Given a ring $R$, a prime ideal $\mathfrak{p}$ of $R$, and an extension ring $S$ (the algebra map $R\to S$ is injective), are there any nontrivial sufficient conditions for the induced map $Spec(S) \to Spec(R)$ to be injective as a map of topological spaces (completely disregarding any of our sheaf structure)? That is, every prime that lifts lifts to a unique prime? If this problem is intractable as-is, I can add more conditions, so please don't add answers unless they are actual answers.
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$\begingroup$ If R--->S is strict epimorhism which is surjective map.Then Spec(S)-->Spec(R) is a closed immersion(as closed subvariety) $\endgroup$– Shizhuo ZhangCommented Feb 5, 2010 at 2:04
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$\begingroup$ A strict epimorphism of rings is surjective, no? $\endgroup$– Harry GindiCommented Feb 5, 2010 at 2:18
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$\begingroup$ Is "strict epimorphism" synonymous with "surjection"? If not, could somebody define the former for me? $\endgroup$– Anton GeraschenkoCommented Feb 5, 2010 at 3:10
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$\begingroup$ Haha, I think it's another incarnation of radiciel but flipped around. $\endgroup$– Harry GindiCommented Feb 5, 2010 at 5:13
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$\begingroup$ It seems to me that there is a disconnect between the title and the body of the question. "When do primes lift uniquely?" seems to be asking "When is it the case that for every prime ideal p of R, there is exactly one prime ideal of S lying over p?" which is a wordier way of asking when the map on spectra is bijective. This is why some people are responsing with radiciel/purely inseparable extensions. On the other hand, in the body it seems to simply be asking for the map on spectra to be injective, which e.g. as VA points out occurs much more commonly... $\endgroup$– Pete L. ClarkCommented Feb 5, 2010 at 5:44
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2 Answers
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Rather than listing various non-trivial sufficient conditions, let me give you a reference. Search EGA for the word "radiciel" (or start reading at Definition 3.5.4 of EGA I). A morphism of schemes $X\to Y$ is said to be radiciel (or universally injective) if for every field $K$, the induced map on $K$-points $X(K)\to Y(K)$ is injective. This generalizes the notion of a purely inseparable field extension.
answered Feb 5, 2010 at 3:01
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$\begingroup$ Are these as functors of points? $\endgroup$ Commented Feb 5, 2010 at 3:03
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$\begingroup$ Yes. Sorry if that was unclear. $\endgroup$ Commented Feb 5, 2010 at 3:05
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$\begingroup$ Oh no, it was clear. Is there a nice translation of this into an algebraic condition? $\endgroup$ Commented Feb 5, 2010 at 3:06
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$\begingroup$ Yes. Look in EGA or in the stacks project. $\endgroup$– EmertonCommented Feb 5, 2010 at 3:13
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1$\begingroup$ I pretty much wasted 4-5 hours trying to avoid a direct check. I feel like a total idiot. $\endgroup$ Commented Feb 5, 2010 at 6:14
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If $S/R$ is purely inseparable then the primes will lift uniquely.
answered Feb 5, 2010 at 2:32
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1$\begingroup$ Could you give a reference? This looks like the condition I needed. Well, also, what do you mean by S/R is purely inseparable? Isn't that a condition on field extensions? $\endgroup$ Commented Feb 5, 2010 at 2:39