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Reference: David Mumford, The Red Book of Varieties and Schemes, Section II.4, Theorem 1.

Let $k_0$ be a field, $k$ its algebraic closure, $R$ a $k_0$-algebra. If it helps, assume $R$ is finitely generated.

Let $P$ be a prime ideal of $R \otimes_{k_0} k$, and let $P_0 = P \cap R$. Let $K = R_{P_0}/P_0R_{P_0}$ be the quotient field of $R/P_0$.

The homomorphism $R \rightarrow K$ induces a homomorphism $R \otimes_{k_0} k \rightarrow K \otimes_{k_0} k$. The book I'm reading claims that (i) the ideal generated by the image of $P$ in $K \otimes_{k_0} k$ is a prime ideal $\mathscr P$ of $K \otimes_{k_0} k$, and that (ii) the contraction of $\mathscr P$ is again $P$. I don't see why either of these things must be true.

I think this is because $K \otimes_{k_0} k$ is supposed to be a localization of $R \otimes_{k_0} k$ at some multiplicatively closed set inside $R \otimes_{k_0} k$, but I am having trouble seeing this.

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    Don't you want to say that $R$ is commutative? – 2017-02-04
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    It is implicit from the context of the question. Algebraic geometry is foremost concerned with commutative rings with identity. – 2017-02-04

0 Answers 0