Let $k$ be a field. I’m trying to do exercise $4.13.a)$ from the book Algebraic Geometry I by Görtz & Wedhorn, which asks to describe the fibers in all points of the following morphism of schemes and see if they're irreducible or reduced:
$$f:\operatorname{Spec} k[T,U]/(TU-1) \rightarrow \operatorname{Spec} k[T]$$
I know that there’s two kind of points in $\operatorname{Spec} k[T] \cong \mathbb{A}^1_k$, depending on whether the associated prime ideal $\mathfrak{p}$ is of the form $(0)$ (the closed point) or $(T-a)$ (for $a\in k$). I've calculated the residue field and the fiber for each of these two kinds of points, and obtained:
For $\mathfrak{p}=(0)$, $\kappa(\mathfrak{p}) \cong k(T)$, and, using that $k[T,U]/(TU-1) \cong k[T,T^{-1}] $,
$$f^{-1}(\mathfrak{p})=\operatorname{Spec}(k[T,T^{-1}] \otimes_{k[T]} k[T])=\operatorname{Spec}k[T,T^{-1}]$$
For $\mathfrak{p}=(T-a)$, $\kappa (\mathfrak{p})\cong k$, and $$f^{-1}(\mathfrak{p})=\operatorname{Spec}(k[T,T^{-1}] \otimes_{k[T]} k).$$
I don't see how to simplify this second expression further or how to make any progress from there. Also, I would like to check if my reasoning so far is correct.
Best Answer
Is $k$ algebraically closed? If not, then there are a lot more closed points.
Assuming it is, then it is important to remember how $k[T]$ acts on $k$. I.e. $T\otimes 1 = 1 \otimes a$.
Then there is a dichotomy about whether or not $a$ is zero. If $a$ is zero, then $T \otimes 1 = 1 \otimes 0$, hence $1 \otimes 1 = (TT^{-1}) \otimes 1 = T^{-1} \otimes 0$, thus $k[T, T^{-1}] \otimes_{k[T]} k = 0$.
If not, then I claim $k[T, T^{-1}] \otimes_{k[T]} k \cong k$, via the map $T \otimes 1 \mapsto 1 \otimes a$. I'll leave the details to you.
Basically what's happening is that this map is just the inclusion of the punctured line into the affine line. If $a \neq 0$, then the "preimage" has only one point. At zero, there is no preimage so it is empty.
Maybe another way to think about this is literally just substitution. If you're looking at the fibre over $T = a\in \mathrm{Spec}k[T]$, then you literally just substitute $T = a$ into $k[T, T^{-1}]$. Again, I'll leave the details to you.