My task is to classify the fibers of $\text{Spec }\mathbb{C}[x,y,t] / \langle xy-t \rangle \to \text{Spec } \mathbb{C}[t]$ (map induced by the inclusion) as reducible or irreducible. Specifically, "Are there reducible fibers? How does the fiber over the unique dense point of $\mathbb{C}[t]$ (the "generic point") compare to a "generic" or "typical" fiber?
The prime ideals of $\mathbb{C}[t]$ are $\langle t-c \rangle, \{0\}.$ As $t = xy$ gets absorbed, the prime ideals of $\text{Spec }\mathbb{C}[x,y,t] / \langle xy-t \rangle$ are $\langle p(x,y) \rangle$ for non-constant irreducible $p$ and the usual $ \langle x-a, y-b \rangle, \{0\}.$ We have $f^{-1}(\{0\}) = \{ \langle p(x,y) \rangle : p \ne xy-c\}$ and $f^{-1}(\langle x-c \rangle) = \{\langle xy-c \rangle \} \cup \{ \langle x-a, y-b \rangle : ab = c\}.$
The closed sets will be $\mathbb{C}[x,y,t] / \langle xy-t \rangle,$ or finite unions of $\{ (x-a, y-b)\}, \{f, (x-a, y-b) : \text{ f irred, } f(a,b) = 0\}.$ This makes the fiber of $\langle x-c \rangle$ not expressible as the union of 2 smaller closed sets. Meanwhile, the fiber of $\{0\}$ is not a closed set (contains no maximal ideals), so how can we judge whether it's reducible or irreducible? By definition as not being of the form $A \cup B$ for $A, B$ disjoint and closed, it's irreducible, but it seems so large it should be reducible under a proper definition.
