Quotient ring local if Ring is local












1














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 $R-R^{times} = {rin R , | r notin R^times }$ 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







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  • 2




    This follows trivially from the ideal correspondence theorem for quotient rings.
    – rschwieb
    2 days ago
















1














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 $R-R^{times} = {rin R , | r notin R^times }$ 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







share|cite|improve this question


















  • 2




    This follows trivially from the ideal correspondence theorem for quotient rings.
    – rschwieb
    2 days ago














1












1








1







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 $R-R^{times} = {rin R , | r notin R^times }$ 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







share|cite|improve this question













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 $R-R^{times} = {rin R , | r notin R^times }$ 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






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










asked 2 days ago









KingDingelingKingDingeling

856




856








  • 2




    This follows trivially from the ideal correspondence theorem for quotient rings.
    – rschwieb
    2 days ago














  • 2




    This follows trivially from the ideal correspondence theorem for quotient rings.
    – rschwieb
    2 days ago








2




2




This follows trivially from the ideal correspondence theorem for quotient rings.
– rschwieb
2 days ago




This follows trivially from the ideal correspondence theorem for quotient rings.
– rschwieb
2 days ago










2 Answers
2






active

oldest

votes


















2














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.






share|cite|improve this answer





















  • Thank you a lot, that really helped.
    – KingDingeling
    2 days ago










  • @King You're welcome.
    – Matt Samuel
    2 days ago



















3














By definition, $R$ is a local ring if and only if it has a unique maximal ideal $mathfrak{m}.$ 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 $mathfrak{m}$. So, $Isubseteq mathfrak{m}$. The lattice isomorphism tells us that the ideals of $R/I$ are all contained in $pi(mathfrak{m})$. $widetilde{mathfrak{m}}=pi(mathfrak{m})$ 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 $pi(mathfrak{m})$ and hence is local.






share|cite|improve this answer





















  • Thanks for taking the time and helping me out, appreciate it.
    – KingDingeling
    2 days ago










  • Not a problem. Good luck
    – Antonios-Alexandros Robotis
    2 days ago











Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














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.






share|cite|improve this answer





















  • Thank you a lot, that really helped.
    – KingDingeling
    2 days ago










  • @King You're welcome.
    – Matt Samuel
    2 days ago
















2














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.






share|cite|improve this answer





















  • Thank you a lot, that really helped.
    – KingDingeling
    2 days ago










  • @King You're welcome.
    – Matt Samuel
    2 days ago














2












2








2






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 2 days ago









Matt SamuelMatt Samuel

37.4k63665




37.4k63665












  • Thank you a lot, that really helped.
    – KingDingeling
    2 days ago










  • @King You're welcome.
    – Matt Samuel
    2 days ago


















  • Thank you a lot, that really helped.
    – KingDingeling
    2 days ago










  • @King You're welcome.
    – Matt Samuel
    2 days ago
















Thank you a lot, that really helped.
– KingDingeling
2 days ago




Thank you a lot, that really helped.
– KingDingeling
2 days ago












@King You're welcome.
– Matt Samuel
2 days ago




@King You're welcome.
– Matt Samuel
2 days ago











3














By definition, $R$ is a local ring if and only if it has a unique maximal ideal $mathfrak{m}.$ 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 $mathfrak{m}$. So, $Isubseteq mathfrak{m}$. The lattice isomorphism tells us that the ideals of $R/I$ are all contained in $pi(mathfrak{m})$. $widetilde{mathfrak{m}}=pi(mathfrak{m})$ 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 $pi(mathfrak{m})$ and hence is local.






share|cite|improve this answer





















  • Thanks for taking the time and helping me out, appreciate it.
    – KingDingeling
    2 days ago










  • Not a problem. Good luck
    – Antonios-Alexandros Robotis
    2 days ago
















3














By definition, $R$ is a local ring if and only if it has a unique maximal ideal $mathfrak{m}.$ 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 $mathfrak{m}$. So, $Isubseteq mathfrak{m}$. The lattice isomorphism tells us that the ideals of $R/I$ are all contained in $pi(mathfrak{m})$. $widetilde{mathfrak{m}}=pi(mathfrak{m})$ 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 $pi(mathfrak{m})$ and hence is local.






share|cite|improve this answer





















  • Thanks for taking the time and helping me out, appreciate it.
    – KingDingeling
    2 days ago










  • Not a problem. Good luck
    – Antonios-Alexandros Robotis
    2 days ago














3












3








3






By definition, $R$ is a local ring if and only if it has a unique maximal ideal $mathfrak{m}.$ 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 $mathfrak{m}$. So, $Isubseteq mathfrak{m}$. The lattice isomorphism tells us that the ideals of $R/I$ are all contained in $pi(mathfrak{m})$. $widetilde{mathfrak{m}}=pi(mathfrak{m})$ 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 $pi(mathfrak{m})$ and hence is local.






share|cite|improve this answer












By definition, $R$ is a local ring if and only if it has a unique maximal ideal $mathfrak{m}.$ 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 $mathfrak{m}$. So, $Isubseteq mathfrak{m}$. The lattice isomorphism tells us that the ideals of $R/I$ are all contained in $pi(mathfrak{m})$. $widetilde{mathfrak{m}}=pi(mathfrak{m})$ 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 $pi(mathfrak{m})$ and hence is local.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 2 days ago









Antonios-Alexandros RobotisAntonios-Alexandros Robotis

9,63241640




9,63241640












  • Thanks for taking the time and helping me out, appreciate it.
    – KingDingeling
    2 days ago










  • Not a problem. Good luck
    – Antonios-Alexandros Robotis
    2 days ago


















  • Thanks for taking the time and helping me out, appreciate it.
    – KingDingeling
    2 days ago










  • Not a problem. Good luck
    – Antonios-Alexandros Robotis
    2 days ago
















Thanks for taking the time and helping me out, appreciate it.
– KingDingeling
2 days ago




Thanks for taking the time and helping me out, appreciate it.
– KingDingeling
2 days ago












Not a problem. Good luck
– Antonios-Alexandros Robotis
2 days ago




Not a problem. Good luck
– Antonios-Alexandros Robotis
2 days ago


















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