 Solution

A fibre product of schemes $Z=X{\times }_{S}Y$ has projection morphisms to $X$ and $Y$ with equal composites to $S$, and for every scheme $T$ the induced map is a bijectionThis universal property determines $Z$ uniquely up to unique isomorphism. On affine charts over $\mathrm{Spec}R$, takeThe tensor product of commutative algebras has the required universal property for ring maps. Localization identifies these constructions on principal-open overlaps. Cover the base and its two inverse images by affine charts and glue the resulting affine products along those overlaps; their local universal properties give the global one.

For a counterexample to preservation of irreducibility, take $X=Y=\mathrm{Spec}\mathbb{C}$ and $S=\mathrm{Spec}\mathbb{R}$ with their usual structure morphisms. All three are integral schemes, but the Chinese remainder theorem givesThe product scheme is the disjoint union of two points, so it is not irreducible.