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.

  • 1
    $\begingroup$ I am not aware of a "commutative algebra-proof". For a topological proof see here. $\endgroup$
    – user26857
    Apr 20 '20 at 20:55
  • $\begingroup$ Related: math.stackexchange.com/questions/1000465 $\endgroup$
    – user26857
    Apr 20 '20 at 20:55
  • $\begingroup$ 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. $\endgroup$ Apr 21 '20 at 23:42
  • $\begingroup$ 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. $\endgroup$ Apr 21 '20 at 23:46

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