Proposition Let $R$ be a PID and let $F = F(X)$ be a free module (generated by $X$). Then every submodule $H$ of $F$ is free with $$\operatorname{rk} H \leq \operatorname{rk} F = |X|.$$
I have seen the proof on page 640 of [Rot], where they use Zermelo principle (every set has a well order). The same proof is poposed on drexel28's blog.
The use of Zermelo principle bothers me because i'm used to use maximality principles: Zorn's lemma or Hausdorff's maximality principle.
Here i try to prove it using Hausdorff principle.
I would appreciate if you were to point out mistakes or give any comment.
- We know that a basis for $H$ cannot be extracted from $X$, consider for example $H = \langle (1, 1) \rangle$ in $R \oplus R = \langle (1, 0), (0, 1)\rangle$.
- Considering free submodules $\langle Y \rangle \subseteq H$ is a bit hopeless since it makes hard to control the rank of limit (union) bases.
This should motivate the choice of going from above. Consider the set
The following proof is nonsense.
I said: do not extract a subset from a basis, and then went doing it.
$$
S = \{Y \subseteq F : H \subseteq \langle Y \rangle \text{ and }
Y \text{ is linearly independent}\}$$
ordered by inclusion.
The second condition in particular gives $|Y| \leq |X|$ since the rank
is well defined and monotone (passing to $F = A/m$-vector spaces).
If not, one could add this condition to the definition of $S$
without effort.
- Since $X \in S$, by Hausdorff there is a maximal chain $\{Y_{\alpha}\}$.
- This chain has a minimum given by $Y = \cap_{\alpha} Y_{\alpha}$.
First note that the intersection $Y$ is in $S$: $$H \subseteq \cap_{\alpha} \langle Y_{\alpha}\rangle \subseteq \langle Y \rangle.$$ The first inclusion is clear, we need to show the second one.
Let $x \in \cap_{\alpha} \langle Y_{\alpha}\rangle$ and consider any two indices $\alpha, \beta$.
Without loss of generality we can assume that $Y_{\alpha} \subseteq Y_{\beta}$ (since this is a chain). Then $$ \begin{align} x &= r_1 \, y^{\alpha}_1 + \ldots + r_n \, y^{\alpha}_n \\ &= s_1 \, y^{\beta}_i + \ldots + s_m \, y^{\beta}_m. \end{align} $$ Since this identity is true in $\langle Y_{\beta} \rangle$, which is free, we must have $\{y^{\alpha}_i\} = \{y^{\beta}_j\}$.
Since this is true for any choice of indices we have that $x \in \langle Y \rangle$.
Finally, by maximality the chain contains $Y$. - To conclude we need to show that $H \supseteq \langle Y \rangle$.
Note that, so far, we have not made use of the fact that $A$ is a PID.
in this step we are also going to use the minimality of $Y$.
Suppose that $$\langle Y \rangle \ni x = r_1 y_1 + \ldots + r_n y_n$$ with $y_i \in Y$ and assume that $x \not\in H$. Then there is an $y_i = y$ which is not in $H$. Set $Y' = Y \setminus \{y\}$. We claim that $$H \subseteq \langle Y' \rangle$$ which gives a contraddiction on the minimality of $Y$.
Since $R$ is a PID, by ($\ast$) we have that $r y \not\in H$ for every $r \neq 0$.
Let $h \in H \subseteq \langle Y \rangle$. Then $$h = r y + r_1 y_1 + \ldots + r_n y_n$$ with $y_i \in Y'$. This gives $r y \in H$, hence $r = 0$ and so $x \in \langle Y' \rangle$.
$\blacksquare$
(*) We have that $H \cap \langle y \rangle$ is a cyclic $R$-module, which is isomorphic either to $0$ or to $R$ (the second case being impossible since $y \not\in H$).
[Rot] Joseph J. Rotman `Advanced modern algebra', GSM second edition, 2010.