Affine fibre products are spectra of tensor products

Given: The objects, hypotheses and conventions in the statement above.

Let $f:X\to S$ and $g:Y\to S$ be morphisms of schemes. A fibre product is a scheme $P$, with projections $p:P\to X$ and $q:P\to Y$, such that $fp=gq$ and, for every scheme $T$ and morphisms $a:T\to X$, $b:T\to Y$ with $fa=gb$, there is exactly one $h:T\to P$ satisfying $ph=a$ and $qh=b$. Thus, naturally in every test scheme $T$, $\mathrm{Hom}(T,P)\cong \mathrm{Hom}(T,X){\times }_{\mathrm{Hom}(T,S)}\mathrm{Hom}(T,Y)\mathrm{.}$ Write $P=X{\times }_{S}Y$. The commutative square with edges $p,q,f,g$ is Cartesian when it has this universal property. Morphisms here are morphisms of locally ringed spaces, as in def-morphism-of-schemes. No existence assertion is part of the definition. (Fibre product of schemes)

For a scheme $X$ and a ring $A$, taking global sections induces a natural bijection $\mathrm{Hom}(X,\mathrm{Spec}A)\cong {\mathrm{Hom}}_{\mathrm{C}\mathrm{R}\mathrm{i}\mathrm{n}\mathrm{g}}(A,\mathrm{\Gamma }(X,{\mathcal{O}}_X))\mathrm{.}$ (Morphisms to an affine scheme and global sections)

Let $A,B,C$ be commutative $R$-algebras. For every pair of $R$-algebra homomorphisms $f:A\to C$ and $g:B\to C$, there is a unique $R$-algebra homomorphism $h:A{\otimes }_{R}B⟶C$ such that $h(a\otimes 1)=f(a)$ and $h(1\otimes b)=g(b)$. It is given by $h(a\otimes b)=f(a)g(b)\mathrm{.}$ Thus $A{\otimes }_{R}B$, with its two canonical maps, is the coproduct of $A$ and $B$ among commutative $R$-algebras. (Universal mapping property of the tensor product of commutative algebras)