The Galois cohomology group $H^1(\mathbb{Q}, \mathbb{Z}/3\mathbb{Z})$ classifies cyclic cubic extensions $K/\mathbb{Q}$ (specifically: the non-trivial elements correspond to Galois cubic field extensions $K/\mathbb{Q}$ together with a choice of isomorphism $\mathrm{Gal}(K/\mathbb{Q}) \cong \mathbb{Z}/3\mathbb{Z}$).

Let $k = \mathbb{Q}(\mu_3)$. There are restriction and corestriction maps $$\mathrm{Res}: H^1(\mathbb{Q}, \mathbb{Z}/3\mathbb{Z}) \to H^1(k, \mathbb{Z}/3\mathbb{Z}), \quad \mathrm{Cores}: H^1(k, \mathbb{Z}/3\mathbb{Z}) \to H^1(\mathbb{Q}, \mathbb{Z}/3\mathbb{Z}).$$ Restriction followed by corestriction is multiplication by $2$ on $H^1(\mathbb{Q}, \mathbb{Z}/3\mathbb{Z})$. As each element is $3$-torsion, it follows that $\mathrm{Res}$ is injective and that $\mathrm{Cores}$ is surjective.

But as $\mu_3 \subset k$, it follows from Kummer theory that $$H^1(k, \mathbb{Z}/3\mathbb{Z}) \cong H^1(k, \mu_3) \cong k^*/k^{*3}.$$ Composing with corestriction, we therefore obtain a surjective map $$f: k^{*}/k^{*3} \to H^1(\mathbb{Q}, \mathbb{Z}/3\mathbb{Z}).$$

Can the map $f$ be made explicit? Namely, given a non-cube $a \in k^*$, what is the cyclic cubic extension of $\mathbb{Q}$ induced by $f$?

I know that the corestriction $H^1(k, \mu_3) \cong k^*/k^{3*} \to \mathbb{Q}^*/\mathbb{Q}^{*3} \cong H^1(\mathbb{Q}, \mu_3)$ is just usual norm map. But this doesn't seem to help here.


1 Answer 1


It's just the map

$$x \mapsto y = \frac{x}{x^{\sigma}},$$

where the corresponding degree three extension of $\mathbb{Q}$ is the degree three subfield of $k(y^{1/3})$. The point is that it is obvious from the restriction map that

$$H^1(\mathbb{Q},\mathbb{Z}/3 \mathbb{Z}) = (k^{\times}/k^{\times 3})^{G = {\chi}},$$

where $\chi$ is the non-trivial character of $G = \mathrm{Gal}(\mathbb{Q}(\zeta_3)/\mathbb{Q}) = \mathbb{Z}/2 \mathbb{Z}$.

(Added Here $M^{\chi}$ means what is says on the tin. If $\sigma \in G$ and $m \in M^{\chi}$, then $\sigma m = \chi(\sigma) m$.)

And basic Kummer theory also says that degree 3 cyclic extensions $K$ of $\mathbb{Q}$ have the form $K(\zeta_3) = \mathbb{Q}(\zeta_3)(\alpha^{1/3})$ where

$$\sigma \alpha = \alpha^{-1} \mod k^{\times}/k^{\times 3}.$$

The same basic structure holds mutatis mutandis with $\mathbb{Q}$ replaced by any number field $F$, and $3$ replaced by $p$, and $G = \chi$ where now $\chi$ is the mod-p cyclotomic character of $G = \mathrm{Gal}(F(\zeta_p)/F)$, which is the canonical (possibly trivial) map $G \rightarrow (\mathbb{Z}/p \mathbb{Z})^{\times}$. And now the map from $k^{\times}/k^{\times p}$ is just the projection to the $\chi$-eigenspace.

Added: If you want an explicit polynomial, you can, of course, use Galois theory to do so. In fact, everything in this question one can (and I do) teach in the introductory undergraduate Galois theory course. To spell out the elementary details, you want an element of $k(y^{1/3})$ which is fixed by the order two element $\sigma \in \mathrm{Gal}(k(y^{1/3})/\mathbb{Q})$ (there is an obvious splitting from $\mathrm{Gal}(k/\mathbb{Q}) \rightarrow \mathrm{Gal}(k(y^{1/3}/\mathbb{Q})$). The obvious element to take is thus

$$z = y^{1/3} + \sigma y^{1/3} = y^{1/3} + y^{-1/3},$$

which is a root of

$$T^3 - 3 T - (y + y^{-1}) = T^3 - 3 T - \left(\frac{x}{x^{\sigma}} + \frac{x^{\sigma}}{x}\right) = T^3 - 3 T - \frac{Tr(x^2)}{N(x)} \in \mathbb{Q}[T].$$

  • $\begingroup$ Thanks for the answer, which I'm currently trying to digest. Can you please clarify what you mean by the notation $(k^*/k^{*3})^{G = \chi}$? Do you mean the set of elements which are invariant under the action of the Galois group $G$? If so, why is this those elements $\alpha$ with $\sigma \alpha \equiv \alpha^{-1} \bmod k^{*3}$, and not $\sigma \alpha \equiv \alpha \bmod k^{*3}$? (I assume that $\sigma$ denotes the non-trivial element of $G$?). $\endgroup$ Sep 18, 2018 at 16:06
  • $\begingroup$ I was also hoping for an explicit polynomial of degree $3$ which defines the cyclic cubic extension of $\mathbb{Q}$. Does your method give this? $\endgroup$ Sep 18, 2018 at 16:06
  • $\begingroup$ Thanks for the extra details. You have defined what $M^{\chi}$ is, but I still don't understand what the notation $M^{G = \chi}$ means. Normally $M^G$ denotes the $G$-invariants for a module with $G$ action, but it seems that you mean something else. $\endgroup$ Sep 18, 2018 at 18:07
  • $\begingroup$ Whilst I appreciate the answer, I don't think I can accept it until the notation is clarified and also it is clarified why $\alpha^{-1}$ appears, rather than $\alpha$. $\endgroup$ Sep 19, 2018 at 21:17
  • 2
    $\begingroup$ @DanielLoughran Going from the name, she probably teaches at Hogwarts. And Pound Sterling must be the reincarnation of quid. $\endgroup$ Sep 21, 2018 at 5:25

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