Let $f : A \rightarrow B$ and $g : A \rightarrow C$ be homomorphisms of Noetherian rings.Suppose
1) $B \otimes_A C$ is Noetherian,
2) $f$ is flat and
3) $g$ is non-degenerate.
Then $1_B \otimes g : B \rightarrow B \otimes C$ is also non-degenerate.
This is a proposition in Mat80, but is left as an exercise. I'm trying to use Theorem 12 in the same book, which states that $\text{Ass}_B(B) = \displaystyle{\bigcup_{\mathfrak{q} \in \text{Ass}(A)}}\text{Ass}_{B}(B/\mathfrak{q}B)$ since $f$ is flat. Using this result, we just need to show that if $\hat{\mathfrak{p}}$ is a contraction of some prime ideal $\mathfrak{p}$ of $B \otimes_A C$, then it is contained in some $\text{Ass}_B(B/\mathfrak{q}B)$. With the non-degeneracy of $g$ and the fact that the flatness of $f$ is preserved by the base change, I can find a prime ideal $\mathfrak{q}$ in $A$ which is the contraction of $\mathfrak{p}$ and one can see that $\mathfrak{q}B \subseteq \mathfrak{p}$. I believe that if I'm able to show that $\text{Ann}_B(B/\mathfrak{q}B) \subseteq \hat{\mathfrak{p}}$, then the proposition is proved. But proving this last arguement seems hard and it seems that the non-degeneracy of $g$ has not been properly used in my argument, so can someone give me a hint about this problem?