So I have recently been studying fibres of points of morphisms of schemes. Hartshorne has an exercise doing a specific example of one of these. The exercise is II 3.10. Suppose we are given a morphism $f: \text{Spec }B \longrightarrow \text{Spec } A$ of affine schemes and $y \in Y$ is some point in Y. Let $\mathfrak{p}$ be the prime ideal of $A$ corresponding to the point $y$. Then the points in the fibre $f^{-1}(y)$ correspond to the prime ideals of $$ B \otimes_{A} \kappa(y) $$ where $\kappa(y)$ is the residue field of $y$. I am comfortable with that fact, and I proved that primarily with commutative algebra and the Going-Up theorem. I am feeling slightly uncomfortable using it in a geometric context though.
The exercise I mentioned above uses the example of a morphism, $$ f: \text{Spec }k[s, t]/(s - t^{2}) \longrightarrow \text{Spec }k[s]. $$ with $k$ an algebraically closed field and the morphism determined by the morphism of rings given by the composotion $$ k[s] \stackrel{\text{inclusion}}{\longrightarrow} k[s, t] \stackrel{\text{quotient}}{\longrightarrow} k[s, t]/(s - t^{2}) $$
I am asked to describe the fibres of the points in $k[s]$ by breaking it into three cases: The point corresponding to the $0$ element of $k$, the closed points corresponding to the non-zero elements of $k$, and the generic point.
Is someone able to see if my argument looks valid? I am a little uncomfortable that I haven't manipulated the tensor products correctly also. The following is my attempt: We will consider the closed points of $\text{Spec } k[s]$ first. These correspond to maximal ideals of $k[s]$ which are of the form $(s - a )$ and which we denote by $\mathfrak{m}_{a}$. Consider first the case of $a \neq 0$. We would like to examine the spectrum of the ring \begin{align*} k[s, t] / (s - t^{2}) \otimes_{k[s]} k &\simeq k[s, t] / (s - t^{2}) \otimes_{k[s]} k[s] / \mathfrak{m}_{a} \\ & \simeq k[t] \otimes_{k[s]} k[s] / \mathfrak{m}_{a} \\ & \simeq k[t] / \left( \mathfrak{m}_{a} k[t] \right) \end{align*} Now the inclusion morphism above fixes $s$, and so the ideal $\mathfrak{m}_{a} k[t]$ is the ideal $$(\pi(s) - a)k[t] \simeq (t^{2}- a)k[t]$$ which is the ideal $(t^{2} - a) \subseteq k[t]$. I claim that the resulting quotient ring $k[t]/(t^{2} - a)$ has two prime ideal. Indeed any such prime ideal will correspond to a prime ideal of $k[t]$ containing $(t^{2} - a)$. Since $k$ is algebraically closed, we have a unique factorization, $$ t^{2} - a = (t - \beta)( t - \gamma), $$ and it is a simple task to check that if $a \neq 0$ then $\beta \neq \gamma$. The points of the fiber then correspond to the prime ideals $(t - \beta)$ and $(t - \gamma)$. Observe also that we didn't invoke the fact that $a \neq 0$ until now, so the same argument as above works for the case $a = 0$ except now we only have the one prime ideal $(t)$. Indeed the fiber is given by the scheme $\text{Spec}\left( k[t] / (t^{2}) \right)$ which is certainly not reduced since $k[t] / (t^{2})$ has a niloptent element.
Finally, we consider the case of the generic point $\eta \in Y$. The residue field is $k(s)$, the field of rational functions in $s$ (the transcendence degree $1$ extension of $k$). Then we have $$ k(s) \otimes_{k[s]} k[s, t] / (s - t^{2}) \simeq k(s)[t] / (s - t^{2}). $$ which is an extension of degree $2$ over $k(s)$. Indeed a basis over $k(s)$ as a vector space is given by $\{1, t \}$.