Ext functor and Projective Module

Question

Let $R$ be a ring and $P$ is an $R$-module. The statement are equivalent:

1-$P$ is projective.

2-For every $R$-module $N$ and for $i\geq 1$ , $Ext^i_R(P,N)=0$

3-For every $R$-module $N$ , $Ext^1_R(P,N)=0$

4-For every finitely presented $R$-module $N$, $Ext^1_R(P,N)=0$

5-For every finitely generated ideal $I$ of $R$, $Ext^1_R(P,R/I)=0$

I can prove 1 $\Rightarrow$ 2 $\Rightarrow$ 3 $\Rightarrow$ 4 $\Rightarrow$ 5 but I dont know how to prove 5 $\Rightarrow$ 1. can anyone help? Thank you.

Can you reverse any of the implications? $3 \Rightarrow 1$ is not so bad. — Qiaochu Yuan, Sep 26 '20 at 07:10
Where did you find this statement? I don’t think it’s true that 4 or 5 imply 1, 2 and 3. — Jeremy Rickard, Sep 26 '20 at 09:10

score 6 · Answer 1 · answered Sep 28 '20 at 08:13

It's not true in general that the fourth or fifth conditions imply the first three.

For example, if $R=\mathbb{Z}_p$, the $p$-adic integers, then since $R$ is a principal ideal domain, every finitely presented $R$-module is a finite direct sum of modules of the form $R/I$, for finitely generated ideals $I$. So the fourth and fifth conditions are clearly equivalent for this $R$.

But $$\operatorname{Ext}^1_{\mathbb{Z}_p}(\mathbb{Q}_p,\mathbb{Z}_p) =\operatorname{Ext}^1_{\mathbb{Z}_p}(\mathbb{Q}_p,\mathbb{Z}_p/p^r\mathbb{Z}_p)=0$$ for all $r$. So $\mathbb{Q}_p$ satisfies the fourth and fifth conditions, but is not projective.

Even if $R=\mathbb{Z}$ it is consistent with $\mathsf{ZFC}$ (the usual axioms of set theory) that there is an abelian group $A$ such that $\operatorname{Ext}^1_{\mathbb{Z}}(A,\mathbb{Z})=0$ but $A$ is not free abelian, and so not projective. (The question of whether such a group exists is the Whitehead Problem.) If $A$ is such a group, then also $\operatorname{Ext}^1_{\mathbb{Z}}(A,\mathbb{Z}/m\mathbb{Z})=0$ for all $m$, so $A$ satisfies the fourth and fifth conditions but not the first three.

Ext functor and Projective Module

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