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Let $A, B, C$ be rings and suppose I have $\phi: C \to A$ and $\psi: C \to B$ so that the tensor product $A \otimes_C B$ makes sense. I want to prove that given prime ideals $P$ of $A$ and $Q$ of $B$ such that $\phi^{-1} (P) = \psi^{-1} (Q)$, there exists a prime ideal $R$ of $A \otimes_C B$ such that $P = i^{-1}(R)$ and $Q = j^{-1} (R)$ where $i: A \to A \otimes_C B$ and $j: B \to A \otimes_C B$. How can I prove that such prime ideal exists? (I am asking this because I want to prove that if I take a fibre product of schemes then I can find a point $z \in X \times Y$ above $x \in X$ and $y \in Y$.)

Johnny T.
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    Points from the tensor product $Frac(A/P) \otimes Frac(B/Q)$ (above $Frac(C/\psi^{-1}(Q))$) will suit your needs. – Aphelli Jul 30 '20 at 14:10
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    Instead of trying to figure out what prime ideals of $A \otimes_C B$ look like in general, for your application it would suffice to find a field $K$ and homomorphisms $A \to K$ and $B \to K$ whose kernels are the prime ideals you are interested in and such that the composites $C \to A \to K$ and $C \to B \to K$ are equal. Then there will be a homomorphism $A \otimes_C B \to K$ whose kernel is prime and pulls back to the prime ideals you started with. – Zhen Lin Jul 30 '20 at 14:11
  • @Mindlack What is the tensor product over? and how do I know this tensor product is non-zero? – Johnny T. Jul 30 '20 at 15:24
  • If $C=\mathbb{Z}, A=\mathbb{Q}$ and $B=\mathbb{F}_p$, then your tensor product will be the zero ring and have no prime ideals. – Pol van Hoften Jul 30 '20 at 18:41
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    @Johnny T. : I already wrote in my comment what the tensor product was above. It’s a tensor product over a field of field extensions of said field (so nonzero vector spaces), so (using eg bases) is nonzero. – Aphelli Jul 31 '20 at 11:26
  • Lemma 26.17.5 in the stacks project gives a relatively explicit description: the prime ideals of $A\otimes_C B$ are in bijection with sets of tuples $(\mathfrak{p},\mathfrak{p}',\mathfrak{q}),$ where $\mathfrak{p}$ (respectively, $\mathfrak{p}'$) is a prime of $A$ (respectively $B$) such that $\phi^{-1}(\mathfrak{p}) = \mathfrak{P} = \psi^{-1}(\mathfrak{p}')$ and $\mathfrak{q}$ is a prime ideal of $\kappa(\mathfrak{p})\otimes_{\kappa(\mathfrak{P})}\kappa(\mathfrak{p}')$ (here, $\kappa(\mathfrak{p}) = A_\mathfrak{p}/\mathfrak{p}.$). – Stahl Aug 04 '20 at 22:01
  • [Continued] This might not be totally satisfying on its own, but it reduces you to thinking about the case when $A,B,$ and $C$ are all fields. This situation is not completely trivial, but has been discussed in several cases which might be helpful for you, for example here, here, ... – Stahl Aug 04 '20 at 22:02
  • [Continued] here, here, or here to start. The proof of the lemma in my first comment might also resolve your question on its own. – Stahl Aug 04 '20 at 22:02

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