Show that $\mathbb Q[t] \otimes_{\mathbb Q[t^2]} \mathbb Q[t] \cong \mathbb Q[x,y]/(x^2-y^2)$ (as $\mathbb Q$--algebras)
So I first tried to show this by first defining the map $\varphi: \mathbb Q[t] \times \mathbb Q[t] \to \mathbb Q[x,y] / (x^2 - y^2)$ by sending $(f(t), g(t))$ to $f(x)g(y) + I$ where $I = (x^2 - y^2)$. Then we can check that $\varphi$ is $R$-balanced. Letting $R:= \mathbb Q[t^2]$, we can give $\mathbb Q[x,y]/I$ an $R$-module structure by defining $h(t^2) \cdot (j(x,y) + I) := h(x^2)j(x,y) + I = h(y^2)j(x,y) + I$. Then we may check that the map $\varphi$ is $R$-bilinear map, so that by the universal property of the tensor product, we have an $R$-module homomorphism $\Phi: \mathbb Q[t] \otimes \mathbb Q[t] \to \mathbb Q[x,y]/(x^2 - y^2)$. We can check that $\Phi$ is surjective since $t \otimes 1 \mapsto x$ and $1 \otimes t \mapsto y$. I am not sure how to prove that the map $\Phi$ is injective. How can I show this?
Someone mentioned to me that there also is another way to solve this problem, and I think his idea was to 'consider $t$ on each size as the square root of $t^2$", which made little sense to me. I am wondering how to prove this question by an alternative method also.
Thank you
You could also define an inverse map. Let $f:\mathbb Q[x,y]\rightarrow \mathbb Q[t]\otimes_{\mathbb Q[t^2]}\mathbb Q[t]$ be given by $x\mapsto t\otimes 1$, $y\mapsto 1\otimes t$. Then we see that $$(x^2-y^2)\mapsto t^2\otimes 1-1\otimes t^2= 1\otimes t^2-1\otimes t^2=0$$ so we get a unique map $\tilde{f}:\mathbb Q[x,y]/\langle x^2-y^2\rangle\rightarrow \mathbb Q[t]\otimes_{\mathbb Q[t^2]}\mathbb Q[t]$. This is clearly the inverse of your map $\mathbb Q[t]\otimes_{\mathbb Q[t^2]}\mathbb Q[t]\rightarrow \mathbb Q[x,y]/\langle x^2-y^2\rangle$, so you have an isomorphism.