I know that if $A$ is contained in $B$ and $B$ is an integral extension of $A$, then the induced map on spectrum of the rings is surjective (and closed). Is it true if $B$ is not assumed to be integral over $A$?
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No. Take any domain $R$ that is not a field, and consider the inclusion $R \hookrightarrow Q$, where $Q$ is the field of fractions of $R$. Then $|\text{Spec}(R)| > 1$, but $|\text{Spec}(Q)| = 1$, so $\text{Spec}(Q) \to \text{Spec}(R)$ cannot be surjective (and as Georges Elencwajg points out, it is not closed either).
zcn
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....nor closed. – Georges Elencwajg Jan 28 '14 at 22:09
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@GeorgesElencwajg: Thank you for pointing that out - it is just as clearly not closed either! – zcn Jan 29 '14 at 06:37