This is exercise 9.3.E in Vakil's Foundations of Algebraic Geometry
Let $k$ be a field of characteristic not 2. Let $X = \operatorname{Spec}(k[t]), Y = \operatorname{Spec}(k[u])$, and consider the morphism $f : X \rightarrow Y$ induced by the ring map $u \rightarrow t^2$. Show that $X \times_Y X$ has two irreducible components.
I made the substitution $u = t^2$ to get $X = \operatorname{Spec}(k[t]) \rightarrow Y = \operatorname{Spec}(k[t^2])$, so the morphism is just induced by inclusion. Thus, the fiber product is $\operatorname{Spec}(k[t] \otimes_{k[t^2]} k[u])$
My approach is to try to find the generic points of the two irreducible components, which correspond to minimal prime ideals. I know that every prime ideal must contain at least one of the two generic points, and no prime ideal is contained in both.
If the zero ideal is prime, then there would only be one irreducible component. Since the question said there are two, that implies zero is not irreducible. There are zero divisors. Since every prime ideal must contain one element of a pair of zero divisors, this should give me elements that every prime ideal must contain.
My first question is : is this correct so far?
My second question is : I found two pairs of zero divisors $(t \otimes t - t^2 \otimes 1) * (t \otimes t + t^2 \otimes 1) = 0$ and $(t \otimes 1 - 1 \otimes t) * (t \otimes 1 + 1 \otimes t) = 0$. I'm not sure how to convert these into the minimal prime ideals that I need to solve this problem. Can you please give me a hint?