Cyclic cubic extensions and Kummer theory

The Galois cohomology group $H^1(mathbbQ, mathbbZ/3mathbbZ)$ classifies cyclic cubic extensions $K/mathbbQ$ (specifically: the non-trivial elements correspond to Galois cubic field extensions $K/mathbbQ$ together with a choice of isomorphism $mathrmGal(K/mathbbQ) cong mathbbZ/3mathbbZ$).

Let $k = mathbbQ(mu_3)$. There are restriction and corestriction maps
$$mathrmRes: H^1(mathbbQ, mathbbZ/3mathbbZ) to H^1(k, mathbbZ/3mathbbZ), quad mathrmCores: H^1(k, mathbbZ/3mathbbZ) to H^1(mathbbQ, mathbbZ/3mathbbZ).$$
Restriction followed by corestriction is multiplication by $2$ on $H^1(mathbbQ, mathbbZ/3mathbbZ)$. As each element is $3$-torsion, it follows that $mathrmRes$ is injective and that $mathrmCores$ is surjective.

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

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

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

1 Answer
1

It's just the map

$$x mapsto y = fracxx^sigma,$$

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

$$H^1(mathbbQ,mathbbZ/3 mathbbZ)
= (k^times/k^times 3)^G = chi,$$

where $chi$ is the non-trivial character of $G = mathrmGal(mathbbQ(zeta_3)/mathbbQ) = mathbbZ/2 mathbbZ$.

(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 $mathbbQ$ have the form $K(zeta_3) = mathbbQ(zeta_3)(alpha^1/3)$ where

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

The same basic structure holds mutatis mutandis with $mathbbQ$ replaced by any number field $F$, and $3$ replaced by $p$, and $G = chi$ where now $chi$ is the mod-p cyclotomic character of $G = mathrmGal(F(zeta_p)/F)$, which is the canonical (possibly trivial) map $G rightarrow (mathbbZ/p mathbbZ)^times$. 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 $sigma in mathrmGal(k(y^1/3)/mathbbQ)$ (there is an obvious splitting from $mathrmGal(k/mathbbQ) rightarrow mathrmGal(k(y^1/3/mathbbQ)$). 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

$$T^3 - 3 T - (y + y^-1) = T^3 - 3 T - left(fracxx^sigma + fracx^sigmaxright)
= T^3 - 3 T - fracTr(x^2)N(x) in mathbbQ[T].$$

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