$\mathrm{Ext}^1_{A}(P,A/I)=0$ for all ideal $I$ of $A$ implies $\mathrm{Ext}^1_{A}(P,N)=0$ for all finitely generated $A$ module $N$.

Question

Let $A$ be commutative ring with $1$ and $P$ be a finitely presented $A$ module. If $\mathrm{Ext}^1_{A}(P,A/I)=0$ for all ideal $I$ of $A$ then how can I show that $\mathrm{Ext}^1_{A}(P,N)=0$ for all finitely generated $A$ module $N$.

Answer 1

If $N$ has $n$ generators $x_1,\ldots,x_n$ we can fit it into an exact sequence $$0\to N'\to N\to N''\to0$$ where $N'$ has one generator $x_1$, and $N''$ has $n-1$ generators ( the images of $x_2,\ldots,x_n$ in $N/N'$). By induction on $n$ we can assume that $\text{Ext}^1(P,N'')=0$; also $\text{Ext}^1(P,N')=0$ as $N\cong A/I$ for some ideal. From the long exact sequence $$\cdots\to \text{Ext}^1(P,N')\to \text{Ext}^1(P,N)\to \text{Ext}^1(P,N'')\to\cdots$$ we get $\text{Ext}^1(P,N)=0$.

This argument works for any half-exact functor.