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I have heard that any submodule of a free module over a p.i.d. is free.

I can prove this for finitely generated modules over a p.i.d. But the proof involves induction on the number of generators, so it does not apply to modules that are not finitely generated.

Does the result still hold? What's the argument?

user26857
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Ben Blum-Smith
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    Rotman has it in Advanced Modern Algebra, page 651! – Dedalus Jun 25 '12 at 18:29
  • (The result still holds) – Dedalus Jun 25 '12 at 18:29
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    For more references, see Rotman’s AMA 2nd Theorem 9.8 (as mentioned by Dedalus) or AMA 3nd Theorem B-2.28 (has a comment on Zorn’s lemma) or Lang’s Algebra , Appendix 2.2 (use Zorn’s lemma rather than well ordering principle) – onRiv Aug 30 '22 at 05:05

1 Answers1

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Let $F$ be a free $R$-module, where $R$ is a PID, and $U \subseteq F$ be a submodule. Then $U$ is also free with $\mathrm{rank}(U) \leq \mathrm{rank}(F)$.

Proof. Let $(e_i)_{i \in I}$ be a basis of $F$. Choose a well-ordering $\leq$ on $I$ (this requires the Axiom of choice). Let $p_i : F \to R$ be the projection on the $i$th coordinate, i.e. $p_j(e_i) = \delta_{ij}$. Let $F_i$ be the submodule of $F$ generated by the $e_j$ with $j \leq i$. Let $U_i = U \cap F_i$. Then $p_i(U_i)$ is a submodule of $R$, that is an ideal, i.e. has the form $R a_i$ for some $a_i \in R$. Choose some $u_i \in U_i$ with $p_i(u_i)=a_i$. If $a_i=0$, we may, and will, choose $u_i=0$.

We claim that the $u_i \neq 0$ constitute a basis of $U$.

We first show that $\langle u_i : i \in I \rangle = U$. Let $u \in U$, so $u \in F$ and we can write $u = r_{i_1} e_{i_1} + \cdots + r_{i_n} e_{i_n}$ with $i_1 < \cdots < i_n$. For $n=0$ there is nothing to do, so let $n \geq 1$. We make an induction on the largest index $j := i_n$. We have $u \in U_j$ and $r_j = p_j(u) \in p_j(U_j) = R a_j$, say $r_j = s a_j = s p_j(u_j)$ for some $s \in R$. Then $p_j(u - s u_j) = 0$, i.e. the $e_j$-coefficient in $u - s u_j$ vanishes. By induction hypothesis, we know $u - s u_j \in \langle u_i : i \in I \rangle$, hence also $u \in \langle u_i : i \in I \rangle$.

To show linear independence, we assume that $0 = r_{i_1} u_{i_1} + \cdots + r_{i_n} u_{i_n} = 0 ~ (\star)$ with $i_1 < \cdots < i_n$. Again consider the largest index $j := i_n$. It suffices to prove $r_j=0$, because then we are done by induction on $n$. Applying $p_j$ to $(\star)$ yields $0 = r_{i_1} 0 + \cdots + r_{i_{n-1}} 0 + r_{j} a_j = r_j a_j$. Since $a_j \neq 0$ (otherwise, we would have $u_j = 0$), we conclude $r_j = 0$. $\checkmark$

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PS: The same proof shows the more general result: If $R$ is a hereditary ring (every ideal of $R$ is projective over $R$), then any submodule of a free $R$-module is a direct sum of ideals of $R$.

Martin Brandenburg
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