Composition of flat morphisms

Suppose I have nice enough schemes $X,Y,Z$ over a field $k$ (I can take $X,Y,Z$ to be smooth projective connected varieties and $k$ to be algebraically closed of characteristic 0 if needed) and morphisms of $k$-schemes $$X \xrightarrow{a} Y \xrightarrow{b} Z$$ such that their composition $b \circ a$ is flat and $a$ is flat.

Can I conclude that $b$ is also flat?

• Suppose $X$ is empty... – Zhen Lin Nov 11 '15 at 9:09

Your hypothesis that $Y$ is connected and $X$ is projective will make $a$ to be faithfully flat if it is flat and then your assumptions will imply that $b$ is flat.
To answer your comment, let me do this in the ring level (after all flatness is a ring concept). So, let $A\to B\to C$ be rings. Your hypothesis is $A\to C$ is flat and $B\to C$ is faithfully flat. Then you wish to show that $A\to B$ is flat. Let $0\to N\to M$ be an exact sequence of $A$-modules. Tensoring, you get $N\otimes B\to M\otimes B$ which you want to show is injective, being the definition of flatness. Let $K$ be its kernel and thus we have $0\to K\to N\otimes B\to M\otimes B$ exact and tensor with $C$ over $B$ to get, $0\to K\otimes C\to N\otimes C\to M\otimes C$ exact. But, flatness of $A\to C$ implies $K\otimes C=0$, which by faithful flatness of $B\to C$ implies $K=0$.
• Thanks. I still don't understand why my hypotheses impy that $B \to C$ is faithfully flat though. If you can expand on this (maybe providing references) I will accept your answer! – lostmathematician Nov 11 '15 at 17:00
• A flat map is faithful if it is surjective. Flat maps are also open. So, your $a$ is open and also closed since $X$ is projective. Since $Y$ is connected, this implies that $a$ is surjective. – Mohan Nov 11 '15 at 17:11