Why does a flat finite type morphism of irreducible noetherian schemes map generic pt to generic pt Basically the title, I came across this statement reading some notes on dimension of fibers of flat maps. (I don't know if Noetherian actually matters for this question)
I think that for maps of irreducible schemes, sending generic point to generic point is the same as $f: X \to Y$ being dominant $\leftrightarrow$ the associated ring map (reducing to the case where everything is affine) is injective, so why should the finitely generated $A$-alg $B$, $f^{\sharp}: A \to B$ flat, imply $f$ is injective?
 A: Any flat morphism $f:X\to Y$ with $X$ and $Y$ irreducible sends the generic point of $X$ to the generic point of $Y$. This is a consequence of the fact that generalizations lift along flat morphisms (of arbitrary schemes), which is itself a global version of the fact that a flat ring map $A\to B$ satisfies going down. 
Here are the relevant Stacks Project references:
https://stacks.math.columbia.edu/tag/03HV (generalizations lift along flat morphisms)
https://stacks.math.columbia.edu/tag/00HS (flat ring maps satisfy going down)
https://stacks.math.columbia.edu/tag/00HW ($A\to B$ satisfies going down if and only if generalizations lift along $\mathrm{Spec}(B)\to\mathrm{Spec}(A)$)
https://stacks.math.columbia.edu/tag/0063 (definition of generalizations lifting along a map)
https://stacks.math.columbia.edu/tag/0061 (definition of generalizations in a topological space)
Now assume $X$ and $Y$ are irreducible with generic points $\eta_X$ and $\eta_Y$, respectively, and let $f:X\to Y$ be flat. Consider the generalization $\eta_Y\rightsquigarrow f(\eta_X)$ (the generic point of an irreducible scheme generalizes every point in the scheme). Since generalizations lift along $f$ (due to $f$ being flat), there is a point $x\in X$ with $f(x)=\eta_Y$ and $x\rightsquigarrow \eta_X$, i.e., $x$ is a generalization of $\eta_X$. Because $\eta_X$ is the generic point of $X$ and the underlying topological space of a scheme is sober (I guess just Kolmogorov is adequate–all that we need is that distinct points have distinct closures), we must have $x=\eta_X$. Thus $f(\eta_X)=f(x)=\eta_Y$.
