This is a recent qual problem that I am struggling with.
Put $M=\mathbb{C}[x]/(x^2+x)$ and $N=\mathbb{C}[x]/(x-1)$.
$(a)$ What is the dimension of $M \otimes_{\mathbb{C}[x]}N$ as a vector space over $\mathbb{C}$?
$(b)$ What is the dimension of $M \otimes_{\mathbb{C}}N$ as a vector space over $\mathbb{C}$?
$(c)$ What is the dimension of $M \otimes_{\mathbb{C}[x^2]}N$ as a vector space over $\mathbb{C}$?
I solved the part $(a)$ and $(b)$. The part $(b)$ is the easiest. Since $M$ (resp. $N$) is a free $\mathbb{C}$-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 $$ \mathbb{C}[x]/(x^2+x,x-1) = 0. $$
I don't know how to solve the part $(c)$. First, I showed $$ M \otimes_{\mathbb{C}[x^2]} N = \{ c(1 \otimes_{\mathbb{C}[x^2]} 1): c \in \mathbb{C} \}. $$ via direct computations. However, this does not guarantee that the dimension is $1$. We need to check that $1 \otimes_{\mathbb{C}[x^2]} 1 \neq 0$. 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 $\beta:M\times N\to\Bbb C, \ (a+bx, \,c) \mapsto (b-a)c$.
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)\cong\Bbb C\ $ as $1$: $$((a+bx)x^2,\, c) = (-ax+bx, \, c) \mapsto (b-a)c\\ (a+bx, \, x^2c) =(a+bx, c) \mapsto (b-a)c$$