Is the category of rings co-well-powered? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T09:35:59Z http://mathoverflow.net/feeds/question/79708 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79708/is-the-category-of-rings-co-well-powered Is the category of rings co-well-powered? Sergei Akbarov 2011-11-01T12:54:12Z 2011-11-03T21:13:44Z <p>Dear colleagues,</p> <p>Can anybody explain me if a category of (associative) rings is co-well-powered (this is the MacLane definition, in Russian literature this is called "locally small from the right side")? I mean, it is well-powered, of course, since for any ring A one can easily find a skeleton in the category Mono(A) (of all monomorphisms with values in A) and this will be a set (the set of all subrings in A). But is it true, that it is co-well-powered, i.e. for any ring A there exists a skeleton in the category Epi(A) (of all epimorphisms from A into other rings) and is this skeleton again a set?</p> <p>Thank you in advance, Sergei Akbarov</p> http://mathoverflow.net/questions/79708/is-the-category-of-rings-co-well-powered/79709#79709 Answer by Guillaume Brunerie for Is the category of rings co-well-powered? Guillaume Brunerie 2011-11-01T12:57:32Z 2011-11-01T12:57:32Z <p>A category is locally small if for all objects $A$ and $B$, $\mathrm{Hom}(A,B)$ is a set. In the category of rings, $\mathrm{Hom}(A,B)$ is a subset of the set of all functions between the underlying sets of $A$ and $B$, so it's a set.</p> <p>I don't understand what you are trying to do with your categories of monomorphisms and epimorphisms (perhaps you meant something else than "locally small"?)</p> http://mathoverflow.net/questions/79708/is-the-category-of-rings-co-well-powered/79711#79711 Answer by Anonymous for Is the category of rings co-well-powered? Anonymous 2011-11-01T13:23:06Z 2011-11-01T13:23:06Z <p>There are at least two different meaninigs of locally small category in mathematics. One is that a category has small hom-sets and other is that set of subobjects is small for all objects. His question is if ring has small set of quotient objects.</p> http://mathoverflow.net/questions/79708/is-the-category-of-rings-co-well-powered/79725#79725 Answer by Buschi Sergio for Is the category of rings co-well-powered? Buschi Sergio 2011-11-01T16:34:04Z 2011-11-03T21:13:44Z <p>Yes, rings as any algebraic theory make a locally presentable category, and this is well- copowred (Adámek and Rosicky, Locally Presentable and Accessible Categories Cambridge University Press, Cambridge, (1994)) </p> <p>About topological case, in the MAria Clementino article "Categorical and topological aspects of semi-abelian theories" (http://www.math.yorku.ca/~tholen/HB07BournClementino.pdf) ther is a call from a Wyler article (see 10.1 in Clementino article), from this follow that: the forgetful funtor from topological rings category to rings category preserve colimits (and these categories have colimits), the it preserve epimorphism, then from the well copoweredness of rings category follow the well cowellpowerness of the topological rings category</p> <p>I seems (and hope) that this work well...</p> <p>(Please, excuse my porr English).</p> <p>About ALgebraic theory modeled on $T_2$ spaces (Hausdorff topological spaces), for example Hausdorff topological rings, or vector (on real or complex numbers) hausdorff topological spaces (with infinite unary operation, one for any number) we have that epimorhpisms in $T_2$ are maps by dense images. Now the $T_2$ topological spaces (and continuous maps) is a cowellpowred category, this follow from the fact that the cardinaliy of a $T_2$ space $Y$ by a dense subset X is bounded by $2^{2^X}$ (take two different point $a, b\in Y$ consider the class of neighborhoods of these points, and their intersection by $X$).</p> <p>THen if the epimorphism in vectorial hausdorf spaces are continuous maps with dense image (of course a such maps is a Ephimorphism) then VEctor topological housdorff spaces are cowellpowred.</p> <p>but I seems that the answere to this question is yes: if a morphism $f: X\to Y$ on linear $T_2$ spaces has a nodense image, let $N\subset Y$ the closure of image of $f$, then we have a non null ($T_2$) quotient $Y/N$ and the two morphism $0, \pi: Y\to Y/N$ that have the some composition (it is $0$) with f. </p> <p>I see that for hausdorff topological groups the answere to the question "has a epimorphism dense image?" is not (http://mathoverflow.net/questions/56453/epimorphisms-have-dense-range-in-tophausgrp)</p> http://mathoverflow.net/questions/79708/is-the-category-of-rings-co-well-powered/79748#79748 Answer by Andrej Bauer for Is the category of rings co-well-powered? Andrej Bauer 2011-11-01T18:57:52Z 2011-11-01T18:57:52Z <p>Sergio Buschi already gave a general answer, and here is a pedestrian way of thinking about it. First, apparently if $e : R \to S$ is epi, then the cardinality $|S|$ of $S$ cannot exceed the cardinality $|R|$ of $R$. This is not obvious, as $e$ need not be surjective. I am no expert on rings, so I simply Googled the fact and found <a href="http://math.columbia.edu/~dejong/wordpress/?p=623" rel="nofollow">this</a>. For each cardinality below $|R|$, there are can be only set-many non-isomorphic rings. So we have a set of sets of candidates, which is again a set. That's co-well-poweredness.</p>