----- > [!theorem] Theorem. ([[submodule of finitely generated free module over a PID is free]]) > **Weak version.** Let $R$ be a [[PID]], let $F$ be a [[submodule generated by a subset|finitely generated]] [[free module|free]] $R$-[[module]], and let $M \subset F$ be a [[submodule]]. Then $M$ is [[free module|free]]. > > **Strengthened version.** There is a [[basis]] $(x_{1},\dots,x_{n})$ of $F$ and elements $a_{1},\dots,a_{m}$ of $R$ (with $m \leq n$) such that $y_{1}=a_{1}x_{1}, \dots, y_{m}=a_{m}x_{m}$ > form a [[basis]] of $M$. Thus, [[module is free iff admits basis|not only]] is $M$ [[free module|free]], but there exist 'compatible bases' of $M$ and $F$. ^theorem > [!proposition] Technical Lemma using the PID Condition. (Splitting off a direct summand from $M$) > Let $R$ be a [[PID]], let $F$ be a [[submodule generated by a subset|finitely generated]] [[free module|free]] $R$-[[module]], and let $M \subset F$ be a nonzero [[submodule]]. Then there exist $a \in R$, $x \in F$, $y \in M$, and [[submodule|submodules]] $F' \subset F$ and $M' \subset M$, such that $y=ax \neq 0$, $M'=F' \cap M$, and $F = \langle x \rangle \oplus F', \ \ \ \ M=\langle y \rangle \oplus M'. $ ^proposition > [!proof] Proof of Technical Lemma that uses PID Condition. > ^proof-1 **Step 1.** For all $\varphi$ in the [[dual vector space]] $\text{Hom}_{R\text{-}\mathsf{Mod}}(F,R)$, $\varphi(M)$ is a [[submodule]] ([[submodule#^basic-example|hence]] an [[ideal]]) of $R$. The family of all these [[ideal|ideals]] is nonempty and $R$ is obviously [[Noetherian module|Noetherian]], thus the [[ascending chain - maximality characterization of Noetherian modules]] guarantees the existence of a [[maximal]] element in the family, say $\alpha(M)$, for a [[linear map|homomorphism]] $\alpha:M \to R$. Since $M \neq 0$, $\varphi(M) \neq 0$, hence $\alpha(M) \neq 0$. Since $R$ a [[PID]], $\alpha(M)$ is a [[principal ideal]], write $\alpha(M)=\langle a \rangle$ for some $a \in R$. Let $y \in \alpha ^{-1}(a)$, $y \neq 0$. These are the elements $a,y$ mentioned in the statement. **Step 2.** The next claim is that $a$ [[divides]] $\varphi(y)$ for all $\varphi$ in $\text{Hom}(F,R)$. Indeed, (using that $R$ is a [[PID]]) let $b$ be a [[ideal generated by a subset|generator]] of $\langle a, \varphi(y) \rangle$ and let $r,s \in R$ such that $b=ra+s \varphi(y)$; consider the [[linear map|homomorphism]] $\psi:=r \alpha+ s \varphi$. Since $a \in \langle b \rangle$, we have $\alpha(M) \subset \langle b \rangle$. On the other hand $b=ra+s \varphi(y)=(ra+s \varphi)(y)=\psi(y) \in \psi(M)$ and so $b \in \psi(M)$. Thus $\alpha(M) \subset \psi(M)$; by maximality, $\alpha(M)=\psi(M)$ and so $\langle a \rangle=\langle b \rangle$; in particular $a | \varphi(y)$ as claimed. **Step 3.** Now we obtain $x$ as mentioned in the statement. Write $y=(s_{1},\dots, s_{n})$ as an element of $F=R^{\oplus n}$. Each $s_{i}$ is the image of $y$ under a [[linear map|homomorphism]] $F \to R$ (e.g., the $i^{th}$ [[projection function|projection]]), so $a$ [[divides]] all of them (by step 2). So there exist $r_{1},\dots,r_{n} \in R$ such that $s_{i}=ar_{i}$; let $x:= (r_{1},\dots,r_{n}) \in F.$ By construction, $y=ax$. And $a=\alpha(y)=\alpha(ax)=a \alpha(x)$; since $R$ is an [[integral domain]] and $a \neq 0$, this implies $\alpha(x)=1$. **Step 4.** With $x,y,a$ all defined, it is time to verify that they fit into the [[direct sum of modules|direct sum]] form provided in the lemma statement. > [!proof] Proof of Weak Version. > ^proof-2 > [!proof] Proof of Strengthened Version. > ^proof ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```