2.5 Module over PID

Definition

Non-zero commutative rings have invariant basis number, and the cardinality of any basis is called the rank of the free module $M$.

Lemma

Let $A$ be any ring. Assume that $f:M\twoheadrightarrow N$ is an epimorphism of $A$-module, with $N$ finite free. Then $M\simeq N\oplus \ker f$.

\begin{proof} When $A=\mathbb{Z}$, $M$ and $N$ are abelian groups and we have proved it before.

We have the short sequence $0\to \ker f\to M\stackrel{f}{\twoheadrightarrow }N\to 0$. ^qtklfv says that $M\simeq \ker f\oplus N$ iff there exists $s:N\to M$ such that $f\circ s={\mathrm{id}}_N$. Because $N$ is finite free, for a basis $\{e_1,\cdots ,e_n\}$ of $N$ we can pick some ${\hat{e}}_i\in M$ such that $f({\hat{e}}_i)=e_i$. Define $s:N\to M,{\sum }_{i=1}^n{a}_i{e}_i={\sum }_{i=1}^n{a}_i{\hat{e}}_i$. It is well-defined because $N$ is free. It is easy to say $f\circ s={\mathrm{id}}_N$.

Alternating proof. Define $g:M\to N\oplus \ker f$ as follows. If $f(m)\in N$ with $f(m)=\sum a_i{e}_i$, then take $g(m)=(f(m),m-\sum a_i{\hat{e}}_i)$. It is easy to show $g$ is a isomorphism. \end{proof}

Theorem

Let $A$ be a PID, and let $M$ be a $A$-module of rank $n$. Suppose $M^{\prime }\subseteq M$ is a submodule, then:

• $M^{\prime }$ is free of rank $q$ for some $0⩽q⩽n$;
• there exists a basis $(e_1,\cdots ,e_n)$ of $M$ and non-zero $a_1\mid a_2\mid \cdots \mid a_q$ such that $(a_1{e}_1,\cdots ,a_q{e}_q)$ is a basis of $M^{\prime }$.

Remark.

• $A=\mathbb{Z}$ is known.
• If $A$ is a domain but not PID, then the theorem fails. Indeed, suppose $I$ is not principal and $M=A$. Then $I\subseteq M$ is free of rank $1$ over $A$, which yields that $I=A\cdot c$ and $I=(c)$, contradiction.

\begin{proof} The proof is similar to the case of $A=\mathbb{Z}$.

i) Use induction on $n$. When $n=1$, $(0)\ne M^{\prime }\subseteq M=A$ yields that $M^{\prime }=A\cdot a$ and we finish the proof. Suppose it holds when $\mathrm{rank}M=n$, Now consider $M^{\prime }\subseteq M=A^{\oplus (n+1)}$. Consider $\pi :M=A\oplus A^n\twoheadrightarrow A^n$, then restrict it to $M^{\prime }$ and we get $\pi {|}_{M^{\prime }}:M^{\prime }\twoheadrightarrow \pi (M^{\prime })\subseteq A^n$ where $\pi (M^{\prime })$ is free of rank $⩽n$ by induction hypothesis. By ^77df33, $M^{\prime }\simeq \pi (M^{\prime })\oplus \ker \pi {|}_{M^{\prime }}$. But $\ker \pi {|}_{M^{\prime }}=(A\oplus 0)\cap M^{\prime }$ is free of rank $⩽1$. Now we finish the proof.

ii) WLOG we can assume $q=n$. When $n=1$, it is trivial because $A$ is a PID. When $n=2$, define $\pi :A^{\oplus n}\to A^{\oplus n-1}$ and $M^{\prime }\simeq N\oplus \ker \pi {|}_{M^{\prime }}$. By $n=1$ case, there exists a basis of $M$ such that $M^{\prime }=⟨a_1{f}_1,a_2{f}_2⟩$. However, $a_1\nmid a_2$ may happen.

Consider $P=\{\text{ideal }(a):\exists b\in A\text{ and basis }(f_1,f_2)\text{ of }M,M^{\prime }=⟨a{f}_1,b{f}_2⟩\}$. Note that $P$ is non-empty by argument above. Since $A$ is a PID and hence is Noetherian, there exists a maximal element $(\alpha )\in P$ such that $M^{\prime }=⟨\alpha f_1,\beta f_2⟩$ such that $M=⟨f_1,f_2⟩$. We want to show $\alpha \mid \beta$, i.e., $(\beta )\subseteq (\alpha )$. Otherwise, there is $(\alpha ,\beta )=(\gamma )⊋(\alpha )$ with $\alpha =t{x}_1$, $\beta =t{x}_2$. Assume that $t=\alpha y_1+\beta y_2$, then we have $x_1{y}_1+x_2{y}_2=1$. We can check

$$M^{\prime }=⟨\alpha f_1,\beta f_2⟩=⟨t(x_1{f}_1+x_2{f}_2),t{x}_1{x}_2(-y_2{f}_1+y_1{f}_2)⟩$$

because RHS equals $\left[\begin{array}{c}\alpha f_1,\beta f_2\end{array}\right]\left[\begin{array}{cc}1& -x_2{y}_2\\ 1& x_1{y}_1\end{array}\right]$ and the matrix is invertible. In addition, $M=⟨x_1{f}_1+x_2{f}_2,-y_2{f}_1+y_1{f}_2⟩=\left[\begin{array}{cc}{f}_1& f_2\end{array}\right]\left[\begin{array}{cc}{x}_1& -y_2\\ x_2& y_1\end{array}\right]$. Therefore, $(t)\in P$, contradicting maximality of $(\alpha )$.

Now suppose ii) holds for $M=A^{\oplus n}$. Now consider $M=A^{\oplus n+1}$. Again $M^{\prime }=\ker ({\pi }_{M^{\prime }})\oplus N=⟨a_1{f}_1⟩\oplus ⟨a_2{f}_2,\cdots ,a_{n+1}{f}_{n+1}⟩$ with $a_2\mid a_3\mid \cdots \mid a_{n+1}$. Define

$$P=\{(a)\subseteq A:(a)\text{ is an ideal, }\exists b_2\mid \cdots \mid b_{n+1},(f_1,\cdots ,f_{n+1})\text{ is a basis of }M,M^{\prime }=⟨a{f}_1,b_2{f}_2,\cdots ,b_{n+1}{f}_{n+1}⟩\}.$$

Since $P\ne \varnothing$ and $A$ is Noetherian, there exists a maximal element $(\alpha )\in P$, that is, $M^{\prime }=⟨\alpha f_1,{\beta }_2{f}_2,\cdots ,{\beta }_{n+1}{f}_{n+1}⟩$. Claim $\alpha \mid {\beta }_2\mid \cdots \mid {\beta }_{n+1}$. First we prove $\alpha \mid {\beta }_{n+1}$. If not, we can change them to $\alpha /x_1$ and $x_1{\beta }_{n+1}$, contradicting maximality of $(\alpha )$. We then prove $\alpha \mid {\beta }_n$. If not, we can change them to $\alpha /x_1$ and ${\beta }_n\cdot x_1$. It deduces that

$$M^{\prime }=⟨\alpha f_1,{\beta }_2{f}_2,\cdots ,{\beta }_n{f}_n,{\beta }_{n+1}{f}_{n+1}⟩=⟨\frac{\alpha }{x_1}{f}_1,{\beta }_2{f}_2,\cdots ,{\beta }_n{x}_1{f}_n,{\beta }_{n+1}{f}_{n+1}⟩.$$

We need to show $\beta x_1\mid {\beta }_{n+1}$. Note that $\alpha =t{x}_1\mid {\beta }_{n+1}$ and ${\beta }_n=t{x}_n\mid {\beta }_{n+1}$, then we have $t{x}_1{x}_n\mid {\beta }_{n+1}$ and so ${\beta }_n{x}_1\mid {\beta }_{n+1}$. Repeat this procedure, we can prove that $\alpha \mid {\beta }_n$ and then $\alpha \mid {\beta }_2$. Now we finish the proof. \end{proof}

structure theorem for finitely generated modules over a PID

Let $A$ be a PID, and let $E$ be a finite generated $A$-module. Then $E\simeq {\prod }_{i=1}^{k}A/{\alpha }_i$ for some ideals ${\alpha }_i$ such that ${\alpha }_1\supseteq {\alpha }_2\supseteq \cdots \supseteq {\alpha }_n$.

\begin{proof} Since $E$ is finite generated, there exists $f:A^n\twoheadrightarrow E$ and so $E\simeq A^n/\ker f\simeq ({\oplus }_{i=1}^{n}A{e}_i)/({\oplus }_{i=1}^{q}A{a}_i{e}_i)\simeq \left({\oplus }_{i=1}^q(A/a_i)e_i\right)\oplus ({\oplus }_{i=q+1}^{n}A{e}_j)$. Let ${\alpha }_i=(a_i)$ for $i⩽q$ and ${\alpha }_i=0$ for $i>q$, then we finish the proof. \end{proof}

Example. Let $V={\mathbb{C}}^n$. For a given linear transform $\mu$, we can view $V$ as a finitely generated module over $\mathbb{C}[x]$, which is defined by $x\cdot v=\mu v$.

• By ^353e4e and the Chinese remainder theorem we have

  $$V\cong {\oplus }_{i=1}^m\mathbb{C}[x]/\left((x-{\lambda }_i{)}^{k_i}\right).$$

• Each module $\mathbb{C}[x]/\left((x-{\lambda }_i{)}^{k_i}\right)$ is indecomposable and corresponds exactly to a Jordan block $J\left({\lambda }_i,k_i\right)$: we can find a basis $v_j=\overline{(x-\lambda {)}^{k-j}}$ for this $k$-dimensional module and so such that the action of $x$ in this basis is precisely the matrix $J(\lambda ,k)$, where $(\mu -\lambda I)v_1=0$ and $(\mu -\lambda I)v_j=v_{j-1}$ for $j>1$.

• This decomposition gives the Jordan normal form of $\mu$. Also see here.

Definition

Let $A$ be a domain and $M$ be an $A$-module. We say $M$ is torsion free, if $am=0$ with $a\in A,m\in M$ then $a=0$ or $m=0$.

Corollary

Let $A$ be a PID, and let $M$ be a finitely generated $A$-module. Then $M$ is finite free iff $M$ is torsion free.

\begin{proof} Apply ^353e4e. \end{proof}