Prime $\mathfrak{p} \in$ Max$(\mathbb{Z}[\sqrt{10}])$ splits completely iff principal

Question

L.S.,

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!

Willem

Answer 1

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.

Answer 2

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.