Do DG-algebras have any sensible notion of integral closure? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T06:45:43Z http://mathoverflow.net/feeds/question/7775 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/7775/do-dg-algebras-have-any-sensible-notion-of-integral-closure Do DG-algebras have any sensible notion of integral closure? Tyler Lawson 2009-12-04T13:47:02Z 2010-01-07T19:46:57Z <p>Suppose R &rarr; S is a map of commutative differential graded algebras over a field of characteristic zero. Under what conditions can we say that there is a factorization R &rarr; R' &rarr; S through an "integral closure" that extends the notion of integral closure in degree zero for connective objects, and respects quasi-isomorphism of the map R to S?</p> <p>I'm willing to accept answers requiring R &rarr; S may need to have some special properties. The motivation for this question is that I'd actually like a generalization of this construction to ring spectra; the hope is that if there is a canonical enough construction on the chain complex level, it has a sensible extension to contexts where it's usually very difficult to construct "ring" objects directly.</p> http://mathoverflow.net/questions/7775/do-dg-algebras-have-any-sensible-notion-of-integral-closure/11064#11064 Answer by Clark Barwick for Do DG-algebras have any sensible notion of integral closure? Clark Barwick 2010-01-07T19:46:57Z 2010-01-07T19:46:57Z <p>I like this question a lot. It deserves an answer, and I really wish I had a good one. Instead, I offer the following idea. Maybe it has some merit?</p> <h3>Background</h3> <p>Let me fix some terminology. Suppose $f:R\to S$ a homomorphism of (classical, commutative) rings.</p> <ol> <li><p>An element $s\in S$ is said to be <em>integral over</em> $R$ if there is a monic polynomial $p\in (R/\ker f)[x]$ such that $s$ is a root of $p$; this is equivalent to saying that the subring $(R/\ker f)[s]\subset S$ is finite over $(R/\ker f)$.</p></li> <li><p>We say that $f$ is <em>integrally closed</em> if it is a monomorphism and if every element of $S$ that is integral over $R$ is in $R$.</p></li> <li><p>At the opposite extreme, we say that $f$ is <em>integrally surjective</em> if every element of $S$ is integral over $R$. (This turns out to be equivalent to being a colimit of proper homomorphisms of finite presentation.)</p></li> <li><p>Among the integrally surjective homomorphisms are the <em>elementary</em> integrally surjective homomorphisms, i.e., homomorphisms of the form $R\to (R/\mathfrak{a})[x]/(p)$, where $R$ is of finite presentation, $\mathfrak{a}\subset R$ is a finitely generated ideal, and $p$ is any monic polynomial.</p></li> </ol> <p>The classical integral closure construction can now be described as a unique factorization of every homomorphism $f:R\to S$ into an integrally surjective homomorphism $R\to\overline{R}$ followed by an integrally closed monomorphism $\overline{R}\to S$.</p> <p>In &sect;3.6 of Mathieu Anel's (really cool!) <a href="http://arxiv.org/abs/0902.1130" rel="nofollow">paper</a>, he describes this factorization system and the "proper topology" constructed from it. In particular, he observes that integrally closed monomorphisms are precisely those morphisms satisfying the unique right lifting property with respect to all elementary integrally surjective homomorphisms.</p> <h3>Making this work for $E_{\infty}$ ring spectra</h3> <p>Since you expressed interest in getting integral closure off the ground for $E_{\infty}$ ring spectra, I'll work in that context.</p> <p>We can use Andre Joyal's theory of factorization systems in &infin;-categories (see &sect;5.2.8 of Jacob Lurie's Higher Topos Theory) to try to play this same game in the &infin;-category $\mathcal{C}$ of connective $E_{\infty}$ ring spectra. (For reasons that will become clear, I'm worried about making this fly for nonconnective $E_{\infty}$ ring spectra.) Once a set $I$ of elementary integrally surjective morphisms is selected, integrally closed monomorphisms are determined as the class of maps that are right orthogonal to $I$, and the integral closure construction is a factorization system on $\mathcal{C}$.</p> <p>So what should $I$ be? I think there might be some flexibility here, depending on your aims, but here's a proposal:</p> <ol> <li><p>Start with the coherent connective $E_{\infty}$ ring spectra that are of finite presentation (over the sphere spectrum). (Concretely, these are the connective $E_{\infty}$ ring spectra $A$ with the following properties: (1) $\pi_0A$ is of finite presentation, (2) for every integer $n$, $\pi_nA$ is a finitely presented module over $\pi_0A$, and (3) the absolute cotangent complex $L_A$ is a perfect $A$-module.) We'll only need these kinds of $E_{\infty}$ ring spectra in our construction of $I$.</p></li> <li><p>Among these $E_{\infty}$ ring spectra, consider the set $I'$ of all morphisms $A\to B$ of finite presentation that induce a surjection on $\pi_0$. Let's call the maps of $I'$ <em>quotients</em>.</p></li> <li><p>Now we need to enlarge $I'$ to allow ourselves morphisms that act as though they are of the form $A\to A[x]/(p)$ for $p$ monic. For any of our connective $E_{\infty}$ ring spectra $A$, we can consider any finitely generated and free $E_{\infty}$-$A$-algebra $A[X]$ (i.e., the symmetric algebra on some free and finitely generated $A$-module), and we can consider quotients (in the sense above) $A[X]\to B$ where $B$ is almost perfect as an $A$-module (equivalently, $\pi_nB$ is finitely presented as an $\pi_0A$-module for every integer $n$); let us add the resulting composites $A\to A[X]\to B$ to our set $I'$ to obtain the set $I$.</p></li> </ol> <p>Now the morphisms that are right orthogonal to $I$ can be called <em>integrally closed monomorphisms</em> of $E_{\infty}$ ring spectra; call the set of them $S_R$. The morphisms that are left orthogonal to that can be called the <em>integrally surjective morphisms</em> of $E_{\infty}$ ring spectra. The integral closure would be a factorization system $(S_L, S_R)$. One shows the existence of a factorization (via a presentation argument), and it follows from general nonsense (more precisely, 5.2.8.17 of HTT) that it is unique.</p> <h3>Three observations</h3> <ol> <li><p>It's not clear that $I$ is big enough for all purposes. One might want to allow shifts of free modules to generate our finitely generated and free $E_{\infty}$-$A$-algebras in the description above. I haven't thought carefully about this.</p></li> <li><p>It's not so obvious how to talk about quotients $A\to B$ when dealing with nonconnective guys. One wants to say that the fiber (in the category of $A$-modules) is "not any more nonconnective than $A$." I'm not quite sure how to formulate this. In any case, that's why I restricted attention to the connective guys above.</p></li> <li><p>Predictably, this is definitely not compatible with the usual integral closure: if I take two classical rings $R$ and $S$ and a ring homomorphism $R\to S$, the integral closure of $HR$ in $HS$ is not in general an Eilenberg-Mac Lane spectrum. If $R$ is a $\mathbf{Q}$-algebra, the two notions are compatible, however.</p></li> </ol> <h3>Making any computation</h3> <p>... seems really hard. But maybe that's not such a big surprise. After all, integral closures are hard to compute classically as well.</p>