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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?

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  • $\begingroup$ What about something like $k[x]/(x^2) \to k$? I think the problem has to come from nilpotents. $\endgroup$ – TTS Jun 22 '13 at 7:46
  • $\begingroup$ 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. $\endgroup$ – Ragib Zaman Jun 22 '13 at 7:59
  • $\begingroup$ See the question math.stackexchange.com/questions/389036/… $\endgroup$ – user314 Jun 22 '13 at 8:49
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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?

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  • $\begingroup$ 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? $\endgroup$ – user52342 Jun 22 '13 at 8:11
  • $\begingroup$ @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. $\endgroup$ – Zev Chonoles Jun 22 '13 at 8:14
  • $\begingroup$ k, thank you ... i'll try $\endgroup$ – user52342 Jun 22 '13 at 8:23
  • $\begingroup$ 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? $\endgroup$ – user52342 Jun 22 '13 at 8:37
  • $\begingroup$ Yup, that's right! $\endgroup$ – Zev Chonoles Jun 22 '13 at 8:38

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