What do epimorphisms of (commutative) rings look like? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T02:54:09Z http://mathoverflow.net/feeds/question/109 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109/what-do-epimorphisms-of-commutative-rings-look-like What do epimorphisms of (commutative) rings look like? Anton Geraschenko 2009-10-05T05:33:42Z 2012-11-14T20:31:34Z <p>(<strong>Background:</strong> In any category, an <em>epimorphism</em> is a morphism $f:X\to Y$ which is "surjective" in the following sense: for any two morphisms $g,h:Y\to Z$, if $g\circ f=h\circ f$, then $g=h$. Roughly, "any two functions on $Y$ that agree on the image of $X$ must agree." Even in categories where you have underlying sets, epimorphisms are not the same as surjections; for example, in the category of Hausdorff topological spaces, $f$ is an epimorphism if its image is dense.)</p> <p>What do epimorphisms of (say commutative) rings look like? It's easy to verify that for any ideal $I$ in a ring $A$, the quotient map $A\to A/I$ is an epimorphism. It's also not hard to see that if $S\subset A$ is a multiplicative subset, then the localization $A\to S^{-1}A$ is an epimorphism. Here's a proof to whet your appetite.</p> <blockquote> <p>If $g,h:S^{-1}A\to B$ are two homomorphisms that agree on $A$, then for any element $s^{-1}a\in S^{-1}A$, we have<br> $$g(s^{-1}a)=g(s)^{-1}g(a)=h(s)^{-1}h(a)=h(s^{-1}a)$$</p> </blockquote> <p>Also, if $A\to B_i$ is a finite collection of epimorphisms, where the $B_i$ have disjoint support as $A$-modules, then $A\to\prod B_i$ is an epimorphism.</p> <p>Is every epimorphism of rings some product of combinations of quotients and localizations? To put it another way, suppose $f: A\to B$ is an epimorphism of rings with no kernel which sends non-units to non-units and such that $B$ has no idempotents. Must $f$ be an isomorphism?</p> http://mathoverflow.net/questions/109/what-do-epimorphisms-of-commutative-rings-look-like/112#112 Answer by Andrew Stacey for What do epimorphisms of (commutative) rings look like? Andrew Stacey 2009-10-05T12:51:09Z 2009-10-06T13:38:39Z <p>A little searching turned up:</p> <p><strong>Ring epimorphisms and C(X)</strong> by Michael Barr, W.D. Burgess and R. Raphael (<a href="http://www.tac.mta.ca/tac/volumes/11/12/11-12abs.html" rel="nofollow">article</a>).</p> <p>They consider this question for rings of the form of continuous functions on a topological space. They quote the following characterisation of epimorphisms in the category of commutative rings:</p> <p><strong>Proposition</strong>: A homomorphism f : A &rarr; B is an epimorphism if and only if for all b &isin; B there exist matrices C, D, E of sizes 1 &times; n, n &times; n, and n &times; 1 respectively, where <i>(i)</i> C and E have entries in B, <i>(ii)</i> D has entries in f(A), <i>(iii)</i> the entries of CD and of DE are elements of f(A) and <i>(iv)</i> b = CDE. (Such a triple is called a <em>zig-zag</em> for b.)</p> <p>This seems a little more complicated than localisation, though I haven't checked the details.</p> <p>They then go on to prove that</p> <p><strong>2.12</strong>: A subspace Y of a perfectly normal first countable space X induces an epimorphism if and only if it is locally closed.</p> <p>If I understand all the terminology correctly, then this implies that</p> <p>C([0,1],&#x0211D;) &rarr; C((0,1),&#x0211D;)</p> <p>is an epimorphism.</p> <p>There are plenty more references in that article, and it would be nice to have an actual zig-zag for this situation. But in the spirit of open-source mathematics, I thought I'd post this and see if someone (possibly me later on) can fill in the details.</p> <p><strong>Added Later:</strong> The example I gave: C([0,1],&#x0211D;) &rarr; C((0,1),&#x0211D;) is a localisation. It is obtained by inverting all functions in C([0,1],&#x0211D;) which are zero <strong>only</strong> at the end-points. Given a function f &isin; C((0,1),&#x0211D;), there will be a function g &isin; C([0,1],&#x0211D;) which is non-zero apart from at 0 and 1 and which goes to 0 at 0 and 1 faster enough that the product g f also goes to 0 at the end-points. Then g f is (the restriction of something in) C([0,1],&#x0211D;) and g becomes invertible in C((0,1),&#x0211D;). So f = g<sup>-1</sup> (g f) is in the specified localisation of C([0,1],&#x0211D;).</p> <p>Indeed, the Barr et. al. paper comments on the fact that in all the examples they consider (function rings), the zig-zag has length 1. I conjecture that if the zig-zags always have length 1 (for a particular function f: A &rarr; B), then B is formed by a localisation on A. A possibly stronger version of this conjecture would be that this is an if-and-only-if. In which case, finding a counter-example to Anton's conjecture would involve finding a case where there was a zig-zag of length 2. I suspect that a universal construction would be the best approach to finding one.</p> <p>In the spirit of wiki-ness and only doing a little at a time, I'll leave this here.</p> <p><strong>Added Even Later:</strong> (Should I timestamp these? I know that the system does so, but is it useful to embed them in the edit?)</p> <p>Here's one direction for my conjecture above.</p> <p>If B = S<sup>-1</sup>A, then for b &isin; B, we have b = s<sup>-1</sup>a for some s &isin; S and a &isin; A. Then we put C = s<sup>-1</sup>, D = s, E = b = s<sup>-1</sup> a. Then CD = 1, DE = a, D &isin; f(A), and CDE = b. So in a localisation, zig-zags have length 1.</p> http://mathoverflow.net/questions/109/what-do-epimorphisms-of-commutative-rings-look-like/139#139 Answer by Anton Geraschenko for What do epimorphisms of (commutative) rings look like? Anton Geraschenko 2009-10-06T14:59:42Z 2009-10-06T14:59:42Z <p>George Bergman gave me a reference (Isbell's "<a href="http://jlms.oxfordjournals.org/cgi/pdf%5Fextract/s2-1/1/265" rel="nofollow">Epimorphisms and dominions, IV</a>") and a very pretty counterexample. In particular, he says that the characterization of epimorphisms Andrew gave us works for non-commutative rings as well:</p> <blockquote> <p>Recall that an inclusion A in B is an epimorphism if and only if the "dominion" of A in B is all of B, where this dominion is defined as the subring of elements b of B which behave the same under all pairs of homomorphisms on B that agree on elements of A.</p> <p>Now the Silver-Mazet-Isbell Zigzag Lemma for rings says that the dominion of A in B consists of those elements of B which can be written XYZ, where X is a row, Y a matrix, and Z a column over B, such that XY and YZ have entries in A. (It is easy to verify that such a product is in the dominion of A -- a generalization of the proof that if Y is in A and has an inverse in B, then this inverse is in the dominion of A.)</p> </blockquote> <p>Let k be a field. Consider the inclusion of k[x, xy, xy<sup>2</sup> - y] into k[x,y]. I claim that this is an epimorphism. Note that it is an inclusion, no non-units become units, and k[x,y] has no idempotents.</p> <p>Suppose f and g are two morphisms from k[x,y] to some other commutative ring which agree on the given subring. Using f(xy)=g(xy) and f(x)=g(x), we see that f(xy<sup>2</sup>)=g(xy<sup>2</sup>):</p> <blockquote> <p>f(yxy) = f(yx)f(y) = g(yx)f(y) = g(y)g(x)f(y) = g(y)f(x)f(y) = g(y)f(xy) = g(y)g(xy) = g(yxy)</p> </blockquote> <p>Since f and g agree on xy<sup>2</sup>-y, they agree on y, so they agree on all of k[x,y].</p> <p>Finally, to see that the inclusion is not an isomorphism, consider the surjective morphism k[x,y] to k[x,x<sup>-1</sup>] sending y to x<sup>-1</sup>. This sends the subring to k[x], which is clearly smaller, so the inclusion of k[x,xy,xy<sup>2</sup>-y] into k[x,y] must be strict.</p> http://mathoverflow.net/questions/109/what-do-epimorphisms-of-commutative-rings-look-like/161#161 Answer by David Rydh for What do epimorphisms of (commutative) rings look like? David Rydh 2009-10-06T23:35:07Z 2009-10-06T23:35:07Z <p>No, not every epimorphism of rings is a composition of localizations and surjections.</p> <p>An epimorphism of commutative rings is the same thing as a <em>monomorphism</em> of affine schemes. Monomorphisms are not only embeddings, e.g., any localization is an epimorphism and the corresponding morphism of schemes is not a locally closed embedding.</p> <p><strong>Example</strong>: Let <em>C</em> be the nodal affine cubic and let <em>X</em> be its normalization. Pick any point <em>x</em> above the node. Then <em>X</em>\{*x*}-><em>C</em> is a monomorphism (see Proposition below). The corresponding homomorphism of rings is injective but not a localization.</p> <p><strong>Proposition</strong> (EGA IV 17.2.6): Let <em>f</em>:<em>X</em>-><em>Y</em> be a morphism <em>locally of finite type</em> between schemes. TFAE:</p> <p>(i) <em>f</em> is a monomorphism.</p> <p>(ii) Every fiber of <em>f</em> is either an isomorphism or empty.</p> <p><strong>Remark</strong>: A <em>flat</em> epimorphism <em>A</em>-><em>B</em> is a localization if <em>A</em> is normal and <strong>Q</strong>-factorial. This is a result by D. Lazard and P. Samuel. [cf. Lazard "Autour de la platitude" (IV, Prop 4.5)]</p> <p><strong>Remark</strong>: There was a <a href="http://www.numdam.org:80/numdam-bin/feuilleter?id=SAC_1967-1968__2_" rel="nofollow">seminar</a> on epimorphisms of rings directed by P. Samuel in 1967-68.</p> http://mathoverflow.net/questions/109/what-do-epimorphisms-of-commutative-rings-look-like/110443#110443 Answer by Torsten Schoeneberg for What do epimorphisms of (commutative) rings look like? Torsten Schoeneberg 2012-10-23T16:23:50Z 2012-11-14T20:31:34Z <p>Here is another perspective on your question. As $\mathbb{Z}$ is the initial object of unital (commutative) rings, one might first of all ask: <strong>What do epimorphisms from $\mathbb{Z}$ look like?</strong> </p> <p>I.e. if $A = \mathbb{Z}$ in the original question, what can $B$ be? The answer to this is known. In fact, these rings $B$ and their classification seem to have been (re)invented several times, as "solid rings" by Bousfield and Kan (see this question: <a href="http://mathoverflow.net/questions/95160/" rel="nofollow">http://mathoverflow.net/questions/95160/</a>), as "T-rings" by R. A. Bowshell and P. Schultz (<em>Unital rings whose additive endomorphisms commute</em>, Math. Ann. 228 (1977), 197-214, <a href="http://eudml.org/doc/162991;jsessionid=07C5F5F5BBD354C0914511776DA20F5E" rel="nofollow">http://eudml.org/doc/162991;jsessionid=07C5F5F5BBD354C0914511776DA20F5E</a>), and the generalisation to Dedekind domains has been done in W. Dicks and W. Stephenson: <em>Epimorphs and Dominions of Dedekind Domains</em>, J. London Math. Soc. (1984) s2-29(2): 224-228, <a href="http://jlms.oxfordjournals.org/content/s2-29/2/224.extract" rel="nofollow">http://jlms.oxfordjournals.org/content/s2-29/2/224.extract</a> . (Also, by Martin Brandenburg and myself this summer, before we found these papers ...)</p> <p>So here is a <strong>positive answer under a restrictive assumption</strong>: If $A \rightarrow B$ is an epimorphism and $A$ is a <strong>Dedekind domain</strong>, then $B$ will be built up from localisations and quotients of $A$ by suitable finite products and direct limits. To make "suitable" more specific, here follows a more concrete description (the literature above mostly says "take colimits/pullbacks"), setting $A = \mathbb{Z}$ for (mostly notational) simplicity: </p> <p>Let $P$ be the set of prime numbers and let $n: P \rightarrow \mathbb{N} \cup \lbrace 0, \infty \rbrace$ be any map (a "supernatural number"). Let $P_{fin}(n)$ be the set of primes $p$ with $n(p) &lt; \infty$. Define</p> <p>$B_n := \lbrace (b_p, b_l) \in \prod_{p \in P_{fin}(n)} \mathbb{Z} / p^{n(p)} \times \mathbb{Z}[P_{fin}(n)^{-1}] :$ $b_p \equiv b_l \text{ mod } p^{n(p)} \text{ for all but finitely many } p \in P_{fin}(n)(b_l) \rbrace$</p> <p>(index "$l$" for "localisation part") where $\mathbb{Z}[P_{fin}(n)^{-1}]$ is the localisation of $\mathbb{Z}$ at the multiplicative set generated by $P_{fin}(n)$, i.e. the subring of $\mathbb{Q}$ generated by $\lbrace p^{-1}: p \in P_{fin}(n) \rbrace$; further, with $v_p$ being the $p$-adic valuation, $P_{fin}(n)(b_l) := \lbrace p \in P_{fin}: v_p(b_l) \ge 0 \rbrace$ and the condition $b_p \equiv b_l \text{ mod } p^{n(p)}$ makes sense and is to be understood in the subring of $\mathbb{Q}$ where only the $p$'s with $v_p(b_l) &lt; 0$ are inverted. Then $B_n$ is in fact a subring of the direct product, and for $n$ ranging over the supernatural numbers, these are all $B$ with injective epimorphisms $\mathbb{Z} \rightarrow B$. (The non-injective ones are just the quotients. With more complicated notation, one could include this case by counting 0 as a prime.)</p> <p>Here are two easy-to-see properties:</p> <ul> <li><p>$B_n$ is noetherian if and only if $|P_{fin}(n) \setminus P_0(n) | &lt; \infty$ if and only if $B_n$ is the direct product of a quotient and a localisation, namely, $\mathbb{Z}/n \times \mathbb{Z}[P_{fin}(n)^{-1}]$ where by abuse of notation $n$ is the natural number $\prod_{p \in P_{fin}(n)} p^{n(p)}$.</p></li> <li><p>The non-zero primes of $B_n$ correspond to the ones in $P \setminus P_0(n)$, where $P_0(n)$ is the set of primes $p$ with $n(p) = 0$. In particular, $B_n$ is artinian if and only if its Krull dimension is 0 if and only if $|P \setminus P_0(n)| &lt; \infty$. Otherwise, its Krull dimension is 1.</p></li> </ul> <p>All this remains true <em>cum grano salis</em> for any Dedekind domain $A$ instead of $\mathbb{Z}$. In particular, as soon as $A$ has infinitely many primes, there are epimorphisms $A \rightarrow B$ where $B$ is non-noetherian. On the other hand, if $A$ has only finitely many primes (which by the way makes it a PID), $B$ will be of the form $A/a \times S^{-1}A$ with $a \in A$ and $S \subseteq A$ multiplicative containing all primes dividing $a$ (and possibly 0). In any case, $B$ will be a colimit of products of localisations and quotients as above, so the answer to the question</p> <blockquote> <p>suppose $f:A \rightarrow B$ is an epimorphism of rings with no kernel which sends non-units to non-units and such that $B$ has no idempotents. Must f be an isomorphism?</p> </blockquote> <p>seems to be <strong>yes</strong> if $A$ is a Dedekind domain: E.g. in the above setting, non-units to non-units implies $P_0(n) = \emptyset$ and $B$ having no idempotents implies $P_{fin}(n) \setminus P_0(n) = \emptyset$.</p> <hr> <p>Further remarks:</p> <p><strong>Remark 1</strong> (cf. David Rydh's first remark): <em>Flat</em> epimorphisms (from <em>any</em> unital ring) are localisations for a certain Gabriel topology and have a kind of a calculus of fractions. For a precise statement, see <em>Quelques observations sur les épimorphismes plats (à gauche) d'anneaux</em> by N. Popescu and T. Spircu, Journ. Alg. vol. 16, no. 1, pp. 40-59, 1970, <a href="http://dx.doi.org/10.1016/0021-8693(70)90039-6" rel="nofollow">http://dx.doi.org/10.1016/0021-8693(70)90039-6</a>, or Bo Stenström's book <em>Rings of Quotients</em>, theorem 2.1 in chapter XI. </p> <p><strong>Remark 2</strong>: Further information might be in the papers of H. H. Storrer, e.g. <a href="http://retro.seals.ch/digbib/view?rid=comahe-002:1973:48::11" rel="nofollow">http://retro.seals.ch/digbib/view?rid=comahe-002:1973:48::11</a> </p> <p><strong>Remark 3</strong>: I have not checked all the details in the generalisation to Dedekind domains, so beware (at least, Martin and I had reached the same result for PIDs). Also, I do not know if there is a generalisation beyond Dedekind domains; I guess Krull domains might be attackable, but I have not seriously tried.</p>