TITLE: Is the fiber product of two irreducible schemes an irreducible scheme ?
QUESTION [5 upvotes]: If $X$ is an integral scheme, is it true that $ X $  x  Spec $Z[t] $ over Spec $Z $ an integral scheme ? I know that it is reduced, but I could not show that it is irreducible.

REPLY [5 votes]: Following Martin Brandenburg's nice notation, let me denote $Y\times_{\text{Spec }\mathbb{Z}}\mathbb{A}^1_\mathbb{Z}$ by $Y[t]$.
As you noted $X[t]$ is obviously reduced. To see it's irreducible it suffices to cover it by irreducible, pairwise intersecting open subschemes. But, take an affine open cover $\{U_\alpha\}$ of $X$. Then, $\{U_\alpha[t]\}$ is an irreducible open cover for $X[t]$. Note though that for all $\alpha,\beta$ we have $U_\alpha\cap U_\beta$ is non-empty (since $X$ is irreducible) and thus $U_\alpha[t]\cap U_\beta[t]$ is also non-empty. Thus, $X[t]$ is irreducible as desired.