If $M$ is a flat $A$-module, then is $M$ a flat $A/I$ module?

Question

Assume that $A$ is a commutative ring with unity and let $M$ be a unital $A$-module. Let $I$ be an ideal of $A$ such that $IM = 0$. Then we can give a natural $A/I$-module structure on $M$ satisfying $\overline{a}.m =a \cdot m$ for any $a \in A$. Just for clarity I will denote $M$ with $A/I$-module structure as $M'$. If $M$ is a flat as an $A$-module then is $M'$ a flat $A/I$-module (described above). I think its true and this is my attempt:
We use extension of scalars via $\pi:A \to A/I$ where $\pi$ is the canonical homomorphism. ie. $A/I \otimes_A M$ is given $A/I$-module structure. Then it is a well known result that $A/I \otimes_A M$ is flat.
If we show $A/I \otimes_A M \cong M'$ (as $A/I$-modules), then we are done. Define a map from $A/I \times M$ to $M'$ given by $(\overline{a},m) \mapsto \overline{a}.m$, so by Universal property of Tensor products, there is a $A/I$-module homomorphism $\phi:A/I \otimes_A M \to M'$ satisfying $\phi(\overline{a} \otimes m) =\overline{a}.m$. Define $\psi:M' \to A/I \otimes_A M$ given by $m \mapsto \overline{1} \otimes m$. Clearly $\psi$ is a right inverse of $\phi$. Now take any elementary tensor $\overline{a} \otimes m$. $$\psi \phi (\overline{a} \otimes m)=\psi(\overline{a}.m)=\overline{1} \otimes \overline{a}.m=\overline{1} \otimes (a \cdot m)=a \cdot (\overline{1} \otimes m)=(\overline{a}\overline{1}) \otimes m=\overline{a} \otimes m$$ (I feel this step was sketchy)
So, $\psi$ is a left inverse of $\phi$. Thus $\phi$ is a bijection. $\; \therefore A/I \otimes_A M \cong M'$ and we are done.
I'm not confident about the step where we show $\psi$ is left inverse of $\phi$. Is my proof correct and how should I make this argument better?

Answer 1

We can do this directly using the universal property. By definition $A/I \otimes_A M$ satisfies

$$\text{Hom}_{A/I}(A/I \otimes_A M, N) \cong \text{Hom}_A(M, N)$$

where $N$ is an $A/I$-module. Now if $M$ is annihilated by $I$ so that it is also an $A/I$-module then we just need to observe that an $A$-module homomorphism $M \to N$ is the same thing as an $A/I$-module homomorphism $M' \to N$, meaning we also have

$$\text{Hom}_{A/I}(M', N) \cong \text{Hom}_A(M, N)$$

and we conclude by the Yoneda lemma.