Let $X$ and $Y$ be scheme. Let $f$ and $g$ be morphism of scheme from $X$ to $Y$. If $f,g$ agree on generic point of $X$, then $f$ and $g$ is the same map on $X$.
Does this hold in general ?
By famous statement of general topology, if $Y$ is Hausdorff, the statement holds.But in Zariski topology $Y$ is often note Hausdorff, so I'm confused.
Thank you in advance.
There are, broadly speaking, two ways for the statement to be wrong. The first one, as in the topological case, is the possibility that $Y$ isn’t separated. In this case, you can consider the two natural injections from $\mathbb{A}^1$ into the “line with two origins”.
The second one, perhaps more subtle, is the possibility that the generic point doesn’t contain all the information about the functions on $X$, and it happens if $X$ is not reduced.
For instance, let $A=k[[t]][u]$ ($k$ field) with $tu=u^2=0$. Consider the two distinct ring maps $A \rightarrow A$ given by the identity and $t \longmapsto t,\,u \longmapsto 0$. Their scheme-theoretic counterparts are the same on the generic point but not equal.