Is $Y$ the scheme theoretic image of $f$?

Question

Let $f:X\to Y$ be an open immersion of schemes, and $X$ is Noetherian, is $Y$ the scheme theoretic image of $f$?

Answer 1

Just to get this out of the unanswered queue. The answer is no, even for pretty well-behaved schemes. Namely, even if $Y$ is connected then $X\hookrightarrow Y$ needn't have scheme-theoretic image all of $Y$. For example, consider $Y=V(xy)\subseteq \mathbb{A}^2_k$. Consider then $X:=D(x)=\mathrm{Spec}(k[x,x^{-1}])$. The map $X\to Y$ corresponds to the ring map $k[x,y]/(xy)\to k[x,x^{-1}]$ given by $x\mapsto x$ and $y\mapsto 0$. Clearly the kernel of this is $(y)$ so that the scheme-theoretic image is $\mathrm{Spec}(k[x,y]/(xy,y))=V(y)\cong\mathbb{A}^1_k$. Of course, $D(x)\ne V(y)$ since they're not even isomorphic.

If you assume that $Y$ is integral then the answer is yes. This is merely because $X$, being an open in $Y$, is dense in $Y$. Since the scheme theoretic image of $X$ has underlying topological space $\overline{X}$ (since our map is q.c.) we deduce that the underlying topological space is $Y$. But, since $Y$ is reduced this implies that the scheme-theoretic image is $Y$.