# Example of a finitely generated module with submodules that are not finitely generated

I'm looking for an example of a finitely generated module with submodules that are not finitely generated.

I've found a similar question dealing with group (i.e. an example of a finitely generated group with subgroups that are not finitely generated). But I can't figure out whether that question do help to this one.

And I actually want to find a more "module-like" example rather than an example derived from a 'strange' group.

• A submodule of a finitely generated module over a noetherian ring is finitely generated. Find a non-noetherian ring. The regular module is finitely generated, but by definition it has submodules that are not. – Jack Schmidt Nov 17 '11 at 15:12
• @JackSchmidt: I'm sorry that I do not know enough examples for such a ring. Could you please give a specific example? Thank you! – Roun Nov 17 '11 at 15:28
• Plus one because over a month ago I proved that every quotient of a fin gen module is fin gen and then made the mistake of assuming sub mods were too. – Prince M Mar 19 '17 at 2:34

Here's a fairly simple example (of a non-Noetherian ring): the ring $R$ of polynomials in one indeterminate $X$ having rational coefficients but with an integer constant term. Its ideal of elements with zero constant term is not finitely generated as an $R$-module.
The ideal $I=\langle X_1,X_2,...,X_n,... \rangle \subset \mathbb R[X_1,X_2,...,X_n,...]=A$ can be seen as a submodule of the free $A$-module of dimension one $A=A^1$, and that module is not finitely generated. Do you see why?
(Hint: even in a polynomial ring with infinitely many indeterminates, each polynomial involves only finitely many variables. In other words $\mathbb R[X_1,X_2,...,X_n,...]=\bigcup_{k\geq 1}\mathbb R[X_1,X_2,...,X_k] \;$ )
• Dear Georges. Can one argue as follows? Assume $I$ was finitely generated. Then $I = \langle X_{n_1}, \dots , X_{n_N} \rangle$. Without loss of generality, we may assume $n_1 < \dots < n_N$. Then the polynomial $p(X_1, X_2, \dots) = X_{n_N + 1}$ is not in $I$. Hence $I$ cannot be finitely generated. – Rudy the Reindeer Jul 23 '12 at 8:51
• Dear Matt, beware that if $I$ were finitely generated there is no reason that it would be generated by monomials $X_{n_i}$ ! – Georges Elencwajg Jul 23 '12 at 9:19
Regarding the "hint" above. Let $$f \in I$$. Then $$f$$ has no constant term and has only finitely many of the variables $$X_i$$. Now suppose that $$f_1,\dots,f_k \in I$$ generate $$I$$ over $$A$$, in other words, suppose that $$I$$ is finitely generated over $$A$$. Let $$n$$ be a large enough natural number so that none of the $$f_i$$ contains the variable $$X_n$$. Now $$X_n \in I$$ therefore $$X_n= p_1f_1 + \ldots + p_kf_k$$. In this expression, set $$X_1= \ldots = X_{n-1}=0$$ and $$X_n=X_{n+1}=\ldots = 1$$. One gets that $$0=1$$, a contradiction.