Let $R$ be a commutative ring with unity and $M$ an $R$-module.

If $M\oplus R^m\cong R^n$ for some integers $m,n \geq 1$ then must $M$ be finitely generated and free?

Can somebody help me with this? Regards.


1 Answer 1


A prominent counter-example is the following: Take $R := {\mathbb R}[x,y,z]/(x^2+y^2+z^2-1)$, the ring of real-valued polynomial functions on the $2$-sphere, and consider the following short exact sequence: $$(\ast)\quad\quad 0\to P\to R\frac{\partial}{\partial x}\oplus R\frac{\partial}{\partial y}\oplus R\frac{\partial}{\partial z}\xrightarrow{\alpha := \begin{pmatrix}\frac{\partial}{\partial x}\mapsto x\\ \frac{\partial}{\partial y}\mapsto y\\ \frac{\partial}{\partial z}\mapsto z\end{pmatrix}} R\to 0.$$ That is, take $P$ to be the kernel of $\alpha$ on the right, and note that since $1 = x^2+y^2+z^2\in (x,y,z)$ in $R$, the map $\alpha$ is surjective. Now, since $R$ is a projective $R$-module, the short exact sequence $(\ast)$ splits, e.g. by $R\mapsto R^{\oplus 3}$, $1\mapsto x\frac{\partial}{\partial x} + y\frac{\partial}{\partial y} + z\frac{\partial}{\partial z}$, and hence $P\oplus R\cong R^{\oplus 3}$. Therefore, $P$ is projective and stably free.

However, $P$ is not free: For this, first note that it suffices to show that $Q := P\otimes_R S$ is not free, where $R\hookrightarrow S := \text{C}({\mathbb S}^2;{\mathbb R})$ is the embedding of the polynomial real valued functions on ${\mathbb S}^2$ into all continuous real valued functions. Since $(\ast)$ is split short exact, $Q$ fits into the short exact sequence $$(\dagger)\quad\quad 0\to Q\to S\frac{\partial}{\partial x}\oplus S\frac{\partial}{\partial y}\oplus S\frac{\partial}{\partial z}\xrightarrow{\alpha := \begin{pmatrix}\frac{\partial}{\partial x}\mapsto x\\ \frac{\partial}{\partial y}\mapsto y\\ \frac{\partial}{\partial z}\mapsto z\end{pmatrix}} S\to 0.$$ Now, we have the following Theorem:

Serre-Swan's theorem: If $X$ is a compact Hausdorff space, then the functor $$\text{VB}_{\mathbb R}(X)\longrightarrow\text{C}(X;{\mathbb R})\text{-proj},\quad {\mathscr E}\mapsto\Gamma(X;{\mathscr E})$$ is well-defined and an equivalence of categories between the category of finite-dimensional real vector bundles over $X$ and finite rank projective $\text{C}(X;{\mathbb R})$-modules. The functor restricts to an equivalence between trivial vector bundles and free $\text{C}(X;{\mathbb R})$-modules.

Under Serre-Swan's equivalence, the split short exact sequence $(\dagger)$ corresponds to a split short exact sequence

$$(\ddagger)\quad\quad 0\to {\mathscr T}\to {\mathscr O}\frac{\partial}{\partial x}\oplus {\mathscr O}\frac{\partial}{\partial y}\oplus {\mathscr O}\frac{\partial}{\partial z}\xrightarrow{\alpha := \begin{pmatrix}\frac{\partial}{\partial x}\mapsto x\\ \frac{\partial}{\partial y}\mapsto y\\ \frac{\partial}{\partial z}\mapsto z\end{pmatrix}} {\mathscr O}\to 0$$

of vector bundles over ${\mathbb S}^2$, where ${\mathscr O}$ denotes the trivial vector bundle. This short exact sequence, however, also defines the tangent bundle on ${\mathbb S}^2$. Hence, $Q$ is free over $S$ if and only if the tangent bundle on ${\mathbb S}^2$ is trivial; however, we have the following classical theorem from topology:

Hairy Ball Theorem The tangent bundle on ${\mathbb S}^2$ does not admit any nowhere vanishing section. In particular, it is not trivial.

We conclude that $Q=P\otimes_R S$ is not free over $S$, so $P$ is not free over $R$.

  • $\begingroup$ Sorry but how do you define the map $(x,y,z): R^3\to R$? $\endgroup$
    – user181834
    Commented Oct 31, 2014 at 23:10
  • $\begingroup$ user181823: I added some details $\endgroup$
    – Hanno
    Commented Nov 1, 2014 at 8:01
  • $\begingroup$ Many thanks- I now understand the first part of your answer- to understand the Serre-Swan theorem I need to learn about vector bundles! $\endgroup$
    – user181834
    Commented Nov 1, 2014 at 20:49

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