Regular elements on $ operatornameExt^1_R(M,R)$

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Let $A$ be a Gorenstein and seminormal $mathbbC$-algebra of finite type, and let $t in A$ be some non-zero divisor.



If $M$ is a finitely generated $A$-module, and $t$ is an $M$-regular element, is $t$ then an $ operatornameExt^1_A(M,A)$-regular element?



One can equivalently formulate this question as such:



If $T : A^oplus n to A^oplus m$ is a linear map such that $tx in operatornameIm(T) Rightarrow x in operatornameIm(T)$ for all $x in A^oplus m$, does the same hold for its transpose $T^t : A^oplus m to A^oplus n$? Here $n$ and $m$ are arbitrary integers.




linear-algebra abstract-algebra algebraic-geometry commutative-algebra




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  • I suggest you try $A=k[t,u]_(t,u)$ and $M=(t,u)$.
    – Mohan
    Jul 18 at 17:53










  • Thanks, that works.
    – Bubbles
    Jul 18 at 18:11














up vote
1
down vote

favorite












Let $A$ be a Gorenstein and seminormal $mathbbC$-algebra of finite type, and let $t in A$ be some non-zero divisor.



If $M$ is a finitely generated $A$-module, and $t$ is an $M$-regular element, is $t$ then an $ operatornameExt^1_A(M,A)$-regular element?



One can equivalently formulate this question as such:



If $T : A^oplus n to A^oplus m$ is a linear map such that $tx in operatornameIm(T) Rightarrow x in operatornameIm(T)$ for all $x in A^oplus m$, does the same hold for its transpose $T^t : A^oplus m to A^oplus n$? Here $n$ and $m$ are arbitrary integers.




linear-algebra abstract-algebra algebraic-geometry commutative-algebra




share|cite|improve this question





















  • I suggest you try $A=k[t,u]_(t,u)$ and $M=(t,u)$.
    – Mohan
    Jul 18 at 17:53










  • Thanks, that works.
    – Bubbles
    Jul 18 at 18:11












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $A$ be a Gorenstein and seminormal $mathbbC$-algebra of finite type, and let $t in A$ be some non-zero divisor.



If $M$ is a finitely generated $A$-module, and $t$ is an $M$-regular element, is $t$ then an $ operatornameExt^1_A(M,A)$-regular element?



One can equivalently formulate this question as such:



If $T : A^oplus n to A^oplus m$ is a linear map such that $tx in operatornameIm(T) Rightarrow x in operatornameIm(T)$ for all $x in A^oplus m$, does the same hold for its transpose $T^t : A^oplus m to A^oplus n$? Here $n$ and $m$ are arbitrary integers.




linear-algebra abstract-algebra algebraic-geometry commutative-algebra




share|cite|improve this question













Let $A$ be a Gorenstein and seminormal $mathbbC$-algebra of finite type, and let $t in A$ be some non-zero divisor.



If $M$ is a finitely generated $A$-module, and $t$ is an $M$-regular element, is $t$ then an $ operatornameExt^1_A(M,A)$-regular element?



One can equivalently formulate this question as such:



If $T : A^oplus n to A^oplus m$ is a linear map such that $tx in operatornameIm(T) Rightarrow x in operatornameIm(T)$ for all $x in A^oplus m$, does the same hold for its transpose $T^t : A^oplus m to A^oplus n$? Here $n$ and $m$ are arbitrary integers.





linear-algebra abstract-algebra algebraic-geometry commutative-algebra





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 18 at 16:40









Bernard

110k635103




110k635103









asked Jul 18 at 16:01









Bubbles

2813




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  • I suggest you try $A=k[t,u]_(t,u)$ and $M=(t,u)$.
    – Mohan
    Jul 18 at 17:53










  • Thanks, that works.
    – Bubbles
    Jul 18 at 18:11
















  • I suggest you try $A=k[t,u]_(t,u)$ and $M=(t,u)$.
    – Mohan
    Jul 18 at 17:53










  • Thanks, that works.
    – Bubbles
    Jul 18 at 18:11















I suggest you try $A=k[t,u]_(t,u)$ and $M=(t,u)$.
– Mohan
Jul 18 at 17:53




I suggest you try $A=k[t,u]_(t,u)$ and $M=(t,u)$.
– Mohan
Jul 18 at 17:53












Thanks, that works.
– Bubbles
Jul 18 at 18:11




Thanks, that works.
– Bubbles
Jul 18 at 18:11















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