# Bijection between $\text{Spec}(A\otimes_R B)$ and $\text{Spec}(A)\times_{\text{Spec}(R)}\text{Spec}(B)$ for $R$-algebras A and B.

I'm looking for a bijection between $$\text{Spec}(A\otimes_R B)$$ and $$\text{Spec}(A)\times_{\text{Spec}(R)}\text{Spec}(B)$$ for $$R$$-algebras $$A$$ and $$B$$.

Where $$\text{Spec}(A)\times_{\text{Spec}(R)}\text{Spec}(B)$$ is the fiber product of sets $$\text{Spec}(A)$$ and $$\text{Spec}(B)$$ in relation with the ring homomorphisms $$\alpha_A:R\to A$$ and $$\alpha_B:R\to B$$ that defines the $$R$$-algebras and $$\tilde{\alpha}_A:\text{Spec}(A)\to \text{Spec}(R)$$ and $$\tilde{\alpha}_B:\text{Spec}(B)\to \text{Spec}(R)$$ are the induced Spectra maps of $$\tilde{\alpha}_A$$ and $$\tilde{\alpha}_B$$, i.e.

$$\text{Spec}(A)\times_{\text{Spec}(R)}\text{Spec}(B)=\{(P,Q)\in \text{Spec}(A)\times \text{Spec}(B)\mid \tilde{\alpha}_A(P)=\tilde{\alpha}_B(Q)\}$$

I've found the function $$\text{Spec}(A\otimes_R B)\to \text{Spec}(A)\times_{\text{Spec}(R)}\text{Spec}(B)$$ defined by $$P\mapsto (\tilde{f}_A(P),\tilde{f}_B(P))$$ where $$f_A:A\to A\otimes_R B$$ is defined by $$a\mapsto a\otimes 1$$ and $$f_B:B\to A\otimes_R B$$ is defined by $$b\mapsto 1\otimes b$$ and $$\tilde{f}_A$$, $$\tilde{f}_B$$ are the respective Spectra maps.

Well, for the inverse i'm looking for a way to construct a prime ideal of $$A\otimes_R B$$ given $$P$$, $$Q$$ prime ideals of $$A$$ and $$B$$ such that $$\tilde{\alpha}_A(P)=\tilde{\alpha}_B(Q)$$.

I know that is a more deeper fact about fiber product of schemes but i need to use elementary facts about tensor product of algebras and ideals.

Any ideas?

Thanks.

• The standard proof is that as functors between $Sch$ and $CRing^{op}$, $\operatorname{Spec}$ is right adjoint to the global sections functor. Right adjoints preserve limits, so preserve fiber products. The fiber product in $CRing^{op}$ is the fiber coproduct in $CRing$, which is the tensor product. – David Lui Jun 1 at 2:45
• If you are asking about the fiber product of underlying sets, this is wrong: take $A=B=\Bbb C$ and $R=\Bbb R$ where the maps are the obvious ones. Then the fiber product of underlying sets is a point while the spectrum of the tensor product has two points. Or have I misinterpreted your question? – KReiser Jun 1 at 2:45
• If $X$ is a scheme, let $sp(X)$ denote the underlying topological space of $X.$ Then by the universal property of the fiber product, there is a canoncial continuous map of topological spaces $f : sp(X\times_S Y)\to sp(X)\times_{sp(S)} sp(Y),$ and this map is always surjective (exercise). However, it is generally not an isomorphism, as KReiser indicated. – Stahl Jun 1 at 23:22
• @KReiser it's true, i verified this. Thanks. – Elvis Torres Pérez Jun 2 at 2:12