projective module - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T18:52:45Z http://mathoverflow.net/feeds/question/25687 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/25687/projective-module projective module ashpool 2010-05-23T16:36:01Z 2012-11-07T16:40:39Z <p>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.</p> http://mathoverflow.net/questions/25687/projective-module/25698#25698 Answer by Torsten Ekedahl for projective module Torsten Ekedahl 2010-05-23T17:41:49Z 2010-05-23T17:41:49Z <p>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.</p> http://mathoverflow.net/questions/25687/projective-module/25733#25733 Answer by Bedini for projective module Bedini 2010-05-24T00:09:12Z 2010-05-24T00:09:12Z <p>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.</p> http://mathoverflow.net/questions/25687/projective-module/57271#57271 Answer by Hicham YAMOUL for projective module Hicham YAMOUL 2011-03-03T17:57:57Z 2011-03-03T17:57:57Z <p>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) $t(P)^2=t(P).$</p> http://mathoverflow.net/questions/25687/projective-module/111739#111739 Answer by eb80 for projective module eb80 2012-11-07T16:40:39Z 2012-11-07T16:40:39Z <p>I think the answer is no. I found the following counterexample.</p> <p>Let $K$ be the field of complex Hahn series with real exponents, i.e. <code>$K = \left\{ f = \sum_{r \in \mathbb{R}} a_r X^r ; \, \operatorname{supp}(f) \textrm{ is a well-ordered subset of } (\mathbb{R}, \ge) \right\},$</code> where <code>$\operatorname{supp}( \sum_{r \in \mathbb{R}} a_r X^r ) := \{r \in \mathbb{R} ; \, a_r \neq 0\}$</code>.</p> <p>Let $R$ be the subring of $K$, defined as <code>$R = \{f \in K ; \, \operatorname{supp}(f) \subseteq \mathbb{R}_{\ge 0}\}$</code>. This is a local ring, with maximal ideal <code>$\mathfrak{m} = \{f \in K ; \operatorname{supp}{f} \subseteq \mathbb{R}_{&gt;0}\}$</code>. The residue field <code>$k= R/m \cong \mathbb{C}$</code> is an $R$-module.</p> <p>I claim that <code>$M = R/\mathfrak{m}$</code> is a counterexample, i.e. <code>$\operatorname{Ext}_R^1(M,R/I) = 0$</code> for all ideals $I$ of $R$, yet $M$ is not a projective $R$-module.</p> <p>Proof:</p> <p>$M$ is not projective: If $M$ were projective, then <code>$0 \to \mathfrak{m} \to R \to M \to 0$</code> 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.</p> <p><code>$\operatorname{Ext}_R^1(M,R/I) = 0$</code>: The short exact sequence $0 \to \mathfrak{m} \to R \to M \to 0$ gives the long exact sequence <code>$\begin{split} 0 &amp;\to \operatorname{Hom}(M, R/I) \to \operatorname{Hom}(R, R/I) \to \operatorname{Hom}(\mathfrak{m}, R/I) \to \\ &amp;\to \operatorname{Ext}^1(M, R/I) \to \operatorname{Ext}^1(R,R/I) \to \dotsm \end{split}$</code></p> <p>Here <code>$\operatorname{Ext}^1(R,R/I) = 0$</code> since $R$ is projective. So it is enough to prove that <code>$R/I = \operatorname{Hom}(R, R/I) \to \operatorname{Hom}(\mathfrak{m}, R/I)$</code> is surjective.</p> <p>The ideals of $R$ are easy to describe: If <code>$c \in \mathbb{R}_{\ge 0}$</code>, then let <code>$I_{\ge c} = (X^c)$</code> and <code>$I_{&gt;c} = (X^r ; \, r &gt; c)$</code>. Then every nonzero ideal is of the form <code>$I_{\ge c}$</code> or <code>$I_{&gt;c}$</code>. In particular <code>$\mathfrak{m} = I_{&gt;0}$</code>.</p> <p>If $I=0$: we need that <code>$R = \operatorname{Hom}(R,R) \to \operatorname{Hom}(\mathfrak{m}, R)$</code> is surjective (in fact it is bijective). This is not hard to check (for <code>$\varphi \in \operatorname{Hom}(\mathfrak{m}, R)$</code>, show that <code>$\varphi(X^r) \in I_{\ge r}$</code>, and <code>$X^{-r} \varphi(X^r) \in R$</code> is independent of <code>$r&gt;0$</code>).</p> <p>If <code>$I=I_{\ge c}$</code>: Let <code>$\varphi \colon \mathfrak{m} \to R/I_{\ge c}$</code> be a homomorphism. We need to show that there is an <code>$h \in R$</code> such that <code>$\varphi(f) = f h + I_{\ge c}$</code>. To prove this, look at <code>$\varphi(X^r)$</code> and take <code>$r \to 0$</code>. If <code>$r&lt;c$</code>, then <code>$\varphi(X^r)$</code> will determine $h$ modulo <code>$I_{\ge c-r}$</code>. To finish the proof, we need that if <code>$E \subseteq [0, c)$</code> such that <code>$E \cap [0,d)$</code> is well-ordered for all <code>$d&lt;c$</code>, then $E$ is also well-ordered.</p> <p>For <code>$I=I_{&gt; c}$</code>, the proof is almost the same.</p>