Vtyjkyuk

This is a recent qual problem that I am struggling with.

Put $M=mathbbC[x]/(x^2+x)$ and $N=mathbbC[x]/(x-1)$.

$(a)$ What is the dimension of $M otimes_mathbbC[x]N$ as a vector space over $mathbbC$?

$(b)$ What is the dimension of $M otimes_mathbbCN$ as a vector space over $mathbbC$?

$(c)$ What is the dimension of $M otimes_mathbbC[x^2]N$ as a vector space over $mathbbC$?

I solved the part $(a)$ and $(b)$. The part $(b)$ is the easiest. Since $M$ (resp. $N$) is a free $mathbbC$-module with basis $1,x $ (resp. $1$), the dimension is $2$. To get the part $(a)$, I used the direct computations:
$$
1 otimes 1 = (1+x^2+x) otimes 1 = 1 otimes (1+x^2+x) = 1 otimes 3 = 3(1 otimes 1).
$$
Thus, we get $1 otimes 1 = 0$, and the dimension is $0$. For another proof of the part $(a)$, we might be able to use that the tensor product is a right-exact functor because
$$
mathbbC[x]/(x^2+x,x-1) = 0.
$$

I don't know how to solve the part $(c)$. First, I showed
$$
M otimes_mathbbC[x^2] N = c(1 otimes_mathbbC[x^2] 1): c in mathbbC .
$$
via direct computations. However, this does not guarantee that the dimension is $1$. We need to check that $1 otimes_mathbbC[x^2] 1 neq 0$. Do you have any ideas? I know the property that the tensor product is a right-exact functor, but I have no idea how to apply it here.

Any comments are welcomed. Thank you in advance.

We can define a bilinear map $beta:Mtimes NtoBbb C, (a+bx, ,c) mapsto (b-a)c$.

Then we only have to check that it's also $Bbb C[x^2]$-bilinear, where $x^2$ acts on $N=Bbb C[x]/(x-1)congBbb C $ as $1$:
$$((a+bx)x^2,, c) = (-ax+bx, , c) mapsto (b-a)c\
(a+bx, , x^2c) =(a+bx, c) mapsto (b-a)c$$