How to show that $Aa/Ja cong Af/Jf$?

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Let $A$ be an algebra, $J= operatornameradA$ and let $e,f$ be primitive idempotents in $A$. Given a non-zero element $a in fJe$. Is there anyway to show that $Aa/Ja cong Af /Jf$? Thanks.



What I have tried is as follows: Suppose $a=fje$ for some $j in J$, define a map $g: Af rightarrow Aa$ such that $g(m)=mje$ for $m in Af$. There is also a canonical surjective map $pi:Aa rightarrow Aa/Ja$. Thus I otain a surjective homomorphism $pi g:Af rightarrow Aa/Ja$. But I do not know how to show $ker(pi g)=Jf$.




modules representation-theory




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  • I think there's more work involved in showing your map is a homomorphism...
    – Steve D
    Jul 26 at 6:16










  • @Steve D Since $g(rm)=rmje=rg(m)$ for $r in A$ and $pi$ is also a homomorphism. Then their composition $pi g$ is a homomorphism.
    – Xiaosong Peng
    Jul 26 at 10:00














up vote
2
down vote

favorite
1












Let $A$ be an algebra, $J= operatornameradA$ and let $e,f$ be primitive idempotents in $A$. Given a non-zero element $a in fJe$. Is there anyway to show that $Aa/Ja cong Af /Jf$? Thanks.



What I have tried is as follows: Suppose $a=fje$ for some $j in J$, define a map $g: Af rightarrow Aa$ such that $g(m)=mje$ for $m in Af$. There is also a canonical surjective map $pi:Aa rightarrow Aa/Ja$. Thus I otain a surjective homomorphism $pi g:Af rightarrow Aa/Ja$. But I do not know how to show $ker(pi g)=Jf$.




modules representation-theory




share|cite|improve this question





















  • I think there's more work involved in showing your map is a homomorphism...
    – Steve D
    Jul 26 at 6:16










  • @Steve D Since $g(rm)=rmje=rg(m)$ for $r in A$ and $pi$ is also a homomorphism. Then their composition $pi g$ is a homomorphism.
    – Xiaosong Peng
    Jul 26 at 10:00












up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
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Let $A$ be an algebra, $J= operatornameradA$ and let $e,f$ be primitive idempotents in $A$. Given a non-zero element $a in fJe$. Is there anyway to show that $Aa/Ja cong Af /Jf$? Thanks.



What I have tried is as follows: Suppose $a=fje$ for some $j in J$, define a map $g: Af rightarrow Aa$ such that $g(m)=mje$ for $m in Af$. There is also a canonical surjective map $pi:Aa rightarrow Aa/Ja$. Thus I otain a surjective homomorphism $pi g:Af rightarrow Aa/Ja$. But I do not know how to show $ker(pi g)=Jf$.




modules representation-theory




share|cite|improve this question













Let $A$ be an algebra, $J= operatornameradA$ and let $e,f$ be primitive idempotents in $A$. Given a non-zero element $a in fJe$. Is there anyway to show that $Aa/Ja cong Af /Jf$? Thanks.



What I have tried is as follows: Suppose $a=fje$ for some $j in J$, define a map $g: Af rightarrow Aa$ such that $g(m)=mje$ for $m in Af$. There is also a canonical surjective map $pi:Aa rightarrow Aa/Ja$. Thus I otain a surjective homomorphism $pi g:Af rightarrow Aa/Ja$. But I do not know how to show $ker(pi g)=Jf$.





modules representation-theory





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 24 at 7:38









Bernard

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asked Jul 24 at 3:24









Xiaosong Peng

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  • I think there's more work involved in showing your map is a homomorphism...
    – Steve D
    Jul 26 at 6:16










  • @Steve D Since $g(rm)=rmje=rg(m)$ for $r in A$ and $pi$ is also a homomorphism. Then their composition $pi g$ is a homomorphism.
    – Xiaosong Peng
    Jul 26 at 10:00
















  • I think there's more work involved in showing your map is a homomorphism...
    – Steve D
    Jul 26 at 6:16










  • @Steve D Since $g(rm)=rmje=rg(m)$ for $r in A$ and $pi$ is also a homomorphism. Then their composition $pi g$ is a homomorphism.
    – Xiaosong Peng
    Jul 26 at 10:00















I think there's more work involved in showing your map is a homomorphism...
– Steve D
Jul 26 at 6:16




I think there's more work involved in showing your map is a homomorphism...
– Steve D
Jul 26 at 6:16












@Steve D Since $g(rm)=rmje=rg(m)$ for $r in A$ and $pi$ is also a homomorphism. Then their composition $pi g$ is a homomorphism.
– Xiaosong Peng
Jul 26 at 10:00




@Steve D Since $g(rm)=rmje=rg(m)$ for $r in A$ and $pi$ is also a homomorphism. Then their composition $pi g$ is a homomorphism.
– Xiaosong Peng
Jul 26 at 10:00















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