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Let Q($\sqrt{d}$) be a number field with discriminant $D$ such that $d \equiv 1,3$ (mod 4). Let $\mathcal{O}$ be its ring of integers and $cl(\mathcal{O})$ the respective class group. Consider an arbitrary class $\overline{I} \in cl(\mathcal{O})$. Then my claim is that either (correct me if I'm wrong):

$N(I)$ is a quadratic residue mod $D$ for any ideal $I \in \overline{I}$. Lets call $\overline{I}$ a QR class.

$N(I)$ is a quadratic non-residue mod $D$ for any ideal $I \in \overline{I}$. Lets call $\overline{I}$ a QNR class.

The reasoning behind this is that any two ideals $I,J \in \overline{I}$ are related by two principal ideals $(a), (b)$ in the following way: $(a)I = (b)J$

I have shown that in these circumstances the norm of any element in $\mathcal{O}$ is a QR mod $D$. Hence $N((a)), N((b))$ are QR's and if $N(I)$ is also a QR, then so is $N(J)$.

I have much computation which leads me to believe that a class is a QR class if and only if it is a square in $cl(\mathcal{O})$. I proved the reverse direction by considering a particular squared ideal $J^2 \in \overline{I}$, where $\overline{I}$ is a square in $cl(\mathcal{O})$. Obviously $N(J^2)$ is a QR and so $\overline{I}$ is a QR class.

But I fail to see a proof of the forward direction. I have not been able to construct a specific square ideal in a QR class and considering the contrapositive has led me nowhere.

I am quite new to Algebraic Number theory and I would be interested in any tools I could learn that would help me solve these kinds of problems. Thank you in advance.

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  • $\begingroup$ If $d \equiv 1 \bmod 4$ then $D =d$ so that $N((\alpha))= N(a+b\sqrt{d}) = a^2+ d b^2 \equiv a^2 \bmod D$. Thus $\left(\frac{N(I)}{D}\right)$ depends only on the class of $I$. $\endgroup$ – reuns Aug 7 '17 at 1:53
  • $\begingroup$ (restricting to the group of (fractional) ideals primes to $D$ might be safer) $\endgroup$ – reuns Aug 7 '17 at 1:58
  • $\begingroup$ The keyword is "genus theory". See the "Introduction to number theory" by Flath or "Primes of the form $x^2 + Ny^2$" by Cox. $\endgroup$ – franz lemmermeyer Aug 7 '17 at 5:59
  • $\begingroup$ @reuns, your first comment, though written immediately after my post, is not quite correct, as explained in my answer. $\endgroup$ – user466572 Aug 7 '17 at 12:26
  • $\begingroup$ @user466572 What is not correct ? The map $I \mapsto (\frac{N(I)}{D})$ is trivial on principal ideals (with norm) coprime to $D$, thus it extends to $\mathcal{I}_D/\mathcal{P}_D$ the group of ideal class coprime to $D$. $\endgroup$ – reuns Aug 7 '17 at 12:44
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$\newcommand\Z{\mathbb{Z}}$ $\newcommand\A{\mathbb{A}}$ $\newcommand\Q{\mathbb{Q}}$ $\newcommand\Gal{\mathrm{Gal}}$ $\newcommand\tpsi{\widetilde{\psi}}$ $\newcommand\Hw{\widetilde{H}}$

Suppose that the discriminant $D$ has $m$ distinct prime divisors. There is a map

$$\psi: (\Z/D \Z)^{\times} \rightarrow (\Z/D \Z)^{\times}/(\Z/D \Z)^{\times 2} = \prod_{p|D} (\Z/2\Z)$$

Note that, if $D$ is even, then the assumptions imply that $4 \| D$ and so $(\Z/4\Z)^{\times}$ modulo squares is $(\Z/2\Z)$. By class field theory, this gives rise to an extension $H/\Q$ unramified outside $D \infty$ whose Galois group is $(\Z/2\Z)^m$. Explicity, it is given by the compositum of the fields $\Q(\sqrt{p^*})$ where, for $p$ odd, $p^* = (-1)^{(p-1)/2} p$, and for $p = 2$, we take $p^* = -1$, and $p$ runs over all primes dividing $D$.

One can also view this as coming from a map on idele class groups:

$$\psi: \Q^{\times} \backslash \A^{\times}_{\Q} \rightarrow \prod_{p | D} (\Z/2 \Z)$$

Explicitly, the map is defined as follows: given an idele $\alpha = (\alpha_v)$, one may --- by an elementary argument that follows from unique factorization in $\Q$ --- multiply by an element in $\Q^{\times}$ so that $\alpha_v \in \Q^{\times}_v$ is a unit for all $v$ and is positive at the real place. Indeed, there is a unique such representative. Then one can project $\alpha$ to $(\alpha_v)_{v \in D}$ and then project to $$\prod_{v|D} (\Z/v \Z)^{\times} = (\Z/D\Z)^{\times}$$ and finally map to the target using the map $\psi$.

On the other hand, one can give an alternate description of $\psi$. Suppose that $\alpha$ is an idele which is a unit for all $v|D$ and is positive at infinity. Then one may consider

$$\prod_{v \ne \infty} |\alpha_v|_v \in (\Z/D\Z)^{\times}$$

and $\psi(\alpha)$ is just the image of this element under $\psi$. This translation is elementary. As an example, suppose that $\alpha_v = 1$ for $v \ne p$ and $\alpha_p = p$ for some $(p,D) = 1$. Then the product above would be $1/p$, and the second description would send $\alpha$ to the image of $1/p$. For the first description, we first multiply $\alpha$ by $1/p$ and then $\alpha_v = 1/p$ for all $v \ne p$ and $\alpha_p = 1$. This is certainly a unit at all places and positive at infinity. But now the element $(1/p,\ldots,1/p) \in \prod_{v|D} \Q^{\times}_v$ projects to $1/p \in (\Z/D \Z)^{\times}$.

Now there is a corresponding map

$$\tpsi:=\psi \circ N_{F/\Q} F^{\times} \backslash \A^{\times}_{F} \rightarrow \prod_{p | D} (\Z/2 \Z) = \Gal(H/\Q).$$

By functoriality (in class field theory), the image of $\tpsi$ is precisely $\Gal(H/F)$, and the corresponding map is exactly the Artin map. On the other hand, the extension $H/F$ is unramified at all finite places by construction. Hence, by class field theory, the map $\psi \circ N_{F/\Q}$ factors through the narrow class group. Moreover, by the second description of $\psi$ above, one can describe the map explicitly as follows: Choose a representative $I$ of the narrow class group $C_{F}$, then take its norm, and then consider the corresponding element in $(\Z/D \Z)^{\times}$ modulo squares. This is very close to what you claim exists in your question, except now we actually have a homomorphism of groups.

One wrinkle is that this map is really a map from the narrow class group of $F$ rather than the class group of $F$. The narrow class group is defined to be non-zero fractional ideals modulo the relation $I \sim J$ when $I = (\alpha) J$ and $\alpha$ is a totally positive element (for a complex quadratic field this is an empty statement, and when $F$ is a real quadratic field it just means that the norm of $\alpha$ is positive). This is because, when $F/\Q$ is real, the extension $H/\Q$ is not always real, and so the extension $H/F$, even though it is unramified at all finite primes, is ramified at infinity. It turns out to be important to work with the narrow class group. As an example, suppose that $F = \mathbf{Q}(\sqrt{21})$ so $D = 21$. In this case, the class group $C_F$ is trivial and the narrow class group $C^{+}_F$ is $\Z/2\Z$. Consider the ideal

$$I:=(\alpha) = \left(\frac{1 + \sqrt{21}}{2}\right).$$

Clearly $I$ is principal. However, the norm of the ideal $I$ is $5$, and this is neither a square modulo $3$ nor modulo $7$. In particular, your claim is not quite correct (since you didn't give the argument, I can't say where you made an error, however). Of course, since the norm of $\alpha$ is $-5$, the class $I$ is different from the trivial class inside the narrow class group.

To recap; we have a map:

$$\tpsi: C^{+}_F \rightarrow \Gal(H/F) \subset \Gal(H/\Q) = \prod_{p | D} (\Z/2 \Z)$$

on the narrow class group, whose image is precisely $\Gal(H/F)$. It certainly factors through $C^{+}_F/2 C^{+}_F$. Consider the extension $\Hw/F$ associated to $C^{+}_F/2 C^{+}_F$ by class field theory. Certainly $\Hw$ contains $H$, and $\Hw$ is Galois over $\Q$, and $\Hw/F$ is unramified at all finite places. On the other hand, the action of $\Gal(F/\Q)$ on $C^{+}_F$ is by $-1$, since $I \cdot \sigma(I) = (N(I))$ is trivial in the narrow class group. Thus the action of $\Gal(F/\Q)$ on the quotient $C^{+}_F/2 C^{+}_F$ is trivial. It follows (again by the compatibility of these constructions by class field theory) that the action of $\Gal(F/\Q)$ on $\Gal(\Hw/F) = C^{+}_F/2 C^{+}_F$ is also trivial. This implies that $\Gal(\Hw/\Q)$ is abelian. But now, using class field theory for $\Q$, this forces $\Hw$ to be contained in $H$, and thus $\Hw = H$. (The point is that since we know the abelian extensions of $\Q$, we can explicitly see that $H$ is the largest such extension unramified over $F$ at all finite places.) Hence the kernel of $\psi$ is precisely $2 C^{+}_F$.

This answers your question in the positive, up to the requirement of working with the narrow class group rather than the class group.

The assumption that $d$ is odd can be modified; if $d$ is even, then instead of asking that the norm be a square modulo $4$, one would ask that either $N(I) \equiv 1,5 \mod 8$ or $1,7 \mod 8$ depending on whether either $F(\sqrt{-2})/F$ or $F(\sqrt{2})/F$ was unramified.

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  • $\begingroup$ Thank you for fixing my claim (+1) and the detailed response. This is the direction I am working towards but it could take a while for me to understand some of these concepts. $\endgroup$ – TCiur Aug 7 '17 at 11:10
  • $\begingroup$ Dear @TCiur, either I answered your question, in which case you should accept the answer, or I did not, in which case you should clarify your question. user franz lemmermeyer has given some references. $\endgroup$ – user466572 Aug 7 '17 at 20:53
  • $\begingroup$ Are you saying that it is not possible to answer my question using undergraduate level maths? $\endgroup$ – TCiur Aug 7 '17 at 22:00

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