Is $\mathbb{C}$ projective as a $\mathbb{C}[x,y]/(x^3,y^3,xy)$-module? Let $R = \mathbb{C}[x,y]$, and $I = (x^3, y^3, xy)$ be the ideal of $R$ generated by $x^3, y^3, xy$. Is it true that the $R/I$-module $\mathbb{C}$ (with the natural action) is a projective module? I think it is, but am not sure how to prove (nor disprove) this. I suspect that in fact $\mathbb{C}$ is a direct summand of $R/I$ and this has been my main line of attack, but I can't prove it (though of course this may not be the case even if $\mathbb{C}$ were a projective module).
 A: You can do this from first principles (undoubtedly there are more enlightening ways of looking at this, but let's go with minimal technology). Let me put the ball on the tee for you.
Let $M$ be the ideal of $R$ generated by $x$ and $y$. The natural way to think of $\Bbb{C}$ as an $R/I$-module is IMHO to think of it as $\Bbb{C}=R/M$. Here $I\subset M$, so $R/M$ is naturally an $R/I$-module.
We have the natural surjection $\pi:R/I\to \Bbb{C}$ defined by $\pi(p(x,y)+I)=p(0,0)$. Explain the following:


*

*$\pi$ is a homomorphism of $R/I$-modules.

*$\Bbb{C}=R/M$ is a projective $R/I$-module if and only if $\pi$ splits. IOW iff there exists a homomorphism of $R/I$-modules $s:\Bbb{C}\to R/I$ such that $\pi\circ s= id_{\Bbb{C}}.$

*Such a homomorphism is fully determined if we know $s(1)=r(x,y)+I$.

*The polynomial $r(x,y)$ must satisfy the constraints: A) $r(0,0)=1$, B) $xr(x,y)\in I$, C) $yr(x,y)\in I$.

*We have $$R/I=\{a_0+a_1x+a_2x^2+b_1y+b_2y^2+I\mid a_0,a_1,a_2,b_1,b_2\in\Bbb{C}\}$$ and all the cosets of $I$ have a unique representative of that form.

*There does not exist an element $r(x,y)+I\in R/I$ satisfying all the constraints 3A/B/C.

*The required splitting homomorphism does not exist.

A: The trivial module is never projective in situations of this form. One can prove that the $\mathbb C$-vector space $\operatorname{Tor}_1^R(\mathbb C,\mathbb C)$ identifies with the vector space of generators of $R$, which in this case has dimension $2$, and is in particular non-zero. In fact, you can show that in your case $\operatorname{Tor}_n^R(\mathbb C,\mathbb C) = 2^n$ for each $n\geqslant 0$. 
