Fiber product of irreducible curves Consider a collection $Y_1,\ldots,Y_n$ of absolutely irreducible curves define over a field $\mathbb{K}$ each equipped with a map to another curve $Z$. Is $X=Y_1\times_Z\ldots\times_Z Y_n$ absolutely irreducible? 
 A: By definition of fiber product of schemes: there exists a canonical morphism $X\to Z$, so $X$ is a scheme over $Z$; but in general $X$ is not irreducible.
Example. Let $\mathbb{K}=\mathbb{Q}$, $Z=\mathrm{Spec}\mathbb{Q}[z],\,Y_1=Y_2=\mathrm{Spec}\mathbb{Q}(\sqrt{2})[x]$ and the morphisms of schemes of $Y$'s over $Z$ are induced by the canonical inclusion $\mathbb{Q}[x]\hookrightarrow\mathbb{Q}(\sqrt{2})[x]$. One has
\begin{gather}
\dim_{Krull}Y_1=\dim_{Krull}Y_2=\dim_{Krull}Z=1,\\
Y_1,\,Y_2\,\text{and}\,Z\,\text{are irreducible (curves) over}\,\mathbb{Q},\\
X=\mathrm{Spec}\left(\mathbb{Q}(\sqrt{2})[x]\otimes_{\mathbb{Q}[z]}\mathbb{Q}(\sqrt{2})[y]\right)\cong\mathrm{Spec}\left(\mathbb{Q}[x,t]_{\displaystyle/(t^2-2)}\otimes_{\mathbb{Q}[z]}\mathbb{Q}(\sqrt{2})[y]\right)\cong\\
\cong\mathrm{Spec}\mathbb{Q}(\sqrt{2})[x,t]_{\displaystyle/(t^2-2)}\cong\mathrm{Spec}\left(\mathbb{Q}(\sqrt{2})[x]\times\mathbb{Q}(\sqrt{2})[y]\right)=\\
=\mathrm{Spec}\left(\mathbb{Q}(\sqrt{2})[x]\right)\coprod\mathrm{Spec}\left(\mathbb{Q}(\sqrt{2})[y]\right)
\end{gather}
that is $X$ is the union of two (irreducible) curves over $\mathbb{Q}$. $(Q.E.D\,\Box)$
