Let $A \to B \to C$ be morphisms of rings (commutative with 1). Suppose $A \to C$ is flat.
Is $B \to C$ flat?
My attempt: let $1 \to M \to N$ be an exact sequence of $B$-modules. Tensor this by $\otimes_B C$: we get an exact sequence of $C$-modules $$1 \to K \to M \otimes_B C \to N \otimes_B C (*),$$ where $K$ is the obvious kernel. My aim is to show that $K = 1$. But since we have a morphism $\phi: A \to B$, $1 \to M \to N$ is also an exact sequence of $A$-modules in a natural way (the action $A \times M$ is given by $(a, m) \mapsto \phi(a) \cdot m$). So let's tensor $1 \to M \to N$ with $\otimes_A C$; by flatness of $A \to C$, we have the exact sequence $$1 \to M \otimes_A C \to N \otimes_A C (**).$$ I want to claim that the sequences $(*)$ and $(**)$ are the same, thus implying that $K = 1$, but I'm not sure how to do this.
Hint: You'll be more successful if instead you try to show that $B\to C$ doesn't have to be flat. To find a counterexample, you might think about cases where $A=C$ and $A\to C$ is the identity map.