Spectrum of a tensor product. I need to compute the spectrum of the following ring,
$$ k[T,U]\big/ (T^2-U^2) \otimes_{k[T]} k[T,U]\big/ (TU-1)$$
where all the maps considered to compute this tensor product are the canonical ones. So I need a simpler description of this ring. In other simple cases I could relate some of the factors with $k[T]/I$ and apply well known properties of the tensor product but in this case I don't know how should I proceed.
 A: In general, if $B$ and $C$ are $A$-algebras and $I\subseteq B$ and $J\subseteq C$ are ideals, then $$B/I\otimes_A C/J\cong (B\otimes_A C)/(I,J),$$ where $(I,J)$ denotes the ideal in $B\otimes_A C$ generated by the images of $I$ and $J$.  In your case with $A=k[T]$, you get $$A[U]/(T^2-U^2)\otimes_A A[V]/(TV-1)\cong A[U,V]/(T^2-U^2,TV-1)$$ (I have renamed one of the $U$s to $V$ to avoid confusion).  So your ring is just $k[T,U,V]/(T^2-U^2,TV-1)$ and you can describe its spectrum as the closed subscheme of $\mathbb{A}^3_k$ cut out by the polynomials $T^2-U^2$ and $TV-1$.
To get an even more explicit description, you can note that the relation $TV-1$ just says $V$ is an inverse to $T$, so you can write the ring as $k[T,T^{-1},U]/(T^2-U^2)$.  Assuming $\operatorname{char} k\neq 2$, $T-U$ and $T+U$ generate the unit ideal in $k[T,T^{-1},U]$ (since $(T-U)+(T+U)=2T$ is a unit), so by the Chinese remainder theorem the ring splits as a product $$k[T,T^{-1},U]/(T-U)\times k[T,T^{-1},U]/(T+U).$$  Each factor is itself isomorphic to $k[T,T^{-1}]$ (by mapping $U$ to $T$ or $-T$, respectively), so the ring is isomorphic to $$k[T,T^{-1}]\times k[T,T^{-1}].$$  Its spectrum is thus a disjoint union of two copies of $\operatorname{Spec} k[T,T^{-1}]=\mathbb{A}^1_k\setminus\{0\}$.
(As mentioned, this is only valid if $\operatorname{char} k\neq 2$.  If $\operatorname{char} k=2$, then $T^2-U^2=(T-U)^2$ and your ring is not reduced.  When you mod out the nilradical you get $k[T,T^{-1},U]/(T-U)\cong k[T,T^{-1}]$, so the spectrum has the same points as $\mathbb{A}^1_k\setminus\{0\}$.)
