Dimension of a tensor product as a vector space over complex number field $\mathbb{C}$ The tensor product $\mathbb{C}[x]/(x^2+x)\otimes_{\mathbb{C}[x^2]}\mathbb{C}[x]/(x^2-1)$ can be viewed as a vector space over $\mathbb{C}$, where $\mathbb{C}[x]/(x^2+x)$ and $\mathbb{C}[x]/(x^2-1)$ are quotient rings of the polynomial ring over field $\mathbb{C}$ and both of them can be viewed as $\mathbb{C}[x^2]$-modules in a natural way since $\mathbb{C}[x^2]$ is a subring of $\mathbb{C}[x]$.
For simplicity, denote $[1]$ and $[x]$ the class of $1$ and $x$ in $\mathbb{C}[x]/(x^2+x)$ (or $\mathbb{C}[x]/(x^2-1)$). I have shown that each element in $\mathbb{C}[x]/(x^2+x)\otimes_{\mathbb{C}[x^2]}\mathbb{C}[x]/(x^2-1)$ can be written as a $\mathbb{C}$-linear combination of $[1]\otimes[1]$, $[1]\otimes[x]$, $[x]\otimes[1]$ and $[x]\otimes[x]$. But in this tensor product
$[1]\otimes[1]=[1]\otimes[x^2]=[x^2]\otimes[1]=-[x]\otimes[1]$,
$[1]\otimes[x]=[1]\otimes[x^3]=[x^2]\otimes[x]=-[x]\otimes[x]$.
This shows that $\mathbb{C}[x]/(x^2+x)\otimes_{\mathbb{C}[x^2]}\mathbb{C}[x]/(x^2-1)$ can be generated by $[1]\otimes[1]$ and $[1]\otimes[x]$ over $\mathbb{C}$. Therefore, the possible dimension of $\mathbb{C}[x]/(x^2+x)\otimes_{\mathbb{C}[x^2]}\mathbb{C}[x]/(x^2-1)$ over $\mathbb{C}$ is $1$ or $2$.
So my question is: whether $[1]\otimes_{\mathbb{C}[x^2]}[1]$ and $[1]\otimes_{\mathbb{C}[x^2]}[x]$ are linearly independent over $\mathbb{C}$ or not?
 A: By Chinese Remainder, $$\mathbb C[x]/(x^2+x)\cong \mathbb C[x]/(x)\oplus\mathbb C[x]/(x+1)\simeq \mathbb C\oplus \mathbb C$$ $$\mathbb C[x]/(x^2-1)\cong\mathbb C[x]/(x+1)\oplus \mathbb C[x]/(x-1)\simeq \mathbb C\oplus\mathbb C$$
These decompositions are $\mathbb C[x]$-module decompositions, therefore $\mathbb C[x^2]$-module decompositions too. Let $\mathbb C_a$ be the $\mathbb C[x^2]\cong\mathbb C[y]$-module such that its underlying $\mathbb C$-module is isomorphic to $\mathbb C$ and $x^2=y$ acts on $\mathbb C$ through multiplication by $a\in\mathbb C$.
We claim $\mathbb C_a\otimes_{\mathbb C[y]}\mathbb C_b =\{0\}$ if $a\not=b$. Indeed we have $$(y\cdot 1)\otimes 1 = a(1\otimes 1) = 1 \otimes (y\cdot 1) = 1\otimes b = b(1\otimes 1)$$ hence $(a-b)(1\otimes 1)=0$ and $1\otimes 1 = 0$ if $a\not=b$.
(P.S. This actually follows from the general fact $R/ I\otimes_R R/J = \{0\}$ if $I+J=(1)$.)
Similarly we also have $\mathbb C_a\otimes_{\mathbb C[y]} \mathbb C_a\cong \mathbb C_a$.
Then the original tensor product is isomorphic to $$(\mathbb C_0\otimes \mathbb C_{1})\oplus (\mathbb C_0\otimes \mathbb C_1) \oplus (\mathbb C_{1}\otimes\mathbb C_{1}) \oplus (\mathbb C_{1}\otimes \mathbb C_1)\cong \mathbb C_1\oplus\mathbb C_1$$ has dimension 2 over $\mathbb C$.
More concretely, it's easy to see that $[x]\otimes [1]$ and $[x]\otimes [x]$ are both fixed by $x^2$, as $[x]\in\mathbb C[x]/(x^2+x)$ and $[1], [x]\in\mathbb C[x]/(x^2-1)$ are all fixed by $x^2$.
