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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.

Can you please help? Thank you!

Community
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Roun
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    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
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    @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 02:34

4 Answers4

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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 $\mathbb R[X_1,X_2,...,X_n,...]=\bigcup_{k\geq 1}\mathbb R[X_1,X_2,...,X_k] \;$ )

Georges Elencwajg
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  • 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 08:51
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    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 09:19
  • Helpful answer! Just a question: what do you mean by $A=A^1$ and why is that module of dimension one/finitely generated? – dahemar May 01 '21 at 18:50
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    @dahemar: I just mean that if $n=1$, then $A^n=A$. In other words the module of $1$-tuples over $A$ is just $A$ seen as an $A$-module. It is finitely generated by the $1$-tuple $(1)$ since any $1$-tuple $(a)$ can be written $(a)=a.(1)$ This is so trivial that it gets confusing and I fear that the more I explain, the more confusing it will get. So I'll just stop here :-) – Georges Elencwajg May 01 '21 at 19:46
  • @GeorgesElencwajg Thank you! I’ve just started learning about modules so I can sometimes find trivial facts confusing! So, just to make sure I understood correctly: you’re just saying that every module $R$ is finitely generated as an $R$-module, right? – dahemar May 01 '21 at 20:06
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    @dahemar: yes, but you should express that as the (not "every") $R$-module $R$ is finitely generated over $R$. There is only one $R$-module called $R$ ! – Georges Elencwajg May 01 '21 at 21:36
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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.

Marc van Leeuwen
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  • OK I figure out why the ideal $I$ is not finitely generated. If $I$ is generated by $f_i=q_iX+...\ (i=1,...,n)$, arbitary $f\in I$ is a form $f=qX+...$ where $q=m_1 q_1+...+m_n q_n\ (m_i\in \mathbb{Z})$. Possibility of denominator of such $q$ is finite, this is contradiction! – user682141 Mar 01 '23 at 07:48
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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.

Romanda de Gore
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I was stuck with the problem as well and I found many examples but they were quite vague (in the sense that there is not a clear reasoning). Now let me restate an example and make it as clear as crystal!

A ring $R$ is a $R$-module itself. Its submodules are just its ideals. Let us consider $R=\mathbb R[x_1,x_2, ..., x_n, ...]$ and $I=\langle x_1,x_2,..,x_n,...,\rangle$.

$R$ is finitely generated by $1_{\mathbb R}$ (as an element in $R$). (Yes! I have stated the generator clearly!)

We claim that $I$ is not finitely generated (by elements in $R$).

Proof:

We first consider the case that $I$ can be generated by a single generator. First we note that, any non-zero constant $c\in \mathbb R$ can not generate $I$ because $1_{\mathbb R} \cdot c=c\notin I.$ Next, we note that for all $j\geq 1$, $x_j$ cannot generate $I$ because there is no $r\in R$ such that $r\cdot x_j =x_k$ for $k\neq j$. Similar reasoning works for all polynomials with degree $1$, by noting that a polynomial has only finitely many terms but we have infinitely many $x_i$'s in $I$. The case that the generator (as a polynomial) has degree $\ge 2$ is even simpler: if the leading term is $a_n x_n^s$ for some $s\ge 2$, then $x_n (\in I)$ can not be generated. So, $I$ cannot be generated by a single generator.

The case of finitely many generators is similar. The essence is that, we have infinitely many $x_i$'s but the generating set is finite. Together with a similar reasoning as above, we can prove the claim.

Bowei Tang
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Sam Wong
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