# If a morphism between affine schemes is dominant, is the corresponding ring morphism injective?

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?

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

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?
• @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 8:14
• 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 8:37