Let A and B be commutative rings with unity. Given a surjective ring homomorphism $$\pi : A \rightarrow B$$ Is every projective B module extended from a projective A module? That is given a B-projective module P, does there exist a projective A-module Q such that $$ Q \otimes_A B = P$$
My attempt: Use the relation for an A-module M $$A/I \otimes_A M \simeq M/IM.$$ I can consider B isomorphic to A/I. Then any B-module P is annihilated by I. Does this mean P is isomorphic to Q/I for a A module Q?
Let $\mathbb{Z}$ be the ring of integers. Then the natural ring homomorphism $\pi:\mathbb{Z}\to\mathbb{Z}/\langle 5\rangle$ is a surjective ring homomorphism. Since $\mathbb{Z}/\langle 5\rangle$ is a field, every $\mathbb{Z}/\langle 5\rangle$-module is free and so is projective. Thus, $\prod_{I} \mathbb{Z}/\langle 5\rangle$ is a projective $\mathbb{Z}/\langle 5\rangle$-module, where $I$ is an infinite set, and since $\mathbb{Z}$ is a PID, every projective $\mathbb{Z}$-module is free, that is, every projective $\mathbb{Z}$-module is of the form $\bigoplus_{J}\mathbb{Z}$ for some set $J$. Also, $(\bigoplus_{J}\mathbb{Z})\bigotimes_\mathbb{Z}\mathbb{Z}/\langle 5\rangle\cong\bigoplus_{J}\mathbb{Z}/\langle 5\rangle$. But if $I$ is an infinite set we have $\prod_{I} \mathbb{Z}/\langle 5\rangle\not\cong\bigoplus_{J} \mathbb{Z}/\langle 5\rangle$ for each set $J$.