Question

Is it true that if $\mbox{Ext}^{1}_{A}(P,A/I)=0 \ \forall I$ then P is projective? Similar statements are true for flat and injective modules, but I'm beginning to suspect that projective modules cannot be characterized soley by ideals.

Answer 1

Answer is yes if A is Noetherian and P is finitely generated. Indeed, your condition implies that Ext^1(P, N)=0 for any finitely generated A-module N, which implies that P is projective.

Answer 2

When $A=\mathbb Z$ the condition is equivalent to $\mathrm{Ext}^1_{\mathbb Z}(A,\mathbb Z)=0$ and the problem as to whether this implies that $A$ is free is the Whitehead problem and was shown by Shelah to be undecidable in ZFC (standard set theory). Hence there is at least one ring for which the problem is difficult.

Answer 3

It is well known that if $mbox{Ext}^1_{A}(P,A/I)=0$ for all $I,$ then $mbox{Ext}^i_{A}(P,A/I)=0$ for all $i$ and for all $I $and $P$ is projective. We can also characterize a projective module $P$ by his trace ideal denoted $t(P),$ we can then for all projective module $P$ the following relations: i) $Pt(P)=P.$ ii) $t(P)^2=t(P).$

Answer 4

I think the answer is no. I found the following counterexample.

Let $K$ be the field of complex Hahn series with real exponents, i.e. \[K = \left\{ f = \sum_{r \in \mathbb{R}} a_r X^r ; \, \operatorname{supp}(f) \textrm{ is a well-ordered subset of } (\mathbb{R}, \ge) \right\},\] where $\operatorname{supp}( \sum_{r \in \mathbb{R}} a_r X^r ) := \{r \in \mathbb{R} ; \, a_r \neq 0\}$.

Let $R$ be the subring of $K$, defined as $R = \{f \in K ; \, \operatorname{supp}(f) \subseteq \mathbb{R}_{\ge 0}\}$. This is a local ring, with maximal ideal $\mathfrak{m} = \{f \in K ; \operatorname{supp}{f} \subseteq \mathbb{R}_{>0}\}$. The residue field $k= R/m \cong \mathbb{C}$ is an $R$-module.

I claim that $M = R/\mathfrak{m}$ is a counterexample, i.e. $\operatorname{Ext}_R^1(M,R/I) = 0$ for all ideals $I$ of $R$, yet $M$ is not a projective $R$-module.

Proof:

$M$ is not projective: If $M$ were projective, then $0 \to \mathfrak{m} \to R \to M \to 0$ would split, hence $\mathfrak{m}$ would be a quotient of $R$, so in particular it could be generated by one element. However $\mathfrak{m}$ is not a finitely generated ideal.

$\operatorname{Ext}_R^1(M,R/I) = 0$: The short exact sequence $0 \to \mathfrak{m} \to R \to M \to 0$ gives the long exact sequence \[ \begin{split} 0 &\to \operatorname{Hom}(M, R/I) \to \operatorname{Hom}(R, R/I) \to \operatorname{Hom}(\mathfrak{m}, R/I) \to \\ &\to \operatorname{Ext}^1(M, R/I) \to \operatorname{Ext}^1(R,R/I) \to \dotsm \end{split} \]

Here $\operatorname{Ext}^1(R,R/I) = 0$ since $R$ is projective. So it is enough to prove that $R/I = \operatorname{Hom}(R, R/I) \to \operatorname{Hom}(\mathfrak{m}, R/I)$ is surjective.

The ideals of $R$ are easy to describe: If $c \in \mathbb{R}_{\ge 0}$, then let $I_{\ge c} = (X^c)$ and $I_{>c} = (X^r ; \, r > c)$. Then every nonzero ideal is of the form $I_{\ge c}$ or $I_{>c}$. In particular $\mathfrak{m} = I_{>0}$.

If $I=0$: we need that $R = \operatorname{Hom}(R,R) \to \operatorname{Hom}(\mathfrak{m}, R)$ is surjective (in fact it is bijective). This is not hard to check (for $\varphi \in \operatorname{Hom}(\mathfrak{m}, R)$, show that $\varphi(X^r) \in I_{\ge r}$, and $X^{-r} \varphi(X^r) \in R$ is independent of $r>0$).

If $I=I_{\ge c}$: Let $\varphi \colon \mathfrak{m} \to R/I_{\ge c}$ be a homomorphism. We need to show that there is an $h \in R$ such that $\varphi(f) = f h + I_{\ge c}$. To prove this, look at $\varphi(X^r)$ and take $r \to 0$. If $r<c$, then $\varphi(X^r)$ will determine $h$ modulo $I_{\ge c-r}$. To finish the proof, we need that if $E \subseteq [0, c)$ such that $E \cap [0,d)$ is well-ordered for all $d<c$, then $E$ is also well-ordered.

For $I=I_{> c}$, the proof is almost the same.