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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.

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  • $\begingroup$ 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. $\endgroup$ – David Lui Jun 1 at 2:45
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    $\begingroup$ 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? $\endgroup$ – KReiser Jun 1 at 2:45
  • $\begingroup$ 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. $\endgroup$ – Stahl Jun 1 at 23:22
  • $\begingroup$ @KReiser it's true, i verified this. Thanks. $\endgroup$ – Elvis Torres Pérez Jun 2 at 2:12

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