Zariski Closure in a Torus and the dimension of the quotient
up vote
2
down vote
favorite
I'm reading the second chapter of Serre's Abelian l-Adic Representations and Elliptic Curves, and need help with the first exercise. Let $K$ be a number field and let $T$ be the Weil Restriction of the multiplicative group $mathbb{G}_m$ over $K$, to the rationals, i.e. $T=R_{K/mathbb{Q}}(mathbb{G}_{m/K})$. Let $E$ be the group of units in the ring of integers of $K$, hence a subgroup of $K^*=T(mathbb{Q})$. Denote $bar{E}$ to be the Zariski closure of $E$ in $T$ and $T_E$ to be the torus $T/bar{E}$. The exercise asks to show that the dimension of $T_E$ is 2 for imaginary quadratic fields and 1 for both real quadratic fields and cubic number fields with one complex place. I have just started reading about algebraic groups and am struggling with finding the Zariski closure of the unit groups in the tori. I know what the group $E$ looks like, with the help of Dirichlet's unit theorem, but am unable to proceed further. Any hints would be appreciated.
algebraic-geometry algebraic-number-theory algebraic-groups zariski-topology
asked Nov 25 at 0:54
wateryeyes
112
add a comment |
up vote
2
down vote
favorite
I'm reading the second chapter of Serre's Abelian l-Adic Representations and Elliptic Curves, and need help with the first exercise. Let $K$ be a number field and let $T$ be the Weil Restriction of the multiplicative group $mathbb{G}_m$ over $K$, to the rationals, i.e. $T=R_{K/mathbb{Q}}(mathbb{G}_{m/K})$. Let $E$ be the group of units in the ring of integers of $K$, hence a subgroup of $K^*=T(mathbb{Q})$. Denote $bar{E}$ to be the Zariski closure of $E$ in $T$ and $T_E$ to be the torus $T/bar{E}$. The exercise asks to show that the dimension of $T_E$ is 2 for imaginary quadratic fields and 1 for both real quadratic fields and cubic number fields with one complex place. I have just started reading about algebraic groups and am struggling with finding the Zariski closure of the unit groups in the tori. I know what the group $E$ looks like, with the help of Dirichlet's unit theorem, but am unable to proceed further. Any hints would be appreciated.
algebraic-geometry algebraic-number-theory algebraic-groups zariski-topology
asked Nov 25 at 0:54
wateryeyes
112
In the case of imaginary quadratic field, Dirichlet's unit theorem says that $E$ has rank $0$, that is $E$ is a finite cyclic group and hence it is already closed in zariski topology(finite union of closed set is closed). Then quotient of a torus by a finite group is again a torus of the same dimension.
– random123
Nov 25 at 9:21
In case of Real quadratic field, E is rank one. That is $E$ is infinite and thus $overline{E}$ is of atleast dimension 1. I claim that $E$ is contained in kernel of the norm map $Res_{k/mathbb{Q}}(mathbb{G}_{m/k}) rightarrow mathbb{G}_{m/mathbb{Q}}$ which is of dimension 1, since dim$Res_{k/mathbb{Q}}(mathbb{G}_{m/k}) = 1$. Thus $overline{E}$ is a closed subset of a zariski close subsets of dimension 1.
– random123
Nov 25 at 9:27
One can also use rank$X(T_E)otimes mathbb{Q} = dim(T_E)$, where the terminology is borrowed from the book mentioned in the question.
– random123
Nov 25 at 9:57
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm reading the second chapter of Serre's Abelian l-Adic Representations and Elliptic Curves, and need help with the first exercise. Let $K$ be a number field and let $T$ be the Weil Restriction of the multiplicative group $mathbb{G}_m$ over $K$, to the rationals, i.e. $T=R_{K/mathbb{Q}}(mathbb{G}_{m/K})$. Let $E$ be the group of units in the ring of integers of $K$, hence a subgroup of $K^*=T(mathbb{Q})$. Denote $bar{E}$ to be the Zariski closure of $E$ in $T$ and $T_E$ to be the torus $T/bar{E}$. The exercise asks to show that the dimension of $T_E$ is 2 for imaginary quadratic fields and 1 for both real quadratic fields and cubic number fields with one complex place. I have just started reading about algebraic groups and am struggling with finding the Zariski closure of the unit groups in the tori. I know what the group $E$ looks like, with the help of Dirichlet's unit theorem, but am unable to proceed further. Any hints would be appreciated.
algebraic-geometry algebraic-number-theory algebraic-groups zariski-topology
asked Nov 25 at 0:54
wateryeyes
112
I'm reading the second chapter of Serre's Abelian l-Adic Representations and Elliptic Curves, and need help with the first exercise. Let $K$ be a number field and let $T$ be the Weil Restriction of the multiplicative group $mathbb{G}_m$ over $K$, to the rationals, i.e. $T=R_{K/mathbb{Q}}(mathbb{G}_{m/K})$. Let $E$ be the group of units in the ring of integers of $K$, hence a subgroup of $K^*=T(mathbb{Q})$. Denote $bar{E}$ to be the Zariski closure of $E$ in $T$ and $T_E$ to be the torus $T/bar{E}$. The exercise asks to show that the dimension of $T_E$ is 2 for imaginary quadratic fields and 1 for both real quadratic fields and cubic number fields with one complex place. I have just started reading about algebraic groups and am struggling with finding the Zariski closure of the unit groups in the tori. I know what the group $E$ looks like, with the help of Dirichlet's unit theorem, but am unable to proceed further. Any hints would be appreciated.
algebraic-geometry algebraic-number-theory algebraic-groups zariski-topology
algebraic-geometry algebraic-number-theory algebraic-groups zariski-topology
asked Nov 25 at 0:54
wateryeyes
112
asked Nov 25 at 0:54
wateryeyes
112
edited Nov 25 at 1:19
asked Nov 25 at 0:54
wateryeyes
112
asked Nov 25 at 0:54
wateryeyes
112
asked Nov 25 at 0:54
wateryeyes
112
112
In the case of imaginary quadratic field, Dirichlet's unit theorem says that $E$ has rank $0$, that is $E$ is a finite cyclic group and hence it is already closed in zariski topology(finite union of closed set is closed). Then quotient of a torus by a finite group is again a torus of the same dimension.
– random123
Nov 25 at 9:21
In case of Real quadratic field, E is rank one. That is $E$ is infinite and thus $overline{E}$ is of atleast dimension 1. I claim that $E$ is contained in kernel of the norm map $Res_{k/mathbb{Q}}(mathbb{G}_{m/k}) rightarrow mathbb{G}_{m/mathbb{Q}}$ which is of dimension 1, since dim$Res_{k/mathbb{Q}}(mathbb{G}_{m/k}) = 1$. Thus $overline{E}$ is a closed subset of a zariski close subsets of dimension 1.
– random123
Nov 25 at 9:27
One can also use rank$X(T_E)otimes mathbb{Q} = dim(T_E)$, where the terminology is borrowed from the book mentioned in the question.
– random123
Nov 25 at 9:57
add a comment |
In the case of imaginary quadratic field, Dirichlet's unit theorem says that $E$ has rank $0$, that is $E$ is a finite cyclic group and hence it is already closed in zariski topology(finite union of closed set is closed). Then quotient of a torus by a finite group is again a torus of the same dimension.
– random123
Nov 25 at 9:21
In case of Real quadratic field, E is rank one. That is $E$ is infinite and thus $overline{E}$ is of atleast dimension 1. I claim that $E$ is contained in kernel of the norm map $Res_{k/mathbb{Q}}(mathbb{G}_{m/k}) rightarrow mathbb{G}_{m/mathbb{Q}}$ which is of dimension 1, since dim$Res_{k/mathbb{Q}}(mathbb{G}_{m/k}) = 1$. Thus $overline{E}$ is a closed subset of a zariski close subsets of dimension 1.
– random123
Nov 25 at 9:27
One can also use rank$X(T_E)otimes mathbb{Q} = dim(T_E)$, where the terminology is borrowed from the book mentioned in the question.
– random123
Nov 25 at 9:57
In the case of imaginary quadratic field, Dirichlet's unit theorem says that $E$ has rank $0$, that is $E$ is a finite cyclic group and hence it is already closed in zariski topology(finite union of closed set is closed). Then quotient of a torus by a finite group is again a torus of the same dimension.
– random123
Nov 25 at 9:21
In the case of imaginary quadratic field, Dirichlet's unit theorem says that $E$ has rank $0$, that is $E$ is a finite cyclic group and hence it is already closed in zariski topology(finite union of closed set is closed). Then quotient of a torus by a finite group is again a torus of the same dimension.
– random123
Nov 25 at 9:21
In case of Real quadratic field, E is rank one. That is $E$ is infinite and thus $overline{E}$ is of atleast dimension 1. I claim that $E$ is contained in kernel of the norm map $Res_{k/mathbb{Q}}(mathbb{G}_{m/k}) rightarrow mathbb{G}_{m/mathbb{Q}}$ which is of dimension 1, since dim$Res_{k/mathbb{Q}}(mathbb{G}_{m/k}) = 1$. Thus $overline{E}$ is a closed subset of a zariski close subsets of dimension 1.
– random123
Nov 25 at 9:27
In case of Real quadratic field, E is rank one. That is $E$ is infinite and thus $overline{E}$ is of atleast dimension 1. I claim that $E$ is contained in kernel of the norm map $Res_{k/mathbb{Q}}(mathbb{G}_{m/k}) rightarrow mathbb{G}_{m/mathbb{Q}}$ which is of dimension 1, since dim$Res_{k/mathbb{Q}}(mathbb{G}_{m/k}) = 1$. Thus $overline{E}$ is a closed subset of a zariski close subsets of dimension 1.
– random123
Nov 25 at 9:27
One can also use rank$X(T_E)otimes mathbb{Q} = dim(T_E)$, where the terminology is borrowed from the book mentioned in the question.
– random123
Nov 25 at 9:57
One can also use rank$X(T_E)otimes mathbb{Q} = dim(T_E)$, where the terminology is borrowed from the book mentioned in the question.
– random123
Nov 25 at 9:57
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3012296%2fzariski-closure-in-a-torus-and-the-dimension-of-the-quotient%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
In the case of imaginary quadratic field, Dirichlet's unit theorem says that $E$ has rank $0$, that is $E$ is a finite cyclic group and hence it is already closed in zariski topology(finite union of closed set is closed). Then quotient of a torus by a finite group is again a torus of the same dimension.
– random123
Nov 25 at 9:21
In case of Real quadratic field, E is rank one. That is $E$ is infinite and thus $overline{E}$ is of atleast dimension 1. I claim that $E$ is contained in kernel of the norm map $Res_{k/mathbb{Q}}(mathbb{G}_{m/k}) rightarrow mathbb{G}_{m/mathbb{Q}}$ which is of dimension 1, since dim$Res_{k/mathbb{Q}}(mathbb{G}_{m/k}) = 1$. Thus $overline{E}$ is a closed subset of a zariski close subsets of dimension 1.
– random123
Nov 25 at 9:27
One can also use rank$X(T_E)otimes mathbb{Q} = dim(T_E)$, where the terminology is borrowed from the book mentioned in the question.
– random123
Nov 25 at 9:57