Existence of maximal free submodules

Question

Let $A$ be a domain and $M$ be a finitely generated module. Is there a free submodule which is maximal among free submodules?

Answer is yes for Noetherian rings, obviously. Also the rank of any such submodule is determined by $M$ by tensoring up with $Q(A)$. I haven't got much further than this.

Answer 1

EDIT: This answer is wrong. I wait to delete it because of the ideas it could give. Thanks to Zoe Allen for pointing the error out.

The answer is positive for any ring by Zorn Lemma. If there exists no free submodule of $M$, there is nothing to show.

Otherwise, consider the poset $$\Sigma = \{ (S, B): \ \ S \subset M \ \textrm{free submodule}, \ \ B \textrm{ basis of } S \} $$ We say that $(S,B) \le (S', B')$ if $S \subset S', B \subset B'$. It is non empty by assumption. Let's see that it respects the ascending chain property. If $(S_{\lambda}, B_{\lambda})$ is an ascending chain, then $S = \bigcup S_{\lambda}, B = \bigcup B_{\lambda}$ is the element required by the ascending chain property. Indeed, the natural map $R^B \to S$ is injective (any linear relationship can be seen in some $S_{\lambda}$) and surjective (same for any expression $s = a_1 b_1 + \ldots + a_n b_n$).

Now consider a maximal element $(S_*, B_*)$ and suppose by contradiction that there exists a free submodule $S_* \subseteq F$. By standard homological arguments, there exists a free submodule $N$ such that $F = S_* \oplus N$. Let $B'$ be a basis of $N$. Then $(S_*,B_*) < (F, B_*\sqcup B')$ is strictly greater in $\Sigma$, a contradiction.

Answer 2

This is not a complete answer, but I would like to prove that there are more sufficient conditions for the existence of maximal free submodules besides noetherianity. In particular, this works for Bezout rings and rings satisfying ascending chain condition on principal ideals (hence, UFDs). These two sufficient conditions cover most (not all) of peculiar examples of rings in literature, that I looked up trying to construct a counterexample. Also the reductions (Lemmas 1&2 below) could be helpful for the solution. $A$ everywhere is supposed to be a commutative unital domain.

For a finitely generated $A$-module $M$ we define $Q(M) = Q(A) \otimes_{A} M$ and $\mathrm{rank} M = \dim_{Q(A)} Q(M)$. As is mentioned in the question, $\mathrm{rank} M$ is the maximal dimension of a free submodule in $M$.

Definition. We say that $M$ satisfies the $MF$-property, if it has a maximal free submodule.

Lemma 1. If there exists a finitely generated $A$-module $M$ that does not satisfy the $MF$-property, then there exists a torsion-free finitely generated $A$-module $N$ that also does not satisfy the $MF$-property.

Proof. Let $T(M) = \{x \in M: \exists\alpha \in A\setminus\{0\}\text{ s.t. } \alpha x = 0 \}$ denote the torsion-submodule of $M$ and let $N = M / T(M)$. By $\pi:M \rightarrow N$ we denote the quotient mapping. Clearly, $N$ is finitely generated. Assume that $F \subset N$ is a maximal free submodule in $N$. Then there exists a submodule $F' \subset M$ such that $\pi$ induces an isomorphism $F' \rightarrow F$ (take any basis in $F$, choose $\pi$-preimages of basis elements, and let $F'$ be generated by these preimages). I claim that $F'$ is maximal in $M$ (among free submodules). If $G \supset F'$ is a bigger free submodule in $M$, then $G \cap T(M) = \{0\}$ and, therefore, $\pi$ induces an isomorphism $G \rightarrow \pi(G)$. Therefore, $\pi(G) = F = \pi(F')$ by maximality of $F$ and $G = F'$ as $\pi$ is injective on $G$. $\square$

Lemma 2. Assume that $M$ is a finitely generated torsion-free $A$-module and let $n = \mathrm{rank} M$. Then $M$ is isomorphic to a submodule in $A^n$.

Proof. Absence of torsion in $M$ implies that the mapping $x \rightarrow 1 \otimes x$ from $M$ to $Q(M)$ is injective. Thus, we can assume that $M$ is a finitely generated submodule in $Q(M) \sim Q(A)^n$. Now we consider a set $\{x_1, \dots, x_m\}$ of generators of $M$. Clearly, there exists $\alpha \in A \setminus \{0\}$ such that $\alpha x_j \in A^n$ for all $j \in \{1, \dots, m\}$ ($\alpha$ is the product of all denominators of all components of generators). Clearly, the mapping $x \rightarrow \alpha x$ is an isomorphism of $M$ onto a submodule in $A^n$. $\square$

Proposition 1. Assume that $A$ satisfies accp. Then all finitely generated $A$-modules satisfy the $MF$-property.

Proof. Assume that there exists a finitely generated $A$-module that does not satisfy the $MF$-property. From lemmas above we can assume that $M$ is a submodule in $A^n$ with $n = \mathrm{rank} M$. Since there is no maximal free submodule in $M$ there is an ascending chain $F_1 \subset F_2 \subset \dots$ of free submodules in $M$ with dimension $n$. Let $B_m \in A^{n \times n}$ denote an $n \times n$ matrix such that $F_m = \{B_m x: x \in A^n\}$. That is, columns of $B_m$ consist of basis vectors in $F_m$. Finally, let $d_m = \det B_m$.

Since $F_m \subset F_{m+1}$, there exists a matrix $C_m \in A^{n \times n}$ such that $B_m = B_{m+1}C_m$. From this we obtain $d_m = d_{m+1} \det C_m$, so $d_1A \subset d_2A \subset \dots$. This chain of principal ideals stabilies by accp, so for large $m$ all ideals $d_m A$ coincide. That means that the matrix $C_m$ has invertible determinant for large $m$, so it is invertible. Finally, we conclude that the sequence $F_m$ stabilizes for large $m$. $\square$

Lemma 3. Assume that $A$ is a Bezout ring (still without zero divisors). Then any finitely generated submodule $M$ in $A^n$ is free (with dimension at most $n$).

Proof. We proceed by induction, the case $n = 1$ being the definition of a Bezout ring. Let $I$ be the ideal of first components of elements of $M$ (i.e. $I = \{\alpha \in A: \exists x \in M \text{ s.t. } \alpha = x_1\}$). Clearly $I$ is a finitely generated ideal, so $I = \alpha A$ for some $\alpha \in A$. Let $z \in M$ be an element such that $z_1 = \alpha$ and let $L$ denote the module generated by $z$. Given $x \in M$ there exists a unique decomposition $x = \beta z + y$, where $y_1 = 0$ and $\beta \in R$. That is, $M$ is a direct sum of $L$ and $M \cap H$, where $H = \{y \in A^n: y_1 = 0\}$. Clearly, $M \cap H$ falls under the induction hypothesis, hence, $M$ is a direct sum of free modules. $\square$

Combining Lemmas 2&3 we obtain that all torsion-free finitely generated $A$-modules are free, provided $A$ is a Bezout domain.

Proposition 2. Assume that $A$ is a Bezout ring. Then every finitely generated $A$-module satisfies the $MF$-property.

Proof. From Lemma 1 we see that it suffices to consider only torsion-free $A$-modules which are free. $\square$