Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 4 / 9E (Groups, Rings and Modules)

An $R$-module is free if it has a basis: a subset $B$ such that every element has a unique expression as a finite $R$-linear combination of elements of $B$.

Choose a maximal ideal $\mathfrak{m}$ of the nonzero ring $R$. If $R^n\cong R^m$, quotienting by $\mathfrak{m}$ givesThese are vector spaces over the field $R/\mathfrak{m}$, so equality of dimension gives $n=m$. This proves invariant basis number for a commutative ring.

A direct summand of a free module need not be free. Take $R=\mathbb{Z}/6\mathbb{Z}$. Its ideals $P=(2)$ and $Q=(3)$ satisfyso $P\oplus Q\cong R$. But $P$ has three elements, whereas a finite-rank free $R$-module has $6^r$ elements; hence $P$ is not free.

Let $P$ be free with basis $(e_i{)}_{i\in I}$. Given a surjection $f:M\to N$ and a map $g:P\to N$, choose $m_i\in M$ satisfying $f(m_i)=g(e_i)$. There is a unique linear map $h:P\to M$ with $h(e_i)=m_i$, and on every basis vectorThus $f\circ h=g$, proving that free modules are projective.

We prove by induction on $n$ that every submodule $N\subseteq R^n$ is free. The case $n=0$ is immediate. Let $\pi :R^n\to R$ be projection onto the first coordinate. Since $R$ is a principal ideal domain,for some $d$. If $d=0$, then $N\subseteq R^{n-1}$ and induction applies. Otherwise choose $x\in N$ with $\pi (x)=d$. For every $y\in N$, write $\pi (y)=rd$; then $y-rx\in \ker (\pi {|}_N)$. Also $Rx\cap \ker (\pi {|}_N)=0$ because $R$ is a domain and $d\ne 0$. ThereforeThe kernel is a submodule of $R^{n-1}$ and is free by induction, while $Rx\cong R$. Hence $N$ is free. This is the submodule theorem for free modules over a principal ideal domain.

If $P$ is finitely generated and projective, choose a surjection $f:R^n\to P$. Projectivity supplies $h:P\to R^n$ with $f\circ h={\mathrm{id}}_P$, soThus $P\cong h(P)$ is a submodule of the finitely generated free module $R^n$, and the result just proved shows that $P$ is free.