Recently I was trying to amuse myself by inventing unusual examples of modules which are projective but not free. One of the candidates I came up with led to a question which I don't know how to answer. Consider the following:

Let $R$ be the ring of continuous real-valued functions defined on $[-1,1]$. Let $E\subseteq R$ be the set of even functions and let $O\subseteq R$ be the set of odd functions.

Since $E$ is a subring of $R$, we can view $R$ as an $E$-module. Unless I am mistaken, $O$ is an $E$-submodule of $R$ and $R$ is isomorphic to $E \oplus_E O$.

I can show that $O$ is not a free $E$-module (details below), but I cannot show the same for $R$. So that's my question: is $R$ a free $E$-module?

Here's my argument for why $O$ is not a free $E$-module:

First of all, notice that no two $f,g\in O$ can be linearly independent over $E$. For if $f,g$ are odd, then $fg$ and $-f^2$ are even, and $(fg)f + (-f^2)g=0$.

So if $O$ were free, there would exist a single $f\in O$ such that any $g\in O$ can be written uniquely as $g=hf$, where $h\in E$. In particular, if $g(x)=x$ for all $x$, we see that $f(x)\neq 0$ for $x\neq 0$.

Since $f$ is odd, so is $f^{1/3}$. Therefore there exists a continuous even function $h$ such that $f^{1/3}=hf$. But then we have $h(x)=\frac{1}{f(x)^{2/3}}$ for $x\neq 0$. Since $f$ is odd, $f(0)=0$. This contradicts the fact that $h$ is continuous at $0$.

  • $\begingroup$ If you want examples of non-free projective modules, check out the Serre-Swan theorem. For instance, just take any nontrivial vector bundle over a smooth manifold. Then the corresponding module (over the ring of smooth functions) of smooth global sections is projective, but not free. $\endgroup$
    – ಠ_ಠ
    Nov 7, 2016 at 23:05

1 Answer 1


Evaluation at $0$ gives a surjective ring homomorphism $e_0:E\to \mathbb{R}$. Write $m$ for the kernel of $e_0$. Evaluation at $0$ also gives a surjective homomorphism of $E$-modules $e_0':R\to \mathbb{R}=E/m$. I claim the kernel of $e_0'$ is $mR$, which implies that $R\otimes_E E/m=R/mR$ has rank $1$ as an $E/m$-module. To see this, suppose $f(x)$ is in the kernel of $e_0'$, i.e. $f(0)=0$. Define $$ \alpha(x) = \frac{f(x) + f(-x)}{2},\;\;\; \beta(x) = \left(\frac{f(x)-f(-x)}{2}\right)^{1/3}\in R. $$ Check that $\alpha$ and $\beta^2$ are in $m$, and that $f=\alpha\cdot 1+\beta^2\cdot \beta$, so that $f\in mR$.

Now, if $R$ is free as an $E$-module, then $R$ must have rank $1$. This is impossible, as $1$ and $x$ are elements of $R$ that are $E$-linearly independent.


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