Let $X$ be a scheme over some field with a map $f: X \to \mathbb{A}^1$ such that for all $t_1, t_2 \in \mathbb{A}^1 \setminus 0$ the fibers $f^{-1} (t_1)$ and $f^{-1} (t_2)$ are isomorphic. The fiber over $0$ may be different. Thus, we have a "generiс" fiber over a nonzero point and a "special" fiber over $0$. My intuition suggests that the fiber over the generic point $\eta \in \mathbb{A}^1$ should have something to do with this "generic" fiber over a nonzero point. Isn't it right? Because in particular examples this intuition fails and may be it is just wrong.

To be concrete: let $f: \mathbb{A}^1 \to \mathbb{A}^1$ be a covering of degree 2 ramified at $0$: namely, $f(x)=x^2$. Thus, this "generic" fiber consists of two points and the "special" fiber consists only of one point. We have a map $k[x] \to k[y]$ where $x$ goes to $y^2$. Let's calculate the preimage of the generic fiber: it is just $k[y] \otimes_{k[x]} k(x)$ which is isomorphic to $k(y)$ and the map $k(x) \to k(y)$ is given by the same formula. Note that $k(y)$ is a ring of function on a point which is not the "generic" fiber of this covering.

Is my intuition wrong or am I missing something?


Let me make a quick terminology note before continuing: we'll use "general fiber" for the fiber over a general point and "generic fiber" for the fiber over the generic point.

It is true that in nice circumstances, many properties of the general and generic fibers are equivalent: that is, for a nice enough map $X\to Y$, the existence of an open locus of $y\in Y$ so that $X_y$ has property $P$ is equivalent to $X_\eta$ having property $P$, where $\eta$ is the generic point of $Y$. For instance, in the presence of mild finiteness assumptions, flatness, (geometric) reducedness, (geometric) connectedness, dimension of fibers, global sections of some fixed dimension $d$, and various other conditions can play the role of $P$.

On the other hand, not all properties work this way. Indeed, you've discovered that cardinality of the fiber is one property that doesn't!