I have a question concerning the issue of whether we can reduce $\operatorname{Tor}_1(M,N)=0$ to $\operatorname{Tor}_1(M,R/I)=0$ and something like that.
Clearly, the case of Tor was instructed in Aytiah, Macdonald's Commutative Algebra.
And the equivalence $N$ is injective $\Leftrightarrow \operatorname{Ext}^1_R(R/I,N)=0$ can be done thanks to Baer's criterion.
My question is, can we do it with projective module, which is $P$ is projective $\Leftrightarrow \operatorname{Ext}^1_R(P,R/I)=0$? ($R$ is commutative here.)
The two resolved cases utilized different methods, or at least I think so. That's the reason I can't solve this, can anyone help? Thank you.
Suppose $N$ is an $R$-module. Then you claim that it is injective iff $\text{Ext}^1(?,N)$ vanishes on every cyclic module $R/I$ with $I$ an ideal of $R$. One direction is immediate, and for the converse you can consider the LES of the SEC $I\to R\to R/I$ to obtain an exact sequence $$\text{Hom}_R(R,N) \to \text{Hom}_R(I,N) \to \text{Ext}^1_R(R/I,N)\to 0$$ Thn Baer's criterion says that if that Ext group vanishes for every ideal then $N$ is injective. If you try to do this to check for projectivity, by using the very SEC and the LES for the other variance, you get an exact sequence
$$\text{Hom}_R(M,R) \to \text{Hom}_R(M,R/I) \to \text{Ext}^1_R(M,I)\to \text{Ext}^1_R(M,R) \to \cdots$$
If it were the case that $R$ is injective we have $\text{Ext}^1_R(M,R)=0$, so you get that every map $M\to R/I$ can be lifted through the surjection $R\to R/I$ for every ideal $I$ if and only if $\text{Ext}^1_R(M,I)=0$ for every ideal $I$.