In an assignment question, I'm a bit stumped by the following:

Let $R$ be a commutative ring, $M$ be an $R$-module and $N$ an $R$-submodule of $M$. I was first asked to prove that if both $N$ and $M/N$ are finitely generated then $M$ is finitely generated, which was fairly straightforward. The second part of the question asks the reader to show that the converse is not true in general, and a hint is given:

"Take, for example, $R:=\mathbf{Z}[X_1,...,X_n,...]$, the polynomial ring over $\mathbf{Z}$ in finitely many variables $X_1,...,X_n,...$, with the obvious operations. Keep in mind that a polynomial in $R$ can involve only finitely many variables, and observe that $R$ is an example of a non-noetherian ring."

I get that if $M$ is finitely generated over $R$ then so is $M/N$, so the counter-example the reader is supposed to find is a submodule that is not finitely generated. The solutions to the assignment use $R$ as in the hint above and define $M:=R$ and so consider $M$ as a module over itself. Clearly it is finitely generated over itself, as given a polynomial $f$ in $\mathbf{Z}[X_1,...,X_n,...]$, it is equal to a copy of itself in $R$, hence $f=rg$ for $r\in R$ and $m\in M$, namely $r=f$ and $g=1$. So the polynomial $1$ generates $M$. The solutions then define $N:=(X_1,...,X_n,...)$ and state that $N$ is a counter-example.

My first problem is that I'm not sure what the notation $(X_1,...,X_n,...)$ means. If it's anything like what it means when talking about ideals, then elements of $N$ are finite $R$-linear combinations of $X_1,...,X_n,...$. Given that this is correct, my second problem is that I don't understand why this isn't finitely generated. I'm not saying that I think it is finitely generated, I just can't tell either way. By the way the solutions are written, I'm gathering that there must be something concrete that prevents $N$ from being finitely generated, but I can't see it.

Edit: I can see that $(X_1,...,X_n,...)$ might be by definition not finitely generated, but couldn't some of the X_i's here be superfluous? I.e. the module generated by a finite subset of elements might be equal to $N$?

Thanks for any replies.

  • $\begingroup$ Or is it by definition not finitely generated? because (X_1,...,X_n,...) generate it and there are infinitely many... $\endgroup$
    – Lammey
    Aug 6, 2014 at 16:24
  • $\begingroup$ Just because $N$ is generated by $X_1,X_2,...,X_n,...$, could it not also be generated by a smaller subset of those generators, i.e. some of the generators could be superfluous? $\endgroup$
    – Lammey
    Aug 6, 2014 at 16:47

1 Answer 1


Just as a heads-up, take any non-Noetherian ring $R$. Then $R$ is itself a finitely-generated $R$-module (generated by $1$), and so is $R/\mathfrak a$ for any ideal $\mathfrak a \trianglelefteq R$ (since it is again generated by $1$), but your ideal $\mathfrak a$ can be picked non-finitely generated if $R$ is picked not Noetherian.

About your example, suppose that $(x_1,\cdots,x_n,\cdots)$ was finitely generated by polynomials $p_1,\cdots,p_m$. These $m$ polynomials involve finitely many variables, so we can pick a variable $x_j$ which is not contained in the expression of $p_1,\cdots,p_m$ ; since the $p_1,\cdots,p_m$ cannot express the polynomial $x_j$ in the form $\sum_{k=1}^m f_k p_k$ (obviously, since all the terms of the polynomials $p_i$ have degree at least $1$ in variables other than $x_j$), this means that $(p_1,\cdots,p_m) \subsetneq (x_1,\cdots,x_n,\cdots)$.

Hope that helps,

  • $\begingroup$ Ah ok I see, thanks a lot. $\endgroup$
    – Lammey
    Aug 6, 2014 at 17:02

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