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If $R$ is a commutative Noetherian ring, then $\mathrm{Hom}_R(X,Y)$ is finitely generated $R$-module whenever $X$ and $Y$ are finite generated $R$-modules.

If $R$ is a commutative non-Noetherian ring I want find an example such that $\mathrm{End}_R(X)$ is not finitely generated $R$-module where $X$ is a finite generated $R$-module.

user26857
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Jian
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1 Answers1

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Let $k$ be a field, $V$ an infinite dimensional vector space over $k$ and $R$ the ring with additive group $k\oplus V$and multiplication $(x,u)(y,v)=(xy,xv+yu)$, so that $V$ is an infinitely generated square zero ideal of $R$.

Take $X=R\oplus R/V$, which is clearly finitely generated.

As an $R$-module, $\text{End}_R(X)$ contains a direct summand $$\text{Hom}_R(R/V,R)\cong V.$$

Jeremy Rickard
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  • So perfect example!I know the example is trivial extension algebra.But I only konw the definition and don't understand.I think it is important as you gave the counter example.How does trivial extension algebra play a role in algebra and representation theory? – Jian Oct 14 '17 at 14:24
  • @Sky You can use a different approach. Take $(R,\mathfrak m)$ a local ring such that $\mathfrak m$ is not finitely generated and $\mathfrak m^2=0$. Then set $M=R\oplus R/\mathfrak m$. (An example of $R$ is $K[X_1,\dots,X_n,\dots]/(X_1,\dots,X_n,\dots)^2$.) – user26857 Oct 14 '17 at 20:27
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    @user26857 yeah.This maybe better.thank you. – Jian Oct 15 '17 at 00:38
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    @Sky Although you're right that this is a (particularly trivial) example of a trivial extension, that's not why I used it. It just happens to be one of my favourite examples (because it's so simple) of a non-Noetherian commutative ring. In fact, if you take the vector space $V$ to have countable dimension, it's exactly the same as user26857's example. – Jeremy Rickard Oct 15 '17 at 06:32