Problem: Show that $$\mathbb{C}[X,Y]/\langle{XY-1}\rangle\otimes_\mathbb{C}\mathbb{C}[X,Y]/\langle{X^2-Y}\rangle\cong \mathbb{C}[X,X^{-1},Z],$$ $$\mathbb{C}[X,Y]/\langle{XY-1}\rangle\otimes_{\mathbb{C}[X,Y]}\mathbb{C}[X,Y]/\langle{X^2-Y}\rangle\cong\mathbb{C}[X]/\langle{X^3}\rangle.$$ Also, find their dimensions over $\mathbb{C}$.
Question: I understand the proofs, but I don't get why $$\mathbb{C}[X,Y]/\langle{XY-1}\rangle\otimes_\mathbb{C}\mathbb{C}[Z,W]/\langle{Z^2-W}\rangle\cong\mathbb{C}[X,Y,Z,W]/\langle{XY-1,Z^2-W}\rangle$$ and $$\mathbb{C}[X,Y]/\langle{XY-1}\rangle\otimes_{\mathbb{C}[X,Y]}\mathbb{C}[X,Y]/\langle{X^2-Y}\rangle\cong\mathbb{C}[X,Y]/\langle{XY,Y-X^2}\rangle.$$ Any thoughts on this? Perhaps could you point me to the theorem that makes this true?
Also, I'm guessing the first dimension is infinite and the other one is $3$. But I don't really know how to argue in that sense. I appreciate any help. Thanks in advance!
Keep in mind that tensor products are coproducts in the category of algebras. This means that the tensor product $A \times_C B$ is characterised by $\mathrm{Hom}(A \times_C B, D) \simeq \mathrm{Hom}(A, D) \times \mathrm{Hom}(B, D)$ where $Hom$ signifies $C$-algebra homomorphisms and the bijection is natural in $D$.
Furthermore, we see that $M = C[x_1, ..., x_n] / (P_1(x_1, ..., x_n), ..., P_m(x_1, ..., x_n))$ is characterized by $\mathrm{Hom}(M, D) \simeq \{(a_1, ..., a_n) \in D^n | P_i(a_1, ...., a_i) = 0$ for all $i\}$ (again the bijection being natural in $D$).
So in general, we have
$\begin{equation} \begin{split} \mathrm{Hom}(C[x_1, ..., x_n] / (P_1(x_1, ..., x_n), ..., P_m(x_1, ..., x_n)) \otimes_C C[y_1, ..., y_p] / (Q_1(y_1, ..., y_p), ..., Q_r(y_1, ..., y_p)), D) \\ \simeq \mathrm{Hom}(C[x_1, ..., x_n] / (P_1(x_1, ..., x_n), ..., P_m(x_1, ..., x_n)), D) \times \mathrm{Hom}(C[y_1, ..., y_p] / (Q_1(y_1, ..., y_p), ..., Q_r(y_1, ..., y_p)), D) \\ \simeq \{(a_1, ..., a_n) \in D^n \mid \forall i,P_i(a_1, ..., a_n) = 0\} \times \{(b_1, ..., b_p) \in D^p \mid \forall j, Q_j(b_1, ..., b_p) = 0\} \\ \simeq \{(a_1, ..., a_n, b_1, ..., b_p) \in D^{n + p} \mid (\forall i, P_i(a_1, ..., a_n) = 0) \land (\forall j, Q_j(b_1,..., b_p) = 0\} \\ \simeq \mathrm{Hom}(C[x_1,..., x_n, y_1,..., y_p] / (P_1(x_1, ..., x_n), ..., P_m(x_1, ..., x_n), Q_1(y_1, ..., y_p), ..., Q_r(y_1, ..., y_p)), D). \end{split} \end{equation}$
So therefore, we have
$C[x_1, ..., x_n] / (P_1(x_1, ..., x_n), ..., P_m(x_1, ..., x_n)) \otimes_C C[y_1, ..., y_p] / (Q_1(y_1, ..., y_p), ..., Q_r(y_1, ..., y_p)) \simeq C[x_1,..., x_n, y_1,..., y_p] / (P_1(x_1, ..., x_n), ..., P_m(x_1, ..., x_n), Q_1(y_1, ..., y_p), ..., Q_r(y_1, ..., y_p))$
by the Yoneda Lemma.
Then we see that as a special case of this, it follows that
$$\mathbb{C}[X, Y] / (XY - 1) \otimes_\mathbb{C} \mathbb{C}[Z, W] / (Z^2 - W) \simeq \mathbb{C}[X, Y, Z, W] / (XY - 1, Z^2 - W)$$
As another case of this, taking the ring to be $\mathbb{C}[X, Y]$, it follows that
$$\mathbb{C}[X, Y] / (XY - 1) \otimes_{\mathbb{C}[X, Y]} \mathbb{C}[X, Y] / (X^2 - Y) \simeq \mathbb{C}[X, Y] / (XY - 1, X^2 - Y)$$
And as $\mathbb{C}$-algebras, we see that $\mathbb{C}[X, Y] / (XY - 1, X^2 - Y) \simeq \mathbb{C}[X] / (X^3 - 1)$ (and we can take the latter a $\mathbb{C}[X, Y]$-algebra by sending $Y$ to $X^2$, making this a $\mathbb{C}[X, Y]$-algebra isomorphism).
Now suppose that there were a $\mathbb{C}$-algebra isomorphism $f : \mathbb{C}[X] / (X^3 - 1) \to \mathbb{C}[X] / (X^3)$. Then write $f(X) = a X^2 + b X + c$. Then it must be the case that $f(X)^3 = f(X^3) = 1$. Therefore, we see that $f(X)^3 = (aX^2 + bX + c)^3 = c^3 + 3c^2 (bX + aX^2) + 3 c b^2 X^2 = 1$. Then we see that $c^3 = 1$. We also see that $3 c^2 b = 0$; then $b = 0$. Then we must also conclude that $a = 0$. But then we see that $f(X) \in \mathbb{C}$, and thus the image of $f$ is a subset of $\mathbb{C}$. Then $f$ is not a bijection. Contradiction.
Thus, we see that your second answer is not correct. There is no $\mathbb{C}$-algebra isomorphism between $\mathbb{C}[X, Y] / (XY - 1) \otimes_{\mathbb{C}[X, Y]} \mathbb{C}[X, Y] / (X^2 - Y)$ and $\mathbb{C}[X] / (X^3)$, so there is certainly no $\mathbb{C}[X, Y]$-algebra isomorphism.