Suppose $\pi:X \rightarrow Y$ is a (locally trivial) fiber bundle $F$, where all spaces are Noetherian. Suppose $F$ and $Y$ are irreducible; show that $X$ is also irreducible.
Here a similar question is discussed; but this question should be simpler, and should have an easier proof.
Thanks in advance.
On
Since $\pi$ is locally trivial, it is an open map. Thus the result follows by http://stacks.math.columbia.edu/tag/004Z which states:
Let $f : X \to Y$ be a continuous map of topological spaces. Assume that (a) $Y$ is irreducible, (b) $f$ is open, and (c) there exists a dense collection of points $y \in Y$ such that $f^{-1}(y)$ is irreducible. Then $X$ is irreducible.
The following proof works if one assumes that the fiber $F$ is proper. I am suspicious it might not be true if the fiber is not proper.
Let $X = Z_1 \cup Z_2$ where each $Z_i$ is closed. Then since each fiber $\iota_y : F \hookrightarrow X$ is irreducible, we have that either $\iota_y^{-1}(Z_i)$ is of the same dimension as $F$ and equal to all of $F$, or has dimension strictly less than that of $F$.
Note that the morphism $X \rightarrow Y$ is proper if and only if $F$ is proper. To see this, properness is local on the target, and restricting to an open cover our map is of the form $U \times F \rightarrow U$. This morphism is proper if and only if $F$ is.
Assume $F$ and hence $X \rightarrow Y$ is proper, by upper-semicontinuity of the dimension of the fiber with respect to the morphism $Z_i \rightarrow Y$, we see that there is a closed subscheme $W_i \subset Y$ where for every closed point $y \in W_i$, $\iota_y^{-1}(Z_i)$ consists of the entire fiber $F$. Since $W_1 \cup W_2 = Y$, it must be that $W_1 = Y$ or $W_2 = Y$. Of course, this implies $Z_1 = X$ or $Z_2 = X$.