Let $text{T} = mathbb{Q}(i)^times/mathbb{Q}^times$ compute $text{T}(mathbb{A})/text{T}(mathbb{Q})$












2












$begingroup$


I have a question about algebraic tori, let $T = mathbb{Q}(i)^times/mathbb{Q}^times$ and let's try to compute $text{T}(mathbb{A})/text{T}(mathbb{Q})$.



One has a product decomposition for the multiplicative group of Gaussian integers:
$$ mathbb{Q}(i)^times simeq mathbb{Z} times prod_{p=4k+1}mathbb{Z}^2 times prod_{p = 4k+3}mathbb{Z}$$
This has to do with Fermat's theorem that $p = a^2 + b^2$ iff $p = 4k+1$. In that case, with we factor out the multiplicative group of $mathbb{Q}$ we'd have something like:
$$ mathbb{Q}(i)^times / mathbb{Q}^times simeq prod_{p=4k+1} mathbb{Z}$$
This does not say much about a topolog of $mathbf{T}$ as a locally compact Abelian group. E.g. this is infinitly generated. We could consider this as a $2 times 2$ matrix group with elements in $mathbb{Q}$:



$$ mathbb{Q}(i)^times / mathbb{Q}^times simeq prod_{p=4k+1} left( begin{array}{cr} a & -b \ b & a end{array} right)^mathbb{Z} $$
This is my best guess, and hopefully that's more informative.



Given all this work, Can we give some kind of matrix way of looking at $mathbf{T}(mathbb{A})/mathbf{T}(mathbb{Q})$ ?






number-theory algebraic-groups locally-compact-groups adeles







share|cite|improve this question









$endgroup$












  • $begingroup$
    Usually to talk about $T(mathbb A)$ and $T(mathbb Q)$, $T$ should be a group scheme over $mathbb Q$.
    $endgroup$
    – D_S
    Dec 18 '18 at 23:32










  • $begingroup$
    Also, I think your description of $mathbb Q(i)^{ast}$ is that of the group of fractional ideals, you should mod out by the units to get it
    $endgroup$
    – D_S
    Dec 18 '18 at 23:35






  • 1




    $begingroup$
    @D_S Try $text{T}=text{Res}_{mathbb{Q}(i)/mathbb{Q}}(mathbb{G_m})/mathbb{G_m}$ where $mathbb{G_m}$ is the multiplicative group functor and and $text{Res}_{K/L}$ is restriction of scalars for number fields $K$ and $L$.
    $endgroup$
    – cactus314
    Dec 19 '18 at 0:52








  • 1




    $begingroup$
    Starting off, $mathbb G_m$ is a closed subgroup of $operatorname{Res}_{mathbb Q(i)/mathbb Q}(mathbb G_m)$ by $x mapsto (x,x)$, so we have an exact sequence $$1 rightarrow mathbb G_m rightarrow operatorname{Res}_{mathbb Q(i)/mathbb Q}(mathbb G_m) rightarrow T rightarrow 1 $$ On points we get an exact sequence $$1 rightarrow mathbb Q^{ast} rightarrow mathbb Q(i)^{ast} rightarrow T(mathbb Q) rightarrow H^1(operatorname{Gal}(overline{mathbb Q}/mathbb Q),overline{mathbb Q}^{ast})$$ So my first thought in approaching this is to determine whether $H^1$ is trivial
    $endgroup$
    – D_S
    Dec 19 '18 at 1:05












  • $begingroup$
    That is, can we identify $T(mathbb Q)$ as $mathbb Q(i)^{ast}/mathbb Q^{ast}$?
    $endgroup$
    – D_S
    Dec 19 '18 at 1:07
















2












$begingroup$


I have a question about algebraic tori, let $T = mathbb{Q}(i)^times/mathbb{Q}^times$ and let's try to compute $text{T}(mathbb{A})/text{T}(mathbb{Q})$.



One has a product decomposition for the multiplicative group of Gaussian integers:
$$ mathbb{Q}(i)^times simeq mathbb{Z} times prod_{p=4k+1}mathbb{Z}^2 times prod_{p = 4k+3}mathbb{Z}$$
This has to do with Fermat's theorem that $p = a^2 + b^2$ iff $p = 4k+1$. In that case, with we factor out the multiplicative group of $mathbb{Q}$ we'd have something like:
$$ mathbb{Q}(i)^times / mathbb{Q}^times simeq prod_{p=4k+1} mathbb{Z}$$
This does not say much about a topolog of $mathbf{T}$ as a locally compact Abelian group. E.g. this is infinitly generated. We could consider this as a $2 times 2$ matrix group with elements in $mathbb{Q}$:



$$ mathbb{Q}(i)^times / mathbb{Q}^times simeq prod_{p=4k+1} left( begin{array}{cr} a & -b \ b & a end{array} right)^mathbb{Z} $$
This is my best guess, and hopefully that's more informative.



Given all this work, Can we give some kind of matrix way of looking at $mathbf{T}(mathbb{A})/mathbf{T}(mathbb{Q})$ ?






number-theory algebraic-groups locally-compact-groups adeles







share|cite|improve this question









$endgroup$












  • $begingroup$
    Usually to talk about $T(mathbb A)$ and $T(mathbb Q)$, $T$ should be a group scheme over $mathbb Q$.
    $endgroup$
    – D_S
    Dec 18 '18 at 23:32










  • $begingroup$
    Also, I think your description of $mathbb Q(i)^{ast}$ is that of the group of fractional ideals, you should mod out by the units to get it
    $endgroup$
    – D_S
    Dec 18 '18 at 23:35






  • 1




    $begingroup$
    @D_S Try $text{T}=text{Res}_{mathbb{Q}(i)/mathbb{Q}}(mathbb{G_m})/mathbb{G_m}$ where $mathbb{G_m}$ is the multiplicative group functor and and $text{Res}_{K/L}$ is restriction of scalars for number fields $K$ and $L$.
    $endgroup$
    – cactus314
    Dec 19 '18 at 0:52








  • 1




    $begingroup$
    Starting off, $mathbb G_m$ is a closed subgroup of $operatorname{Res}_{mathbb Q(i)/mathbb Q}(mathbb G_m)$ by $x mapsto (x,x)$, so we have an exact sequence $$1 rightarrow mathbb G_m rightarrow operatorname{Res}_{mathbb Q(i)/mathbb Q}(mathbb G_m) rightarrow T rightarrow 1 $$ On points we get an exact sequence $$1 rightarrow mathbb Q^{ast} rightarrow mathbb Q(i)^{ast} rightarrow T(mathbb Q) rightarrow H^1(operatorname{Gal}(overline{mathbb Q}/mathbb Q),overline{mathbb Q}^{ast})$$ So my first thought in approaching this is to determine whether $H^1$ is trivial
    $endgroup$
    – D_S
    Dec 19 '18 at 1:05












  • $begingroup$
    That is, can we identify $T(mathbb Q)$ as $mathbb Q(i)^{ast}/mathbb Q^{ast}$?
    $endgroup$
    – D_S
    Dec 19 '18 at 1:07














2












2








2


2



$begingroup$


I have a question about algebraic tori, let $T = mathbb{Q}(i)^times/mathbb{Q}^times$ and let's try to compute $text{T}(mathbb{A})/text{T}(mathbb{Q})$.



One has a product decomposition for the multiplicative group of Gaussian integers:
$$ mathbb{Q}(i)^times simeq mathbb{Z} times prod_{p=4k+1}mathbb{Z}^2 times prod_{p = 4k+3}mathbb{Z}$$
This has to do with Fermat's theorem that $p = a^2 + b^2$ iff $p = 4k+1$. In that case, with we factor out the multiplicative group of $mathbb{Q}$ we'd have something like:
$$ mathbb{Q}(i)^times / mathbb{Q}^times simeq prod_{p=4k+1} mathbb{Z}$$
This does not say much about a topolog of $mathbf{T}$ as a locally compact Abelian group. E.g. this is infinitly generated. We could consider this as a $2 times 2$ matrix group with elements in $mathbb{Q}$:



$$ mathbb{Q}(i)^times / mathbb{Q}^times simeq prod_{p=4k+1} left( begin{array}{cr} a & -b \ b & a end{array} right)^mathbb{Z} $$
This is my best guess, and hopefully that's more informative.



Given all this work, Can we give some kind of matrix way of looking at $mathbf{T}(mathbb{A})/mathbf{T}(mathbb{Q})$ ?






number-theory algebraic-groups locally-compact-groups adeles







share|cite|improve this question









$endgroup$




I have a question about algebraic tori, let $T = mathbb{Q}(i)^times/mathbb{Q}^times$ and let's try to compute $text{T}(mathbb{A})/text{T}(mathbb{Q})$.



One has a product decomposition for the multiplicative group of Gaussian integers:
$$ mathbb{Q}(i)^times simeq mathbb{Z} times prod_{p=4k+1}mathbb{Z}^2 times prod_{p = 4k+3}mathbb{Z}$$
This has to do with Fermat's theorem that $p = a^2 + b^2$ iff $p = 4k+1$. In that case, with we factor out the multiplicative group of $mathbb{Q}$ we'd have something like:
$$ mathbb{Q}(i)^times / mathbb{Q}^times simeq prod_{p=4k+1} mathbb{Z}$$
This does not say much about a topolog of $mathbf{T}$ as a locally compact Abelian group. E.g. this is infinitly generated. We could consider this as a $2 times 2$ matrix group with elements in $mathbb{Q}$:



$$ mathbb{Q}(i)^times / mathbb{Q}^times simeq prod_{p=4k+1} left( begin{array}{cr} a & -b \ b & a end{array} right)^mathbb{Z} $$
This is my best guess, and hopefully that's more informative.



Given all this work, Can we give some kind of matrix way of looking at $mathbf{T}(mathbb{A})/mathbf{T}(mathbb{Q})$ ?






number-theory algebraic-groups locally-compact-groups adeles




number-theory algebraic-groups locally-compact-groups adeles






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 18 '18 at 23:11









cactus314cactus314

15.5k42269




15.5k42269












  • $begingroup$
    Usually to talk about $T(mathbb A)$ and $T(mathbb Q)$, $T$ should be a group scheme over $mathbb Q$.
    $endgroup$
    – D_S
    Dec 18 '18 at 23:32










  • $begingroup$
    Also, I think your description of $mathbb Q(i)^{ast}$ is that of the group of fractional ideals, you should mod out by the units to get it
    $endgroup$
    – D_S
    Dec 18 '18 at 23:35






  • 1




    $begingroup$
    @D_S Try $text{T}=text{Res}_{mathbb{Q}(i)/mathbb{Q}}(mathbb{G_m})/mathbb{G_m}$ where $mathbb{G_m}$ is the multiplicative group functor and and $text{Res}_{K/L}$ is restriction of scalars for number fields $K$ and $L$.
    $endgroup$
    – cactus314
    Dec 19 '18 at 0:52








  • 1




    $begingroup$
    Starting off, $mathbb G_m$ is a closed subgroup of $operatorname{Res}_{mathbb Q(i)/mathbb Q}(mathbb G_m)$ by $x mapsto (x,x)$, so we have an exact sequence $$1 rightarrow mathbb G_m rightarrow operatorname{Res}_{mathbb Q(i)/mathbb Q}(mathbb G_m) rightarrow T rightarrow 1 $$ On points we get an exact sequence $$1 rightarrow mathbb Q^{ast} rightarrow mathbb Q(i)^{ast} rightarrow T(mathbb Q) rightarrow H^1(operatorname{Gal}(overline{mathbb Q}/mathbb Q),overline{mathbb Q}^{ast})$$ So my first thought in approaching this is to determine whether $H^1$ is trivial
    $endgroup$
    – D_S
    Dec 19 '18 at 1:05












  • $begingroup$
    That is, can we identify $T(mathbb Q)$ as $mathbb Q(i)^{ast}/mathbb Q^{ast}$?
    $endgroup$
    – D_S
    Dec 19 '18 at 1:07


















  • $begingroup$
    Usually to talk about $T(mathbb A)$ and $T(mathbb Q)$, $T$ should be a group scheme over $mathbb Q$.
    $endgroup$
    – D_S
    Dec 18 '18 at 23:32










  • $begingroup$
    Also, I think your description of $mathbb Q(i)^{ast}$ is that of the group of fractional ideals, you should mod out by the units to get it
    $endgroup$
    – D_S
    Dec 18 '18 at 23:35






  • 1




    $begingroup$
    @D_S Try $text{T}=text{Res}_{mathbb{Q}(i)/mathbb{Q}}(mathbb{G_m})/mathbb{G_m}$ where $mathbb{G_m}$ is the multiplicative group functor and and $text{Res}_{K/L}$ is restriction of scalars for number fields $K$ and $L$.
    $endgroup$
    – cactus314
    Dec 19 '18 at 0:52








  • 1




    $begingroup$
    Starting off, $mathbb G_m$ is a closed subgroup of $operatorname{Res}_{mathbb Q(i)/mathbb Q}(mathbb G_m)$ by $x mapsto (x,x)$, so we have an exact sequence $$1 rightarrow mathbb G_m rightarrow operatorname{Res}_{mathbb Q(i)/mathbb Q}(mathbb G_m) rightarrow T rightarrow 1 $$ On points we get an exact sequence $$1 rightarrow mathbb Q^{ast} rightarrow mathbb Q(i)^{ast} rightarrow T(mathbb Q) rightarrow H^1(operatorname{Gal}(overline{mathbb Q}/mathbb Q),overline{mathbb Q}^{ast})$$ So my first thought in approaching this is to determine whether $H^1$ is trivial
    $endgroup$
    – D_S
    Dec 19 '18 at 1:05












  • $begingroup$
    That is, can we identify $T(mathbb Q)$ as $mathbb Q(i)^{ast}/mathbb Q^{ast}$?
    $endgroup$
    – D_S
    Dec 19 '18 at 1:07
















$begingroup$
Usually to talk about $T(mathbb A)$ and $T(mathbb Q)$, $T$ should be a group scheme over $mathbb Q$.
$endgroup$
– D_S
Dec 18 '18 at 23:32




$begingroup$
Usually to talk about $T(mathbb A)$ and $T(mathbb Q)$, $T$ should be a group scheme over $mathbb Q$.
$endgroup$
– D_S
Dec 18 '18 at 23:32












$begingroup$
Also, I think your description of $mathbb Q(i)^{ast}$ is that of the group of fractional ideals, you should mod out by the units to get it
$endgroup$
– D_S
Dec 18 '18 at 23:35




$begingroup$
Also, I think your description of $mathbb Q(i)^{ast}$ is that of the group of fractional ideals, you should mod out by the units to get it
$endgroup$
– D_S
Dec 18 '18 at 23:35




1




1




$begingroup$
@D_S Try $text{T}=text{Res}_{mathbb{Q}(i)/mathbb{Q}}(mathbb{G_m})/mathbb{G_m}$ where $mathbb{G_m}$ is the multiplicative group functor and and $text{Res}_{K/L}$ is restriction of scalars for number fields $K$ and $L$.
$endgroup$
– cactus314
Dec 19 '18 at 0:52






$begingroup$
@D_S Try $text{T}=text{Res}_{mathbb{Q}(i)/mathbb{Q}}(mathbb{G_m})/mathbb{G_m}$ where $mathbb{G_m}$ is the multiplicative group functor and and $text{Res}_{K/L}$ is restriction of scalars for number fields $K$ and $L$.
$endgroup$
– cactus314
Dec 19 '18 at 0:52






1




1




$begingroup$
Starting off, $mathbb G_m$ is a closed subgroup of $operatorname{Res}_{mathbb Q(i)/mathbb Q}(mathbb G_m)$ by $x mapsto (x,x)$, so we have an exact sequence $$1 rightarrow mathbb G_m rightarrow operatorname{Res}_{mathbb Q(i)/mathbb Q}(mathbb G_m) rightarrow T rightarrow 1 $$ On points we get an exact sequence $$1 rightarrow mathbb Q^{ast} rightarrow mathbb Q(i)^{ast} rightarrow T(mathbb Q) rightarrow H^1(operatorname{Gal}(overline{mathbb Q}/mathbb Q),overline{mathbb Q}^{ast})$$ So my first thought in approaching this is to determine whether $H^1$ is trivial
$endgroup$
– D_S
Dec 19 '18 at 1:05






$begingroup$
Starting off, $mathbb G_m$ is a closed subgroup of $operatorname{Res}_{mathbb Q(i)/mathbb Q}(mathbb G_m)$ by $x mapsto (x,x)$, so we have an exact sequence $$1 rightarrow mathbb G_m rightarrow operatorname{Res}_{mathbb Q(i)/mathbb Q}(mathbb G_m) rightarrow T rightarrow 1 $$ On points we get an exact sequence $$1 rightarrow mathbb Q^{ast} rightarrow mathbb Q(i)^{ast} rightarrow T(mathbb Q) rightarrow H^1(operatorname{Gal}(overline{mathbb Q}/mathbb Q),overline{mathbb Q}^{ast})$$ So my first thought in approaching this is to determine whether $H^1$ is trivial
$endgroup$
– D_S
Dec 19 '18 at 1:05














$begingroup$
That is, can we identify $T(mathbb Q)$ as $mathbb Q(i)^{ast}/mathbb Q^{ast}$?
$endgroup$
– D_S
Dec 19 '18 at 1:07




$begingroup$
That is, can we identify $T(mathbb Q)$ as $mathbb Q(i)^{ast}/mathbb Q^{ast}$?
$endgroup$
– D_S
Dec 19 '18 at 1:07










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