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Suppose we have $\phi$ a ring morphism from $A$ to $B$, let $X=\operatorname{Spec}A$ , $Y=\operatorname{Spec}B$ and $\psi$ is the induced morphism of affine schemes. Is it true that if $\psi$ dominant, than $\phi$ is injective?

Zev Chonoles
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user52342
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  • What about something like $k[x]/(x^2) \to k$? I think the problem has to come from nilpotents. – TTS Jun 22 '13 at 07:46
  • This result is true if $X$ and $Y$ are quasi-projective, so prove it for that case and try to see where it could go wrong when you try to generalize. – Ragib Zaman Jun 22 '13 at 07:59
  • See the question http://math.stackexchange.com/questions/389036/dominant-morphism-between-affine-varieties-induces-injection-on-coordinate-rings/389060#389060 –  Jun 22 '13 at 08:49
  • Adding to TTS's comment : conversely, in the case where $A$ is reduced, dominance of $\psi$ implies injectivity of $\phi$. The idea is that in this case $\psi^{-1}D(s)=D(\phi s)$ is non-empty for each non-nilpotent hence for each non-zero $s\in A$, making each $\phi s$ non-nilpotent hence non-zero; it is quite often to see this being implicitly used in some arguments. – Wang Samuel Sep 10 '21 at 03:11

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Hint: If $A$ and $B$ each only have one prime ideal, then $X$ and $Y$ are singletons, so we would have to have $\psi(Y)=X$, hence $\psi(Y)$ is dense in $X$, hence $\psi$ is dominant. Can you think of a ring morphism between two such rings $A$ and $B$ that is not injective?

Zev Chonoles
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  • I'm think about a morphism that goes from all the element of one rings, with only a prime ideal, to the zero ,the only prime ideal of the other. Is this right? – user52342 Jun 22 '13 at 08:11
  • @user: No, I'm afraid not; remember, ring morphisms have to send $1$ to $1$, so the only way a ring morphism could send "everything to zero" is if $0=1$ in $B$, i.e. if the ring $B$ was the zero ring, but the zero ring has no prime ideals. The user TTS gave a good example above though, try that. – Zev Chonoles Jun 22 '13 at 08:14
  • k, thank you ... i'll try – user52342 Jun 22 '13 at 08:23
  • ok the example that TTS make me, let $\psi$ to be dominant, because the kernel of $\phi$ that goes from $K[X]/x^2$ to $K$ is contained in the nihilradical of $K[X]/x^2$, but it's not iniective as we can see. Is this right now? – user52342 Jun 22 '13 at 08:37
  • Yup, that's right! – Zev Chonoles Jun 22 '13 at 08:38
  • ok i found this characterizaton for morphism of affine schemes to be to be dominant in your forum, bur can you give some good reference book to look it at ? – user52342 Jun 22 '13 at 08:43
  • I'm not sure what you're after; this fact occurs as exercise 21, part v in Chapter 1 of Atiyah & Macdonald, if that helps. Any introductory book on algebraic geometry that talks about schemes would probably mention this fact or have it as an exercise somewhere, I imagine. – Zev Chonoles Jun 22 '13 at 08:46