Serre's Trick for flatness of a morphism of schemes I'm reading some exercises on abelian varieties and I came across the following claim:


Claim (Serre's Trick): Let $X,Y,S$ be schemes and suppose that $X \times_S Y$ is flat over $S$. If $X(S) \neq \emptyset$ then $Y$ is flat over $S$.


Evidently to prove the claim we may reduce to the case where $X = \operatorname{Spec} C$, $S = \operatorname{Spec} A$ and $Y= \operatorname{Spec} B$ where $A,B,C$ are local rings and the maps $A \to B$, $A \to C$ are local homomorphisms.
Let $\mathfrak{m}_A, \mathfrak{m}_B$ and $\mathfrak{m}_C$ be the maximal ideals of $A,B,C$ respectively. 


(**) Suppose for the moment that the tensor product $B \otimes_A C$ contains a maximal ideal $\mathfrak{m}$ that contracts simultaneously to $\mathfrak{m}_A$ and $\mathfrak{m}_B$.


By localizing at $\mathfrak{m}$, we may assume that $B \otimes_A C$ is local, and the composition
$$A \to B \to B \otimes_A C$$
is faithfully flat (flat + local homomorphism implies faithfully flat).
With this I think we can prove that $B$ is flat over $A$ as follows. Let
$M \to N$ be an injection of $A$-modules. Let $K$ be the kernel of the map
$M \otimes_A B \to N \otimes_A B$. Tensoring with $C$, we get that 
$$K \otimes_A C  = 0$$
using flatness of $B \otimes_A C$ over $A$. Now tensor with $B$ to get
$$ K \otimes_A (C \otimes_A B) = 0$$
and conclude that $K= 0$ by faithful flatness.


My question is: A necessary condition for (**) to be true is that the product $B \otimes_A C$ is not zero. How can I get this just from the fact that there is a map $C \to A$? Indeed the example
    $$ \Bbb{Z}/2 \otimes_{\Bbb{Z}} \Bbb{Q} = 0$$
    shows that the condition on the existence of a section is really needed.\


Edit: I was tired from travelling and stupidly concluded that $-\otimes_A C$ was injective.

Let $f:C\rightarrow A$ be a  map (which is non zero, I assume you have units in the ring involved), you have a non zero $A$-bilinear map
$h:C\times B\rightarrow B$ defined by $h(c,b)=f(c)b$, by the universal property of the tensor product, $h$ factors by a map $\bar h:C\otimes_AB\rightarrow B$, henceforth, $C\otimes_AB$ is not trivial.
