/5) = -1$, so this equation is only solvable if $(2/p) = (5/p) = -1$ by quadratic reciprocity. On the other hand, solvability of $x^2 - 10y^2 = p$ implies $(p/5) = +1$, so $(2/p) = (5/p) = +1$ by quadratic reciprocity.

  • $\begingroup$ Thank you very much Franz! I understand your first part. (also, when $(2/p) \neq (5/p) = \pm 1$, we have that $p$ inert in $K$, and thus is principal and splits. I tried to use reciprocity for $(2/p) = (5/p) = 1$ to obtain solvability for $x^2 - 10y^2 = p$ (1). I got that $(5/p) = (p/5)$, and then modulo 5 indeed (1) has a solution. But how would one go the other way around? And does solvability of $(1)$ imply non-solvability of $x^2 - 10y^2 = 2p$? We have only done a couple of mordell-weil equations in class! Thanks again! $\endgroup$ – Willem Beek Jan 23 '16 at 14:07
  • $\begingroup$ thank you very much!! (somewhere i think 19 should be 10) $\endgroup$ – Willem Beek Jan 25 '16 at 9:45

There are various ways to approach this problem, but from a class field theoretic perspective, the problem is essentially equivalent to the fact that the class number of $\mathbb Q(\sqrt{10})$ is two. So one can use this fact to solve the problem. It is somewhat involved, though.

Let $\mathfrak p$ denote the prime ideal $(2,\sqrt{10}).$ This is a non-pricipal ideal of norm $2$. Now what you need to verify is that if $\mathfrak q$ is any non-principal prime ideal, then $\mathfrak p \mathfrak q$ is principal. (You can do this in various ways: one way is to choose an element of $\mathfrak q$ of least norm. This doesn't generate all of $\mathfrak q$, since the latter is not principal; but, by employing some kind of division algorithm, you can prove that it generates an ideal of index $2$ in $\mathfrak q$, which has to be $\mathfrak p \mathfrak q$. Or, if you know the theory of class groups, you can just prove that the class number is two.)

Now suppose that $q$ splits in $\mathbb Z[\sqrt{10}],$ but non-principally, i.e. $q = \mathfrak q \mathfrak q'$ with $\mathfrak q$ and $\mathfrak q'$ non-principal. Then $\mathfrak p \mathfrak q$ is principal, say equal to $(\alpha)$. We see that $q = N(\mathfrak q) = N(\alpha)/N(\mathfrak p)= N(\alpha)/2.$

Now it's not hard to see that $\langle 2,\sqrt{10}\rangle$ actually forms a basis for the ideal $\mathfrak p$ as a $\mathbb Z$-module, and so we may write $\alpha = 2x + \sqrt{10}y.$ We then have $N(\alpha)/2 = 2 x^2 - 5y^2$, and thus we find that $q = 2 x^2 - 5y^2$.

In conclusion, if $q$ splits principally we may write $q = x^2 - 10 y^2$, while if $q$ splits non-principally we may write $q = 2 x^2 - 5 y^2$, which is what you needed.

  • $\begingroup$ thank you very very much tracing for your explanation! unfortunately I can accept only one answer, and I just chose to just accept the first. $\endgroup$ – Willem Beek Jan 25 '16 at 9:46
  • 1
    $\begingroup$ You're welcome, and no worries re the green check mark. This is the kind of concrete application of algebraic number theory which historically was one of the driving engines for developing the theory, but which is surprisingly hard to find discussed in contemporary accounts. If you haven't read it, H. Cohn's Advanced number theory is one text that does discuss the relationship between class groups and quadratic forms, and which you might want to look at. $\endgroup$ – tracing Jan 25 '16 at 12:01

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