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Let $K$ be a number field and let $G$ be the group of automorphisms of $K$ over $\mathbf Q$. The group $G$ acts in a natural way on the ideal class group of $K$. I would like to know if there are any results giving a formula for the number of orbits of this action (or equivalently a formula for the number of ideal classes that are fixed by some element of $G$). In particular, I would like to compare the number of orbits to the class number of $K$.

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    $\begingroup$ There is some literature about Galois module structure of K-groups. I think what you are asking can be reformulated in terms of the Galois module structure of $K_0$, so that literature might contain the answer to your question. Unfortunately, I am not a suitable guide through that literature. $\endgroup$ Commented Jul 30, 2014 at 13:11
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    $\begingroup$ look up ambiguous class number formula and genus theory. $\endgroup$ Commented Aug 6, 2014 at 17:52
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    $\begingroup$ @MatthiasWendt: the Galois group of a quadratic field $K$ can act trivially on the ideal class group of $K$ even if the class group is nontrivial. For example, let $K = {\mathbf Q}(\sqrt{-5})$, with class group of order 2 generated by the ideal class of the prime $\mathfrak p = (2,1+\sqrt{-5})$. This prime ideal is fixed by the Galois group, so its ideal class is fixed. Therefore the Galois action on the ideal class group is trivial. More generally, if all elements of the class group have order 1 or 2 then the Galois action on ideal classes is trivial since ideal classes of primes are fixed. $\endgroup$
    – KConrad
    Commented Aug 28, 2014 at 21:16
  • $\begingroup$ @KConrad: you're right, of course. I removed my comment. $\endgroup$ Commented Aug 29, 2014 at 8:18
  • $\begingroup$ Is there an elliptic curve analogue of the ambiguous class number formula in literature, i.e. an analogous result for say the Selmer group or p-part of Selmer groups? $\endgroup$
    – debanjana
    Commented Nov 6, 2019 at 3:55

3 Answers 3

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I assume you want $K$ to be Galois over $\mathbb{Q}$. More generally, let $L/K$ be a Galois extension of number fields. The the class group $C_K$ of $K$ maps to $C_L^{G_{L/K}}$, the part of $C_L$ fixed by the Galois group of $L/K$, and you seem to be asking what the quotient $C_L^{G_{L/K}}/C_K$ looks like.

Taking cohomology of the exact sequences $$ 1\to R_L^*\to L^*\to L^*/R_L*\to1 \quad\text{and}\quad 1\to L^*/R_L* \to I_L \to C_L \to 1 $$ gives (if I'm not mistaken) exact sequences $$ 0 \to H^1(G_{L/K},L^*/R_L*) \to H^2(G_{L/K},R_L^*) \to \text{Br}(L/K) $$ and $$ 0 \to C_K \to C_L^{G_{L/K}} \to H^1(G_{L/K},L^*/R_L*), $$ so the quotient that you're interested in naturally injects $$ C_L^{G_{L/K}}/C_K \hookrightarrow \text{Ker}\Bigl(H^2(G_{L/K},R_L^*) \to \text{Br}(L/K)\Bigr). $$ The Galois structure of unit groups has been much studied. You might look at some of Ted Chinburg's papers (http://www.math.upenn.edu/~ted/CVPubs9-10-07.html)

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As Franz Lemmermeyer suggested, one should consider the so-called Ambigous Class Number Formula: you find it, for instance, in Gras' book "Class Field Theory", II.6.2.3.

It says that if $L/K$ is a finite cyclic Galois extension with group $G$ and if we denote by $Cl_L,E_L$ the class group and the unit group of $L$ respectively (and likewise for $K$), then $$ \vert Cl_L^G\vert=\frac{\vert Cl_K\vert\cdot Ram(L/K)}{[L:K]\cdot [E_K:E_K\cap\mathrm{Norm}_{L/K}(L^\times)]} $$ where $Ram(L/K)$ is the product of all ramification indexes of finite and infinite primes of $K$ in the extension $L/K$.

So, in case $K=\mathbb{Q}$, you get simply $$ \vert Cl_L^G\vert=\frac{Ram(L/\mathbb{Q})}{[L:\mathbb{Q}]}\quad\text{or}\quad \vert Cl_L^G\vert=\frac{Ram(L/\mathbb{Q})}{2[L:\mathbb{Q}]} $$ according as whether $L^\times$ contains or not an element of norm $-1$.

The proof is basically cohomological following the one suggested by Joe Silverman: the term $Ram(L/K)$ comes from fixed ideals in $I_L^G$ and the index of the norm of units of $L$ inside those of $K$ controls some capitulation kernel - but you can look in Gras' book for details.

In case of dihedral extensions, you can look at the question Class groups in dihedral extensions - some sort of Spiegelungssatz?

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  • $\begingroup$ For $\mathbb{Q}(i)/\mathbb{Q}$ this gives $1/2$. Is there perhaps a missing factor of $2^{r_2}$? $\endgroup$
    – MT_
    Commented Apr 15, 2018 at 21:59
  • $\begingroup$ Thanks, my formula was wrong! I should consider finite and infinite primes, so that $Ram(\mathbb{Q}(i)/\mathbb{Q})=4$. $\endgroup$ Commented May 2, 2018 at 13:15
  • $\begingroup$ You should assume that $L/K$ is cyclic. Consider $\mathbb{Q}(\sqrt{-3},\sqrt{13})/\mathbb{Q}$. Also replacing $13$ with any prime $p$ satisfies $p\equiv 1 \bmod 12$ is a class of counter-examples. $\endgroup$
    – J.Li
    Commented Jul 16, 2019 at 6:41
  • $\begingroup$ @J.Li Oh, thanks, sure! Corrected. $\endgroup$ Commented Jul 24, 2019 at 10:42
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Since the map $j$ from $C_K$ to $C_L^{\operatorname{Gal}(L/K)}$ is not always injective, we are looking at the cokernel of this map. $\operatorname{coker} j$ will map onto $$H^1(\operatorname{Gal}(L/K),L^\times/R_{L^\times})= \ker\left(H^2(\operatorname{Gal}(L/K),R_{L^\times}) \to \operatorname{Br}(L/K)\right)$$ with kernel equal to the natural image of $H^1(\operatorname{Gal}(L/K),U_L)$ in $\operatorname{coker} j$, $U_L$ the idele units. In particular we have an isomorphism when $L/K$ is unramified. The ambiguous class number formula applies only to cyclic extensions $L/K$.

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