# Example of a finitely generated module X such that End(X) is not finitely generated [closed]

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.

## closed as off-topic by Thomas, Daniel W. Farlow, Namaste, Xander Henderson, user26857Oct 14 '17 at 18:28

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Thomas, Daniel W. Farlow, Namaste, Xander Henderson, user26857
If this question can be reworded to fit the rules in the help center, please edit the question.

• What is the meaning of 'End X' here? – nurun nesha Oct 10 '17 at 19:43
• @mathiu_lady $Hom_R(X,X)$ – Sky Oct 11 '17 at 0:18

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 $$\text{Hom}_R(R/V,R)\cong V.$$
• @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
• @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 6:32