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If $f:X\rightarrow Y$ is a surjective morphism of schemes and $g:X'\rightarrow Y$ is another morphism of schemes, one can show that $p_{2}:X\times_{Y}X'\rightarrow X'$ is also surjective. The proof that I have seen (in Mumford's Red Book, for instance) goes as follows: Given $P\in X'$, we can find points $R\in Y$ and $P\in X$ and inclusions $k\left(R\right)\hookrightarrow k\left(P\right)$ and $k\left(R\right)\hookrightarrow k\left(Q\right)$. The desired result can then be established if we can say that there exists a field $F$ containing each of $k\left(P\right)$ and $k\left(Q\right)$. However, I don't see how we can conclude that such a field exists.

It is easy to just give a different proof without looking at a field extensions of $k\left(P\right)$ and $k\left(Q\right)$, but I wanted to know how the sketch of a proof above can be used. Thanks!

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Form the tensor product $k(P)\otimes_{k(R)} k(Q)$. Since $k(R)$ is a field, and each of $k(P)$ and $k(Q)$ is non-zero, this tensor product is non-zero. Thus it contains at least one maximal ideal, say $\mathfrak m$. Write $F :=(k(P)\otimes_{k(R)} k(Q))/\mathfrak m.$

By construction $F$ is a field extension of $k(R)$, and it admits $k(R)$-algebra morphisms $k(P)\to F$ and $k(Q) \to F$, which must be embeddings, since their domains are fields. Thus $F$ is the desired field.

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