Localisation of $mathbb{Z}/(p^k)$.
$begingroup$
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 '18 at 12:52
roi_saumonroi_saumon
51028
$endgroup$
add a comment |
$begingroup$
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 '18 at 12:52
roi_saumonroi_saumon
51028
$endgroup$
1
$begingroup$
I assume they meant "the only non-trivial way".
$endgroup$
– Tobias Kildetoft
Dec 3 '18 at 12:56
$begingroup$
Your addition does not really make sense. How are you supposed to write these elements as pairs of numbers?
$endgroup$
– Tobias Kildetoft
Dec 3 '18 at 13:08
add a comment |
$begingroup$
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 '18 at 12:52
roi_saumonroi_saumon
51028
$endgroup$
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 '18 at 12:52
roi_saumonroi_saumon
51028
asked Dec 3 '18 at 12:52
roi_saumonroi_saumon
51028
edited Dec 3 '18 at 13:03
roi_saumon
asked Dec 3 '18 at 12:52
roi_saumonroi_saumon
51028
asked Dec 3 '18 at 12:52
roi_saumonroi_saumon
51028
asked Dec 3 '18 at 12:52
roi_saumonroi_saumon
51028
51028
1
$begingroup$
I assume they meant "the only non-trivial way".
$endgroup$
– Tobias Kildetoft
Dec 3 '18 at 12:56
$begingroup$
Your addition does not really make sense. How are you supposed to write these elements as pairs of numbers?
$endgroup$
– Tobias Kildetoft
Dec 3 '18 at 13:08
add a comment |
1
$begingroup$
I assume they meant "the only non-trivial way".
$endgroup$
– Tobias Kildetoft
Dec 3 '18 at 12:56
$begingroup$
Your addition does not really make sense. How are you supposed to write these elements as pairs of numbers?
$endgroup$
– Tobias Kildetoft
Dec 3 '18 at 13:08
1
1
$begingroup$
I assume they meant "the only non-trivial way".
$endgroup$
– Tobias Kildetoft
Dec 3 '18 at 12:56
$begingroup$
I assume they meant "the only non-trivial way".
$endgroup$
– Tobias Kildetoft
Dec 3 '18 at 12:56
$begingroup$
Your addition does not really make sense. How are you supposed to write these elements as pairs of numbers?
$endgroup$
– Tobias Kildetoft
Dec 3 '18 at 13:08
$begingroup$
Your addition does not really make sense. How are you supposed to write these elements as pairs of numbers?
$endgroup$
– Tobias Kildetoft
Dec 3 '18 at 13:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 '18 at 15:49
NatalioNatalio
334111
$endgroup$
add a comment |
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$begingroup$
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 '18 at 15:49
NatalioNatalio
334111
$endgroup$
add a comment |
$begingroup$
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 '18 at 15:49
NatalioNatalio
334111
$endgroup$
add a comment |
$begingroup$
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 '18 at 15:49
NatalioNatalio
334111
$endgroup$
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 '18 at 15:49
NatalioNatalio
334111
answered Dec 4 '18 at 15:49
NatalioNatalio
334111
answered Dec 4 '18 at 15:49
NatalioNatalio
334111
answered Dec 4 '18 at 15:49
NatalioNatalio
334111
334111
add a comment |
add a comment |
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$begingroup$
I assume they meant "the only non-trivial way".
$endgroup$
– Tobias Kildetoft
Dec 3 '18 at 12:56
$begingroup$
Your addition does not really make sense. How are you supposed to write these elements as pairs of numbers?
$endgroup$
– Tobias Kildetoft
Dec 3 '18 at 13:08