Is the fiber product of two irreducible schemes an irreducible scheme ?

Question

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.

Answer 1

In general no. For instance, $\mathbf C \otimes_{\mathbf R} \mathbf C$ is not a domain.

Answer 2

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.

Answer 3

Yes. In the affine case, this is the well-known result:

• If $R$ is an integral domain, then $R[t]$ is an integral domain.

Now also observe that:

• If $R \to S$ is injective, then $R[t] \to S[t]$ is injective.

Geometrically, this means: If $U \subseteq V$ are two non-empty open affines in $X$ (since $X$ is integral, this implies $\mathcal{O}(V) \hookrightarrow \mathcal{O}(U)$!), then the induced inclusion $U[t] \to V[t]$ preserves the generic points. By looking at intersections, it follows that $U[t]$ has the same generic point as $V[t]$ for every pair of non-empty open affines $U,V$ in $X$. This unique point is a generic point of $X[t]$. Hence, $X[t]$ is irreducible.