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Let $A,B$ be two $R-$algebra. (i.e. we have two homomorphisms: $f:R\rightarrow A$ and $g:R\rightarrow B$) We denote two canonical homomorphism $A\rightarrow A\otimes_R B$ and $B\rightarrow A\otimes_R B$ by $u,v$ respectively.

Suppose $p\in \operatorname{Spec}A$, $q\in \operatorname{Spec}B$ and $f^{-1}(p)=g^{-1}(q)$. How to find a $t\in \operatorname{Spec}(A\otimes_R B) $ such that $u^{-1}(t)=p$ and $v^{-1}(t)=q$.

Richard
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    duplicate of https://math.stackexchange.com/questions/1629695/constructing-prime-ideal-of-tensor-product-from-two-prime-ideals – Pig Jun 08 '21 at 07:30
  • Effectively, the point is you can reduce $A, B, R$ to a field by 1. localizing at $p$ for $A$, $q$ for $B$, and their common preimage for $R$, 2. mod out the maximal ideal.

    For the field case (where $A,B,R$ are fields, and $p,q = 0$.), you can then just pick any maximal ideal of the tensor product - they must pullback to a prime ideal of $A$ and $B$, which then must be the zero ideal.

     – Pig Jun 08 '21 at 07:38

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By the universal property of the tensor product, from the maps $R\to A \to Frac(A/p)$ and $R\to B\to Frac(B/q)$ we get a map $$A\otimes_R B\to Frac(A/p)\otimes_R Frac(B/q)\cong Frac(A/p)\otimes_{Frac(R/f^{-1}(p))} Frac(B/q)$$ and the pullback of any prime ideal in this ring suffices. To show this ring has prime ideals, it suffices to show that it's nonzero, but this is clear because it's a tensor product of nonzero vector spaces over a field.

Hank Scorpio
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