Suppose $P$ is finitely generated projective $A$-module then is it true that there is a exact sequence of form $0 \to K \to A^m \to P \to 0$ where $K$ is finitely generated ?
I've shown there is a exact sequence of form $A^n \to A^n \to P$ but stuck now.
If $P$ is generated by $p_1,\ldots, p_m$, we obtain a homomorphism $f\colon A^m\to P$ by sending the $(a_1,\ldots, a_m)$ to $a_1p_1+\ldots + a_mp_m$. This homomorphism is onto, giving rise to a short exact sequence $$0\to K\to A^m\to P\to 0 $$ As $P$ is projective, we can pull the identity $P\to P$ to a homomorphism $g\colon P\to A^m$ such that $f\circ g=\operatorname{id}_P$. The sequence is said to split. A sequence split on the right also splits on the left (this holds in more general context): Given $x\in A^n$, we see that $f(x-g(f(x)))=0$, hence $x\mapsto x-g(f(x))$ is a map $A^m\to K$, and is in fact an epimorphism.