Tensor product of quotient rings of polynomial rings

Question

I am reading Algebraic Geometry, Vol 1, Kenji Ueno. My problem is that $$k\left[ x,y,t\right]/\left(xy-t\right)\otimes_{k\left[t\right]}k\left[t\right]/\left(t-a\right) \simeq k\left[x,y\right]/\left(xy-a\right) $$ where $k$ is a field and $a$ is an element in $k$. I don't understand how it works. I tried to use the result $$R\otimes_A A/I =R/IR,$$ where $A$ and $R$ are commutative rings and $I$ is an ideal of $A$. Then I obtain $$LHS=\left[k\left[ x,y,t\right]/\left(xy-t\right)\right]/\left[\left(t-a\right)k\left[ x,y,t\right]/\left(xy-t\right)\right].$$ But now it seems be hard to manage. Please help me. Thank you very much for helps.

Answer 1

First check the isomorphism $$k[x,y,t]\otimes_{k[t]} k[t]\simeq k[x,y,t]$$ where the isomorphism is $$a\otimes b\mapsto ab.$$

Next note that the ideal $(t-a)$ on the right hand side corresponds to the ideal $(1\otimes (t-a))$ on the left. This gives the isomorphism

$$k[x,y,t]\otimes_{k[t]} k[t]/(t-a)\simeq k[x,y]$$

where the map is $$p(x,y,t)\otimes 1\mapsto p(x,y,a)$$

And finally the ideals $((xy-t)\otimes 1)$ and $(xy-a)$ correspond.

Answer 2

Let $R=k[x,y,t]/(xy-t)$.

Consider the surjective map $R \to k[x,y]/(xy-a)$ given by $x \mapsto x, y \mapsto y, t \mapsto a$.

We have to show that the kernel is the principal ideal $(t-a)R$, because this shows $R/(t-a)R \cong k[x,y]/(x-a)$ as desired.

Clearly $t-a$ is contained in the kernel. On the other hand, let $f \in R$ be contained in the kernel, i.e. $f(x,y,a) \in (xy-a)$. Division with remainder yields $f(x,y,t) = (t-a)g(x,y,t) + r(x,y)$. Evaluate at $t=a$ to get $r(x,y) \in (xy-a)$. Thus we have $f \in (t-a,xy-a)=(t-a,xy-t)$, i.e. $f \in (t-a)R$.