Example or classification of existentially closed modules - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T23:43:15Z http://mathoverflow.net/feeds/question/44202 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/44202/example-or-classification-of-existentially-closed-modules Example or classification of existentially closed modules Andrew Parker 2010-10-30T01:09:22Z 2010-11-01T13:57:36Z <p>In the language of modules, it suffices to restrict our view to <em>positive-primitive</em> formulas - that is to say, formulas with one existential quantifier and no negation.</p> <p>And I mean <em>existentially closed</em> in the sense that any witness to a particular positive-primitive formula over the module is already in the module itself.</p> <p>My intuition is that I should look along the lines of injective modules. Can anyone help by offering direction or perhaps an example / verification?</p> http://mathoverflow.net/questions/44202/example-or-classification-of-existentially-closed-modules/44442#44442 Answer by arsmath for Example or classification of existentially closed modules arsmath 2010-11-01T13:57:36Z 2010-11-01T13:57:36Z <p>Do primitive positive formulas only allow conjunctions, not disjunctions? If so, then injective modules are existentially closed, but are a strictly stronger concept in that the equivalent for infinite conjunctions are allowed.</p> <p>The concept you're after is that of <a href="http://en.wikipedia.org/wiki/Algebraically_compact_module" rel="nofollow">algebraically compact</a> module (also known as pure inective module). Here's a link to a paper by <a href="http://eprints.ma.man.ac.uk/1148/01/covered/MIMS_ep2008_83.pdf" rel="nofollow">Prest</a> that talks about some their properties.</p> <p><a href="http://en.wikipedia.org/wiki/Injective_module#Baer.27s_criterion" rel="nofollow">Baer's criterion</a> asserts existential closure under infinite conjunctions. Let $$(\exists m) r_1 m = m_1 \wedge r_2 m = m_2 \wedge \cdots$$ be a particular formula, where $R$ is your coefficient ring, $M$ a module over $R$, with $r_i \in R$ and $m_i \in M$. Then consider the ideal $I$ generated by $r_i$ in $R$. There the map that sends $r_i$ to $m_i$ is a homomorphism from $I$ to $M$ if and only if that formula is consistent. (There's a homomorphism if and only if the formula does not imply $s = s'$ for two distinct elements $s, s' \in R$.)</p>