Quotient ring local if Ring is local
Clash Royale CLAN TAG#URR8PPP
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I want to show: If $R$ is local and $Ineq R$ an ideal, then $R/I$ is also local.
We already know: A Ring $R$ is local if and only if $R-R^times = rin R , $ is an ideal. We also know that if $I$ is an maximal Ideal, $R/I$ is a field.
I have no idea how to show that other than using the above Lemma.
abstract-algebra ring-theory factoring quotient-group local-rings
asked Jan 5 at 21:10
KingDingelingKingDingeling
1366
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add a comment |
$begingroup$
I want to show: If $R$ is local and $Ineq R$ an ideal, then $R/I$ is also local.
We already know: A Ring $R$ is local if and only if $R-R^times = rin R , $ is an ideal. We also know that if $I$ is an maximal Ideal, $R/I$ is a field.
I have no idea how to show that other than using the above Lemma.
abstract-algebra ring-theory factoring quotient-group local-rings
asked Jan 5 at 21:10
KingDingelingKingDingeling
1366
$endgroup$
2
$begingroup$
This follows trivially from the ideal correspondence theorem for quotient rings.
$endgroup$
– rschwieb
Jan 5 at 21:21
add a comment |
$begingroup$
I want to show: If $R$ is local and $Ineq R$ an ideal, then $R/I$ is also local.
We already know: A Ring $R$ is local if and only if $R-R^times = rin R , $ is an ideal. We also know that if $I$ is an maximal Ideal, $R/I$ is a field.
I have no idea how to show that other than using the above Lemma.
abstract-algebra ring-theory factoring quotient-group local-rings
asked Jan 5 at 21:10
KingDingelingKingDingeling
1366
$endgroup$
I want to show: If $R$ is local and $Ineq R$ an ideal, then $R/I$ is also local.
We already know: A Ring $R$ is local if and only if $R-R^times = rin R , $ is an ideal. We also know that if $I$ is an maximal Ideal, $R/I$ is a field.
I have no idea how to show that other than using the above Lemma.
abstract-algebra ring-theory factoring quotient-group local-rings
abstract-algebra ring-theory factoring quotient-group local-rings
asked Jan 5 at 21:10
KingDingelingKingDingeling
1366
asked Jan 5 at 21:10
KingDingelingKingDingeling
1366
asked Jan 5 at 21:10
KingDingelingKingDingeling
1366
asked Jan 5 at 21:10
KingDingelingKingDingeling
1366
asked Jan 5 at 21:10
KingDingelingKingDingeling
1366
1366
2
$begingroup$
This follows trivially from the ideal correspondence theorem for quotient rings.
$endgroup$
– rschwieb
Jan 5 at 21:21
add a comment |
2
$begingroup$
This follows trivially from the ideal correspondence theorem for quotient rings.
$endgroup$
– rschwieb
Jan 5 at 21:21
2
2
$begingroup$
This follows trivially from the ideal correspondence theorem for quotient rings.
$endgroup$
– rschwieb
Jan 5 at 21:21
$begingroup$
This follows trivially from the ideal correspondence theorem for quotient rings.
$endgroup$
– rschwieb
Jan 5 at 21:21
add a comment |
2 Answers
2
active
oldest
votes
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The easiest characterization of local rings here is that a ring is local if and only if it has exactly one maximal ideal. Let $M$ be the maximal ideal of $R$ and $f:Rto R/I$ the surjection. If $R/I$ has a maximal ideal $M'neq f(M) $, then $f^-1(M')neq M$ is a maximal ideal of $R$, which is a contradiction.
answered Jan 5 at 21:22
Matt SamuelMatt Samuel
37.7k63665
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Thank you a lot, that really helped.
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– KingDingeling
Jan 5 at 21:26
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@King You're welcome.
$endgroup$
– Matt Samuel
Jan 5 at 21:27
add a comment |
$begingroup$
By definition, $R$ is a local ring if and only if it has a unique maximal ideal $mathfrakm.$ The lattice isomorphism theorem for commutative rings states that the quotient homomorphism $pi:Rto R/I$ induces an inclusion preserving correspondence between ideals of $R$ containing $I$ and ideals of $R/I$.
Every ideal $Isubseteq R$ is contained in a (unique) maximal ideal. $R$ is local so that the only maximal ideal is $mathfrakm$. So, $Isubseteq mathfrakm$. The lattice isomorphism tells us that the ideals of $R/I$ are all contained in $pi(mathfrakm)$. $widetildemathfrakm=pi(mathfrakm)$ is a maximal ideal in $R/I$ (check this) so that it is also the unique maximal ideal in $R/I$ since any other maximal ideal is contained in it by the lattice isomorphism theorem.
So, $R/I$ has unique maximal ideal $pi(mathfrakm)$ and hence is local.
answered Jan 5 at 21:19
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,75241640
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$begingroup$
Thanks for taking the time and helping me out, appreciate it.
$endgroup$
– KingDingeling
Jan 5 at 21:27
$begingroup$
Not a problem. Good luck
$endgroup$
– Antonios-Alexandros Robotis
Jan 5 at 21:37
add a comment |
Your Answer
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
The easiest characterization of local rings here is that a ring is local if and only if it has exactly one maximal ideal. Let $M$ be the maximal ideal of $R$ and $f:Rto R/I$ the surjection. If $R/I$ has a maximal ideal $M'neq f(M) $, then $f^-1(M')neq M$ is a maximal ideal of $R$, which is a contradiction.
answered Jan 5 at 21:22
Matt SamuelMatt Samuel
37.7k63665
$endgroup$
$begingroup$
Thank you a lot, that really helped.
$endgroup$
– KingDingeling
Jan 5 at 21:26
$begingroup$
@King You're welcome.
$endgroup$
– Matt Samuel
Jan 5 at 21:27
add a comment |
$begingroup$
The easiest characterization of local rings here is that a ring is local if and only if it has exactly one maximal ideal. Let $M$ be the maximal ideal of $R$ and $f:Rto R/I$ the surjection. If $R/I$ has a maximal ideal $M'neq f(M) $, then $f^-1(M')neq M$ is a maximal ideal of $R$, which is a contradiction.
answered Jan 5 at 21:22
Matt SamuelMatt Samuel
37.7k63665
$endgroup$
$begingroup$
Thank you a lot, that really helped.
$endgroup$
– KingDingeling
Jan 5 at 21:26
$begingroup$
@King You're welcome.
$endgroup$
– Matt Samuel
Jan 5 at 21:27
add a comment |
$begingroup$
The easiest characterization of local rings here is that a ring is local if and only if it has exactly one maximal ideal. Let $M$ be the maximal ideal of $R$ and $f:Rto R/I$ the surjection. If $R/I$ has a maximal ideal $M'neq f(M) $, then $f^-1(M')neq M$ is a maximal ideal of $R$, which is a contradiction.
answered Jan 5 at 21:22
Matt SamuelMatt Samuel
37.7k63665
$endgroup$
The easiest characterization of local rings here is that a ring is local if and only if it has exactly one maximal ideal. Let $M$ be the maximal ideal of $R$ and $f:Rto R/I$ the surjection. If $R/I$ has a maximal ideal $M'neq f(M) $, then $f^-1(M')neq M$ is a maximal ideal of $R$, which is a contradiction.
answered Jan 5 at 21:22
Matt SamuelMatt Samuel
37.7k63665
answered Jan 5 at 21:22
Matt SamuelMatt Samuel
37.7k63665
answered Jan 5 at 21:22
Matt SamuelMatt Samuel
37.7k63665
answered Jan 5 at 21:22
Matt SamuelMatt Samuel
37.7k63665
37.7k63665
$begingroup$
Thank you a lot, that really helped.
$endgroup$
– KingDingeling
Jan 5 at 21:26
$begingroup$
@King You're welcome.
$endgroup$
– Matt Samuel
Jan 5 at 21:27
add a comment |
$begingroup$
Thank you a lot, that really helped.
$endgroup$
– KingDingeling
Jan 5 at 21:26
$begingroup$
@King You're welcome.
$endgroup$
– Matt Samuel
Jan 5 at 21:27
$begingroup$
Thank you a lot, that really helped.
$endgroup$
– KingDingeling
Jan 5 at 21:26
$begingroup$
Thank you a lot, that really helped.
$endgroup$
– KingDingeling
Jan 5 at 21:26
$begingroup$
@King You're welcome.
$endgroup$
– Matt Samuel
Jan 5 at 21:27
$begingroup$
@King You're welcome.
$endgroup$
– Matt Samuel
Jan 5 at 21:27
add a comment |
$begingroup$
By definition, $R$ is a local ring if and only if it has a unique maximal ideal $mathfrakm.$ The lattice isomorphism theorem for commutative rings states that the quotient homomorphism $pi:Rto R/I$ induces an inclusion preserving correspondence between ideals of $R$ containing $I$ and ideals of $R/I$.
Every ideal $Isubseteq R$ is contained in a (unique) maximal ideal. $R$ is local so that the only maximal ideal is $mathfrakm$. So, $Isubseteq mathfrakm$. The lattice isomorphism tells us that the ideals of $R/I$ are all contained in $pi(mathfrakm)$. $widetildemathfrakm=pi(mathfrakm)$ is a maximal ideal in $R/I$ (check this) so that it is also the unique maximal ideal in $R/I$ since any other maximal ideal is contained in it by the lattice isomorphism theorem.
So, $R/I$ has unique maximal ideal $pi(mathfrakm)$ and hence is local.
answered Jan 5 at 21:19
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,75241640
$endgroup$
$begingroup$
Thanks for taking the time and helping me out, appreciate it.
$endgroup$
– KingDingeling
Jan 5 at 21:27
$begingroup$
Not a problem. Good luck
$endgroup$
– Antonios-Alexandros Robotis
Jan 5 at 21:37
add a comment |
$begingroup$
By definition, $R$ is a local ring if and only if it has a unique maximal ideal $mathfrakm.$ The lattice isomorphism theorem for commutative rings states that the quotient homomorphism $pi:Rto R/I$ induces an inclusion preserving correspondence between ideals of $R$ containing $I$ and ideals of $R/I$.
Every ideal $Isubseteq R$ is contained in a (unique) maximal ideal. $R$ is local so that the only maximal ideal is $mathfrakm$. So, $Isubseteq mathfrakm$. The lattice isomorphism tells us that the ideals of $R/I$ are all contained in $pi(mathfrakm)$. $widetildemathfrakm=pi(mathfrakm)$ is a maximal ideal in $R/I$ (check this) so that it is also the unique maximal ideal in $R/I$ since any other maximal ideal is contained in it by the lattice isomorphism theorem.
So, $R/I$ has unique maximal ideal $pi(mathfrakm)$ and hence is local.
answered Jan 5 at 21:19
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,75241640
$endgroup$
$begingroup$
Thanks for taking the time and helping me out, appreciate it.
$endgroup$
– KingDingeling
Jan 5 at 21:27
$begingroup$
Not a problem. Good luck
$endgroup$
– Antonios-Alexandros Robotis
Jan 5 at 21:37
add a comment |
$begingroup$
By definition, $R$ is a local ring if and only if it has a unique maximal ideal $mathfrakm.$ The lattice isomorphism theorem for commutative rings states that the quotient homomorphism $pi:Rto R/I$ induces an inclusion preserving correspondence between ideals of $R$ containing $I$ and ideals of $R/I$.
Every ideal $Isubseteq R$ is contained in a (unique) maximal ideal. $R$ is local so that the only maximal ideal is $mathfrakm$. So, $Isubseteq mathfrakm$. The lattice isomorphism tells us that the ideals of $R/I$ are all contained in $pi(mathfrakm)$. $widetildemathfrakm=pi(mathfrakm)$ is a maximal ideal in $R/I$ (check this) so that it is also the unique maximal ideal in $R/I$ since any other maximal ideal is contained in it by the lattice isomorphism theorem.
So, $R/I$ has unique maximal ideal $pi(mathfrakm)$ and hence is local.
answered Jan 5 at 21:19
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,75241640
$endgroup$
By definition, $R$ is a local ring if and only if it has a unique maximal ideal $mathfrakm.$ The lattice isomorphism theorem for commutative rings states that the quotient homomorphism $pi:Rto R/I$ induces an inclusion preserving correspondence between ideals of $R$ containing $I$ and ideals of $R/I$.
Every ideal $Isubseteq R$ is contained in a (unique) maximal ideal. $R$ is local so that the only maximal ideal is $mathfrakm$. So, $Isubseteq mathfrakm$. The lattice isomorphism tells us that the ideals of $R/I$ are all contained in $pi(mathfrakm)$. $widetildemathfrakm=pi(mathfrakm)$ is a maximal ideal in $R/I$ (check this) so that it is also the unique maximal ideal in $R/I$ since any other maximal ideal is contained in it by the lattice isomorphism theorem.
So, $R/I$ has unique maximal ideal $pi(mathfrakm)$ and hence is local.
answered Jan 5 at 21:19
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,75241640
answered Jan 5 at 21:19
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,75241640
answered Jan 5 at 21:19
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,75241640
answered Jan 5 at 21:19
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,75241640
9,75241640
$begingroup$
Thanks for taking the time and helping me out, appreciate it.
$endgroup$
– KingDingeling
Jan 5 at 21:27
$begingroup$
Not a problem. Good luck
$endgroup$
– Antonios-Alexandros Robotis
Jan 5 at 21:37
add a comment |
$begingroup$
Thanks for taking the time and helping me out, appreciate it.
$endgroup$
– KingDingeling
Jan 5 at 21:27
$begingroup$
Not a problem. Good luck
$endgroup$
– Antonios-Alexandros Robotis
Jan 5 at 21:37
$begingroup$
Thanks for taking the time and helping me out, appreciate it.
$endgroup$
– KingDingeling
Jan 5 at 21:27
$begingroup$
Thanks for taking the time and helping me out, appreciate it.
$endgroup$
– KingDingeling
Jan 5 at 21:27
$begingroup$
Not a problem. Good luck
$endgroup$
– Antonios-Alexandros Robotis
Jan 5 at 21:37
$begingroup$
Not a problem. Good luck
$endgroup$
– Antonios-Alexandros Robotis
Jan 5 at 21:37
add a comment |
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$begingroup$
This follows trivially from the ideal correspondence theorem for quotient rings.
$endgroup$
– rschwieb
Jan 5 at 21:21