Localisation of $mathbb{Z}/(p^k)$.
up vote
0
down vote
favorite
I was looking at the wiki that explains localization. It says that the only way to localize $mathbb{Z}/(p^k)$ is ${0}$. The argument is that the elements of $mathbb{Z}/(p^k)$ are either units or nilpotents elements.
So if $x in S$ (multiplicatively closed subset) is a nilpotent (and $x^n=0$), than clearly $0 in S$ so the only localization is $0$. But if $u in S$ is a unit, does this imply that $0 in S$ in general? Why?
Why if $S={(1,0),(1,1)}$ is the localization $mathbb{Z}/a mathbb{Z}$?
commutative-algebra localization
asked Dec 3 at 12:52
roi_saumon
35517
add a comment |
up vote
0
down vote
favorite
I was looking at the wiki that explains localization. It says that the only way to localize $mathbb{Z}/(p^k)$ is ${0}$. The argument is that the elements of $mathbb{Z}/(p^k)$ are either units or nilpotents elements.
So if $x in S$ (multiplicatively closed subset) is a nilpotent (and $x^n=0$), than clearly $0 in S$ so the only localization is $0$. But if $u in S$ is a unit, does this imply that $0 in S$ in general? Why?
Why if $S={(1,0),(1,1)}$ is the localization $mathbb{Z}/a mathbb{Z}$?
commutative-algebra localization
asked Dec 3 at 12:52
roi_saumon
35517
1
I assume they meant "the only non-trivial way".
– Tobias Kildetoft
Dec 3 at 12:56
Your addition does not really make sense. How are you supposed to write these elements as pairs of numbers?
– Tobias Kildetoft
Dec 3 at 13:08
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was looking at the wiki that explains localization. It says that the only way to localize $mathbb{Z}/(p^k)$ is ${0}$. The argument is that the elements of $mathbb{Z}/(p^k)$ are either units or nilpotents elements.
So if $x in S$ (multiplicatively closed subset) is a nilpotent (and $x^n=0$), than clearly $0 in S$ so the only localization is $0$. But if $u in S$ is a unit, does this imply that $0 in S$ in general? Why?
Why if $S={(1,0),(1,1)}$ is the localization $mathbb{Z}/a mathbb{Z}$?
commutative-algebra localization
asked Dec 3 at 12:52
roi_saumon
35517
I was looking at the wiki that explains localization. It says that the only way to localize $mathbb{Z}/(p^k)$ is ${0}$. The argument is that the elements of $mathbb{Z}/(p^k)$ are either units or nilpotents elements.
So if $x in S$ (multiplicatively closed subset) is a nilpotent (and $x^n=0$), than clearly $0 in S$ so the only localization is $0$. But if $u in S$ is a unit, does this imply that $0 in S$ in general? Why?
Why if $S={(1,0),(1,1)}$ is the localization $mathbb{Z}/a mathbb{Z}$?
commutative-algebra localization
commutative-algebra localization
asked Dec 3 at 12:52
roi_saumon
35517
asked Dec 3 at 12:52
roi_saumon
35517
edited Dec 3 at 13:03
asked Dec 3 at 12:52
roi_saumon
35517
asked Dec 3 at 12:52
roi_saumon
35517
asked Dec 3 at 12:52
roi_saumon
35517
35517
1
I assume they meant "the only non-trivial way".
– Tobias Kildetoft
Dec 3 at 12:56
Your addition does not really make sense. How are you supposed to write these elements as pairs of numbers?
– Tobias Kildetoft
Dec 3 at 13:08
add a comment |
1
I assume they meant "the only non-trivial way".
– Tobias Kildetoft
Dec 3 at 12:56
Your addition does not really make sense. How are you supposed to write these elements as pairs of numbers?
– Tobias Kildetoft
Dec 3 at 13:08
1
1
I assume they meant "the only non-trivial way".
– Tobias Kildetoft
Dec 3 at 12:56
I assume they meant "the only non-trivial way".
– Tobias Kildetoft
Dec 3 at 12:56
Your addition does not really make sense. How are you supposed to write these elements as pairs of numbers?
– Tobias Kildetoft
Dec 3 at 13:08
Your addition does not really make sense. How are you supposed to write these elements as pairs of numbers?
– Tobias Kildetoft
Dec 3 at 13:08
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Let $A$ be your ring. For what you have said, it is clear that the problem is when $S$ has a nilpotent element. If $S$ has not nilpotent elements then it is only composed by units and the localization $ S^{-1}A$ is isomorphic to $A$ becuse you are not adding new units.
So you are right, $uin S$ does not implies $ 0in S$ so ${0} $ is not the only way of localizing $ A$. There are in fact two localizations of the quotient by a primary ideal (which is what you have there) i.e. ${0} $ and the ring it self.
answered Dec 4 at 15:49
Natalio
332111
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let $A$ be your ring. For what you have said, it is clear that the problem is when $S$ has a nilpotent element. If $S$ has not nilpotent elements then it is only composed by units and the localization $ S^{-1}A$ is isomorphic to $A$ becuse you are not adding new units.
So you are right, $uin S$ does not implies $ 0in S$ so ${0} $ is not the only way of localizing $ A$. There are in fact two localizations of the quotient by a primary ideal (which is what you have there) i.e. ${0} $ and the ring it self.
answered Dec 4 at 15:49
Natalio
332111
add a comment |
up vote
0
down vote
Let $A$ be your ring. For what you have said, it is clear that the problem is when $S$ has a nilpotent element. If $S$ has not nilpotent elements then it is only composed by units and the localization $ S^{-1}A$ is isomorphic to $A$ becuse you are not adding new units.
So you are right, $uin S$ does not implies $ 0in S$ so ${0} $ is not the only way of localizing $ A$. There are in fact two localizations of the quotient by a primary ideal (which is what you have there) i.e. ${0} $ and the ring it self.
answered Dec 4 at 15:49
Natalio
332111
add a comment |
up vote
0
down vote
up vote
0
down vote
Let $A$ be your ring. For what you have said, it is clear that the problem is when $S$ has a nilpotent element. If $S$ has not nilpotent elements then it is only composed by units and the localization $ S^{-1}A$ is isomorphic to $A$ becuse you are not adding new units.
So you are right, $uin S$ does not implies $ 0in S$ so ${0} $ is not the only way of localizing $ A$. There are in fact two localizations of the quotient by a primary ideal (which is what you have there) i.e. ${0} $ and the ring it self.
answered Dec 4 at 15:49
Natalio
332111
Let $A$ be your ring. For what you have said, it is clear that the problem is when $S$ has a nilpotent element. If $S$ has not nilpotent elements then it is only composed by units and the localization $ S^{-1}A$ is isomorphic to $A$ becuse you are not adding new units.
So you are right, $uin S$ does not implies $ 0in S$ so ${0} $ is not the only way of localizing $ A$. There are in fact two localizations of the quotient by a primary ideal (which is what you have there) i.e. ${0} $ and the ring it self.
answered Dec 4 at 15:49
Natalio
332111
answered Dec 4 at 15:49
Natalio
332111
answered Dec 4 at 15:49
Natalio
332111
answered Dec 4 at 15:49
Natalio
332111
332111
add a comment |
add a comment |
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%2f3024024%2flocalisation-of-mathbbz-pk%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
1
I assume they meant "the only non-trivial way".
– Tobias Kildetoft
Dec 3 at 12:56
Your addition does not really make sense. How are you supposed to write these elements as pairs of numbers?
– Tobias Kildetoft
Dec 3 at 13:08