1
$\begingroup$

Bear with me, this question is long. Suppose you're thinking about the field $K=\mathbb{Q}(\sqrt{-5})$, and you wonder what the ray class field of this with modulus $(2)$ is. A genie comes by and hands you the field $L=K\left(\sqrt{1+2i}\right)$. How would you verify that this is in fact the ray class field?

One thing to do is to show that all principal prime ideals generated by elements $1\mod 2$ in $K$ split in $L$. This is easy enough for primes of $\mathbb{Q}$ that are inert in $K$, but it is more difficult to do this for primes that split going from $\mathbb{Q}$ to $K$. For any single prime, e.g. $(3+2\sqrt{-5})$, it's easy to show that this splits as

$(3+2\sqrt{-5})=(3+2\sqrt{-5}, \sqrt{1+2i}+5)(3+2\sqrt{-5}, \sqrt{1+2i}+8)(3+2\sqrt{-5}, \sqrt{1+2i}-8)(3+2\sqrt{-5}, \sqrt{1+2i}-5)$.

But given any prime, you need to be able to show that it factorizes. This ends up being the same as solving $x^4-2x^2+5\equiv0\mod p$, as we did for the prime in $K$ over $29$; $x^4-2x^2+5$ factorizes $\mod29$ as $(x+5)(x+8)(x-5)(x-8)$.

This method is infeasible, and is actually sort of the end goal for me, so let's look at something else. The ray class group of $K$ with modulus $(2)$ has 4 elements; nonprincipal ideals have index 2 inside the group of ideals, and principal ideals are generated by either $\sqrt{-5}+2\alpha$ or $1+2\alpha$. So there are 2 principal ideal classes, those in the identity element of the group and those that are not but whose squares are. And we can take any nonprincipal element, say $(3, 1+\sqrt{-5})$, and square it to get $(3, 1+\sqrt{-5})^2=(2-\sqrt{-5})$, so there is some nonprincipal ideal class which, when squared, is not the identity element in this group. So we must have the ray class group cyclic of order $4$. So the Galois group $Gal(L/K)$ must be too.

Now $K(i)$ is the Hilbert class field of $K$, and so it must be inside $L$. Thus this is the unique quadratic extension of $K$ in $L$, and $L$ is some quadratic extension of $K(i)$, say $K(\sqrt{a+bi})$ with $a, b\in K$. The nontrivial element of $Gal(K(i)/K)$ extends to an element of $Gal(L/K)$ which does not fix $K(i)$, so it is a generator of $Gal(L/K)$; so the element of the Galois group sending $\sqrt{a+bi}$ to $\sqrt{a-bi}$ is a generator. It also sends $\sqrt{a-bi}$ to $-\sqrt{a+bi}$.

Now we have to determine what $a$ and $b$ are. If $\sqrt{a-bi}=(c+di)+(e+fi)\sqrt{a+bi}$, then applying the generator twice gives $-\sqrt{a-bi}=(c+di)-(e+fi)\sqrt{a+bi}$; then $c=d=0$ and $\sqrt{a-bi}=(e+fi)\sqrt{a+bi}$. And $\sqrt{a^2+b^2}=\sqrt{a+bi}\cdot\sqrt{a-bi}$ is in $L$, and $K(\sqrt{a^2+b^2})$ is a quadratic extension of $K$, so $K(\sqrt{a^2+b^2})=K(i)$, so $\sqrt{a^2+b^2}=vi$ for some $v\in K$. That is, we have $a^2+b^2+v^2=0$.

Now we need $K(\sqrt{a+bi})$ to be unramified away from $(2)$, so we look at some options. I found $a=r, b=\sqrt{-5}r, v=2r$ works or permutations, and so we get extensions $K(\sqrt{(1+\sqrt{5})r})$ or $K(\sqrt{(1+2i)r})$. Taking $r=\frac{1}{2}$ in the first or $r=1$ in the second gives the same extension of $K$, since $\left(-1+i\cdot\left(\frac{1-\sqrt{5}}{2}\right)\right)\cdot\sqrt{\dfrac{1+\sqrt{5}}{2}}=\sqrt{1+2i}$. So these are feasible extensions, and for all primes that I checked, the polynomial $x^4-2x^2+5$ factorizes as it should. But what else can I do? How do I eliminate all other possible triples $(a, b, v)$?

$\endgroup$
  • $\begingroup$ Is there a text or lecture notes that you are working from ? $\endgroup$ – Rene Schipperus Nov 13 '16 at 4:07
  • $\begingroup$ @ReneSchipperus Well, I learned all this stuff from a combination of "Primes of the Form x^2+ny^2" by David Cox, and Milne's notes on class field theory here: jmilne.org/math/CourseNotes/CFT.pdf It's a little bit more of a question my advisor asked. $\endgroup$ – Bob Jones Nov 13 '16 at 4:17
  • $\begingroup$ Verify that $L/{\mathbb Q}$ is normal and check that the Galois group is dihedral. This allows you to choose $a$ and $b$ from the rational integers. $\endgroup$ – franz lemmermeyer Nov 13 '16 at 19:14
  • $\begingroup$ @franzlemmermeyer Sorry, but I'm not sure I see why any of those three assertions are true. If we assume that $L/\mathbb{Q}$ is Galois, then L contains 3 quadratic extensions of $\mathbb{Q}$ and a degree 4, so it must be either $\mathbb{Z}/2\times\mathbb{Z}/4$ or $D_8$, but I'm not sure why it can't be abelian. And why does that allow us to choose $a,b$ to be in $\mathbb{Z}$? Thanks for your help. $\endgroup$ – Bob Jones Nov 15 '16 at 23:00
  • $\begingroup$ The Artin isomorphism respects the Galois action, so the automorphism $\tau$ of the base extension acts on the Artin symbol $\sigma$, and you can compute the commutator of these groups rather easily. $\endgroup$ – franz lemmermeyer Nov 16 '16 at 7:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.