Does dominant morphism of integral schemes is injective on sheaves?

Question

Let $f:X \to Y$ be a dominant morphism of integral schemes. Is it true that it is equivalent to the fact that $\mathcal O_Y \to f_* \mathcal O_X$ is injective? Or does one imply another? It's quite easy that they are equivalent in the case of affine morphism, but I can't understand are they equivalent in the general case?

Answer 1

$\newcommand{\Ohol}{{\mathcal O}}$ $\newcommand{\sheaf}[1]{{\mathcal #1}}$ $\newcommand{\ideal}[1]{{\mathfrak #1}}$ $\newcommand{\spec}[1]{{{\mathrm{spec}}(#1)}}$ $\newcommand{\PSP}{{\mathbb P}}$

When $f:X \to Y$ is in addition quasicompact and separated then we have an exact sequence of quasi-coherent sheafs $$0 \to \sheaf{I} \to \Ohol_Y \to f_*\Ohol_X$$ and the scheme theoretic image of $f$, call it $Z$, is $V(\sheaf{I})$. It is $Z= \overline{f(X)}$. So if $\Ohol_Y \to f_* \Ohol_X$ is injective then $\sheaf{I} = 0$ and $f$ is dominant. Otherway, if $V(\sheaf{I}) = Y$ then $\sheaf{I}$ must be zero as $Y$ is integral.

In fact for $X,Y$ integral schemes one does not need assumptions on $f$. For any scheme $X$ and $s \in \Ohol_X(X)$ call $X_s = \{x \in X \mid s_x \notin \ideal{m}_x\}$ where $\ideal{m}_x$ is the maximal ideal of $\Ohol_{X,x}$ and $s_x$ is the image of $s$ under the canonical map $\Ohol_X(X) \to \Ohol_{X,x}$.

Now assume first that $Y=\spec{A}$ affine. Then we have $f^{-1}(Y_a) = X_{\phi(a)}$ where $a \in A$ and $$\phi=f^\sharp:\Ohol_Y(Y) \to (f_*\Ohol_X)(Y) = \Ohol_X(X).$$

Now we have the equivalences $f:X \to Y$ dominant. $\Leftrightarrow$ $f^{-1}(D(a)) \neq \emptyset$ for all $a \in A$ with $a \neq 0$ $\Leftrightarrow$ $X_{\phi(a)} \neq \emptyset$ $\Leftrightarrow$ $\phi(a) \neq 0$ in $\Ohol_X(X)$.

The last equivalence can easily be seen by considering affine open subsets of $X$. Let $X_s$ with $s \in \Ohol_X(X)$ be given, then for all $U=\spec{B} \subseteq X$ we have $X_s \cap U = U_t$ where $t$ is the image of $s$ in $B$. Now if $X_s = \emptyset$ we have $U_t = \emptyset$ and as $B$ is integral $t$ is zero, so considering all $U$ also $s$ is zero. On the other hand if $s$ is zero then obviously $X_s = \emptyset$.

Now for the general case where $Y$ needs not be affine we have the implications:

$f:X \to Y$ dominant $\Leftrightarrow$ $f:f^{-1}(V) \to V$ dominant for all $V \subseteq Y$ affine. $\Leftrightarrow$ $\Ohol_Y(V) \to \Ohol_X(f^{-1}(V))$ injective for all $V \subseteq Y$ affine. $\Leftrightarrow$ $\Ohol_Y \to f_*\Ohol_X$ injective.

It may be noted that the whole proof above for the integral case follows quite closely the lines of the proof for $X,Y$ affine with $X_s$ taking the place of the $D(s)$ in the affine case.