# Infinite linear independent family in a finitely generated $A$-module

So I'm stuck with this problem.

Let $A$ be a nonzero commutative ring (with unit). I have several questions that are really close to each other.

1) Let $M$ be a finitely generated module over $A$. Can we have $\{x_n, n \in \mathbb{N}\} \subset M$ a linear independent family in $M$ ?

2) Let $M$ be a module over $A$ generated by $f_1, \ldots, f_n \in M$. Can we have $x_1,\ldots, x_{n+1} \in M$ a linear independent family ? (Obviously a negative answer to the latter implies a negative answer to the former.)

3) Let $N$ be a free module over $A$, and $M$ a module such that there exists an exact sequence $0 \to N \to M$ and another exact sequence $N \to M \to 0$. Is $M$ a free module as well (is $M \simeq N$)? If not, can we add $N$ of finite rank to get a positive answer ?

4) Let $M$ be finitely generated by $f_1, \ldots, f_n$, and $x_1, \ldots, x_n$ be a linear independent family in $M$. Is $M$ free?

Thank you for your help!

• In (3), are you assuming that the modules are finitely generated? Since if they aren't, the statement must not necessarily be true. To see this, take as $N$ a free $\mathbb{Z}$-module with countable base and for $M$ a $\mathbb{Q}$-vector space with countable base. – Tomasz Lenarcik Nov 22 '13 at 9:15

Well, the key words are onto endomorphisms are isomorphisms. (Actually, the paper proves more than the title says!)

2) The answer is negative and as you remarked this also solves 1). Let $F$ be the submodule of $M$ generated by $x_1,\dots, x_{n+1}$. We have an isomorphism $F\to A^{n+1}$. Since $M$ is generated by $n$ elements there exists $K\le A^n$ such that $A^n/K\to M$ is an isomorphism. Now take the canonical projection $p:A^{n+1}\to A^n$ which sends $(a_1,\dots,a_{n+1})$ to $(a_1,\dots,a_{n})$, and thus we get an onto homomorphism $F\to M$. This must be an isomorphism, and we get that $p$ is injective, a contradiction.

4) Yes, it is! Use similar arguments as above to get that $M\cong A^n$.

3) This follows immediately from Orzech's Theorem.

Remark. A proof of Orzech's result can be found on M.SE in this answer.

(Another proof of these questions, based on McCoy's Theorem, can be found in this paper.)

• Sorry I didn't checked the answer yet, I had a busy weekend... by onto homomorphism, do you mean surjective $A$-linear map? So because M is generated by $n$ elements I have a surjective map $A^{n}$ and because $F$ is isomorph to $A^{n+1}$ I get another surjective map from $F$ to $A^{n }$ I compose the two and that's your onto homomorphism from $F$ to $M$ ? I can't understand why you tell me about $K$... thanks for your quick answer though :) – jeanmfischer Nov 25 '13 at 14:56
• @jeanmfischer As you could have seen in the first link I gave you, onto = epi = surjective. On the other side, I don't understand your trouble with $K$ as long as a module is finitely generated iff there exists a surjective (or onto, if you like it) homomorphism $A^n\to M$ (for some $n$) iff $M$ is isomorphic to a quotient of a free module of finite rank, that is, $M\cong A^n/K$ (for some $n$ and some $K$). – user89712 Nov 25 '13 at 20:46