Mixing
Cyclic cubic extensions and Kummer theory
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
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 an isomorphism and that $mathrmCores$ is surjective (possibly it is even an isomorphism?).
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.
nt.number-theory galois-theory kummer-theory
asked 2 hours ago
Daniel Loughran
10.3k22365
add a comment |Â
up vote
3
down vote
favorite
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 an isomorphism and that $mathrmCores$ is surjective (possibly it is even an isomorphism?).
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.
nt.number-theory galois-theory kummer-theory
asked 2 hours ago
Daniel Loughran
10.3k22365
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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 an isomorphism and that $mathrmCores$ is surjective (possibly it is even an isomorphism?).
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.
nt.number-theory galois-theory kummer-theory
asked 2 hours ago
Daniel Loughran
10.3k22365
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 an isomorphism and that $mathrmCores$ is surjective (possibly it is even an isomorphism?).
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.
nt.number-theory galois-theory kummer-theory
nt.number-theory galois-theory kummer-theory
asked 2 hours ago
Daniel Loughran
10.3k22365
asked 2 hours ago
Daniel Loughran
10.3k22365
asked 2 hours ago
Daniel Loughran
10.3k22365
asked 2 hours ago
Daniel Loughran
10.3k22365
asked 2 hours ago
Daniel Loughran
10.3k22365
10.3k22365
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
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 $G = mathrmGal(mathbbQ(zeta_3)/mathbbQ) = mathbbZ/2 mathbbZ$.
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.
answered 1 hour ago
Margerie Mumblecrust
18612
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
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 $G = mathrmGal(mathbbQ(zeta_3)/mathbbQ) = mathbbZ/2 mathbbZ$.
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.
answered 1 hour ago
Margerie Mumblecrust
18612
add a comment |Â
up vote
4
down vote
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 $G = mathrmGal(mathbbQ(zeta_3)/mathbbQ) = mathbbZ/2 mathbbZ$.
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.
answered 1 hour ago
Margerie Mumblecrust
18612
add a comment |Â
up vote
4
down vote
up vote
4
down vote
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 $G = mathrmGal(mathbbQ(zeta_3)/mathbbQ) = mathbbZ/2 mathbbZ$.
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.
answered 1 hour ago
Margerie Mumblecrust
18612
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 $G = mathrmGal(mathbbQ(zeta_3)/mathbbQ) = mathbbZ/2 mathbbZ$.
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.
answered 1 hour ago
Margerie Mumblecrust
18612
answered 1 hour ago
Margerie Mumblecrust
18612
answered 1 hour ago
Margerie Mumblecrust
18612
answered 1 hour ago
Margerie Mumblecrust
18612
18612
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f310803%2fcyclic-cubic-extensions-and-kummer-theory%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
