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user91931
user91931

Let $K$ be a field of characteristic $0$ and let $R$ be a (noncommutative) Noetherian $K$-algebra. Let $M$ and $N$ be simple left $R$-modules and assume futherfurther the following conditions on $R$:

  • (Stably Noetherian property) For any field extension $L$ of $K$, the ring $R\otimes_KL$ is again Noetherian.
  • (Nullstellensatz property) For any simple left $R$-module $P$, the endomorphism ring $End_R(P)$ is an algebraic ring extension of $K$.

I have shown that under thisthese conditions, $Hom_R(M,N)$ is a finite dimensional vector space over $K$. I would like to transfer the finite dimensionality result to any $Ext_R^n(M,N)$ for $n\geq0$. I failed to prove this by a dimension shift.

Is there any result from homological algebra which guarantees the finite dimensionality of $Ext$ under the condition that it is already shown for $Hom$? If no, are there any (weak) additional requirements to the algebra $R$ which would guarantee the finite dimensionality?

Let $K$ be a field of characteristic $0$ and let $R$ be a (noncommutative) Noetherian $K$-algebra. Let $M$ and $N$ be simple left $R$-modules and assume futher the following conditions on $R$:

  • (Stably Noetherian property) For any field extension $L$ of $K$, the ring $R\otimes_KL$ is again Noetherian.
  • (Nullstellensatz property) For any simple left $R$-module $P$, the endomorphism ring $End_R(P)$ is an algebraic ring extension of $K$.

I have shown that under this conditions, $Hom_R(M,N)$ is a finite dimensional vector space over $K$. I would like to transfer the finite dimensionality result to any $Ext_R^n(M,N)$ for $n\geq0$. I failed to prove this by a dimension shift.

Is there any result from homological algebra which guarantees the finite dimensionality of $Ext$ under the condition that it is already shown for $Hom$? If no, are there any (weak) additional requirements to the algebra $R$ which would guarantee the finite dimensionality?

Let $K$ be a field of characteristic $0$ and let $R$ be a (noncommutative) Noetherian $K$-algebra. Let $M$ and $N$ be simple left $R$-modules and assume further the following conditions on $R$:

  • (Stably Noetherian property) For any field extension $L$ of $K$, the ring $R\otimes_KL$ is again Noetherian.
  • (Nullstellensatz property) For any simple left $R$-module $P$, the endomorphism ring $End_R(P)$ is an algebraic ring extension of $K$.

I have shown that under these conditions, $Hom_R(M,N)$ is a finite dimensional vector space over $K$. I would like to transfer the finite dimensionality result to any $Ext_R^n(M,N)$ for $n\geq0$. I failed to prove this by a dimension shift.

Is there any result from homological algebra which guarantees the finite dimensionality of $Ext$ under the condition that it is already shown for $Hom$? If no, are there any (weak) additional requirements to the algebra $R$ which would guarantee the finite dimensionality?

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user91931
user91931

Finite dimensionality of Ext(M,N)

Let $K$ be a field of characteristic $0$ and let $R$ be a (noncommutative) Noetherian $K$-algebra. Let $M$ and $N$ be simple left $R$-modules and assume futher the following conditions on $R$:

  • (Stably Noetherian property) For any field extension $L$ of $K$, the ring $R\otimes_KL$ is again Noetherian.
  • (Nullstellensatz property) For any simple left $R$-module $P$, the endomorphism ring $End_R(P)$ is an algebraic ring extension of $K$.

I have shown that under this conditions, $Hom_R(M,N)$ is a finite dimensional vector space over $K$. I would like to transfer the finite dimensionality result to any $Ext_R^n(M,N)$ for $n\geq0$. I failed to prove this by a dimension shift.

Is there any result from homological algebra which guarantees the finite dimensionality of $Ext$ under the condition that it is already shown for $Hom$? If no, are there any (weak) additional requirements to the algebra $R$ which would guarantee the finite dimensionality?