# Proving that the tangent bundle of the sphere is projective but not free

In my Commutative Algebra exercises I have the following exercise:

"Let $$A = \frac{\mathbb{R}[x,y,z]}{\left\langle x^2+y^2+z^2-1 \right\rangle}$$ and $$M = \left\lbrace (\varphi, \chi, \eta) \in A \oplus A \oplus A : \varphi x + \chi y + \eta z =0 \right\rbrace$$. Prove that M is a projective A-module but it is not a free A-module."

Seeing that it is projective is easy because $$M$$ is the kernel of the homomorphism $$f: A \oplus A \oplus A \rightarrow A$$ sending $$(\varphi, \chi, \eta)$$ to $$\varphi x + \chi y + \eta z$$. $$f$$ is surjective because $$1=f(x,y,z)$$ and so the sequence $$0 \rightarrow M \rightarrow A \oplus A \oplus A \rightarrow A \rightarrow 0$$is exact, but $$A$$ is a projective $$A$$-module so the sequence splits and $$A \oplus A \oplus A \cong M \oplus A$$. Hence M is projective.

However, I can't figure out how to prove that its not free. I get the idea that $$M$$ something like the algebraic tangent bundle of the sphere and we know from algebraic topology that the tangent bundle of the sphere is not trivial and since trivial bundle <-> globally free bundle, M shouldn't be free.

Even though I look for a commutative algebra-proof of the fact, any insights to the questions are welcome.

• I am not aware of a "commutative algebra-proof". For a topological proof see here. Apr 20 '20 at 20:55
• Apr 20 '20 at 20:55
• I would expect that this result is equivalent to the fact that $S^2$ is not parallelizable. One direction is easy. Since $C(S^2)\otimes_A M \cong \Gamma(TS^2)$, where $C(S^2)$ is the ring of continuous functions of $S^2$ and $\Gamma(TS^2)$ is the $C(S^2)$-module of global sections of the tangent bundle $TS^2$. Thus, if $M$ is free, then $TS^2$ would be a trivial bundle. I don't have any evidence for the other implication other than my gut. Apr 21 '20 at 23:42
• This is certainly not true for general algebraic manifolds and vector bundles, not even in the case of line bundles since the Picard group may not be discrete. Apr 21 '20 at 23:46