Rings in which every non-unit is a zero divisor - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T15:56:09Z http://mathoverflow.net/feeds/question/42647 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42647/rings-in-which-every-non-unit-is-a-zero-divisor Rings in which every non-unit is a zero divisor lhf 2010-10-18T14:48:06Z 2013-03-19T17:33:50Z <p>Is there a special name for the class of (commutative) rings in which every non-unit is a zero divisor? The main example is $\mathbf{Z}/(n)$. Are there other natural or interesting examples?</p> http://mathoverflow.net/questions/42647/rings-in-which-every-non-unit-is-a-zero-divisor/42650#42650 Answer by Pete L. Clark for Rings in which every non-unit is a zero divisor Pete L. Clark 2010-10-18T15:01:39Z 2010-10-18T15:20:11Z <p>I don't know the name for this class of commutative rings. Two quick examples:</p> <p>1) Any finite ring: then for all $x$ there exist $0 &lt; k &lt; l$ such that $x^k = x^l$, so $x^k(x^{l-k}-1) = 0$. This shows that $x$ is a zero divisor unless $x^{l-k}-1 = 0$, i.e., $x^{l-k} = 1$, in which case $x$ is a unit. </p> <p>2) Any Boolean ring, i.e., each element is an idempotent: if $x^2 = x$, then $x(1-x) = 0$. </p> <p><b>Added</b>: Charles Staats's comment gives another important class of rings satifying the desired condition. To flesh it out, first note that the OP's class of rings includes</p> <p>3) Any local Artinian ring: each nonunit is a member of the unique maximal ideal, which is nilpotent (e.g. Theorem 82 of <a href="http://math.uga.edu/~pete/integral.pdf" rel="nofollow">http://math.uga.edu/~pete/integral.pdf</a>).</p> <p>Also</p> <p>4) The OP's class is closed under finite products. Indeed, let $R = R_1 \times \ldots \times R_n$, where the $R_i$ are in the OP's class. Let $x = (x_1,\ldots,x_n)$ be a nonunit of $R$. This happens iff for at least one $i$, $x_i$ is a nonunit in $R_i$. Without loss of generality say $i = 1$. Then there exists a nonzero $y_1$ in $R_1$ such that $x_1 y_1 = 0$. Putting $y = (y_1,0,\ldots,0)$, we get $xy = 0$.</p> <p>It follows that any finite product of local Artinian rings is in the OP's class. But every Artinian ring is a finite product of local Artinian rings (e.g. Theorem 86 of <a href="http://math.uga.edu/~pete/integral.pdf" rel="nofollow">http://math.uga.edu/~pete/integral.pdf</a>), so every Artinian ring is in the OP's class.</p> http://mathoverflow.net/questions/42647/rings-in-which-every-non-unit-is-a-zero-divisor/42651#42651 Answer by Todd Trimble for Rings in which every non-unit is a zero divisor Todd Trimble 2010-10-18T15:05:23Z 2010-10-18T15:12:56Z <p>I don't know whether this concept has been named. But other examples include Boolean rings and products of endomorphism (matrix) algebras, and rings such as $\mathbb{C}[x]/(x^2)$, or more generally the total algebra of a graded algebra which is bounded in degree, as for example, the cohomology algebra of a space homotopy equivalent to a finite CW complex (with coefficients in a field). </p> <p>Edit: Sorry, I meant a connected graded algebra (where the degree 0 part is the field of coefficients). </p> http://mathoverflow.net/questions/42647/rings-in-which-every-non-unit-is-a-zero-divisor/42656#42656 Answer by Chris Leary for Rings in which every non-unit is a zero divisor Chris Leary 2010-10-18T15:27:39Z 2010-10-18T15:27:39Z <p>From a more positive perspective, you are looking at rings with the property that every regular element is a unit. I have done some work with these rings. If the ring is commutative, the condition is equivalent to the ring being a quoring, i.e., it's its own classical ring of quotients. Non-commutative rings with the property must be quorings, but the converse is not necessarily true, as seen with von Neumann regular rings. I called rings with every regular element a unit Dedekind finite because they are characterized as R-modules by the property that every monic endomorphism is an isomorphism (the Dedekind definition of finite set; I picked up the name from L. N. Stout). This is not standard terminology I believe (see Lam's book "Lectures on Modules and Rings).</p> http://mathoverflow.net/questions/42647/rings-in-which-every-non-unit-is-a-zero-divisor/42657#42657 Answer by Martin Brandenburg for Rings in which every non-unit is a zero divisor Martin Brandenburg 2010-10-18T15:30:18Z 2012-05-18T10:18:51Z <p>A commutative ring $A$ has the property that every non-unit is a zero divisor if and only if the canonical map $A \to T(A)$ is an isomorphism, where $T(A)$ denotes the total ring of fractions of $A$. Also, every $T(A)$ has this property. Thus probably there will be no special terminology except "total rings of fractions".</p> <p>Artinian rings provide examples: If $x \in A$, the chain $... \subseteq (x^2) \subseteq (x) \subseteq A$ is stationary, say $x^k = y x^{k+1}$ for some minimal $k \geq 0$. If $k=0$, $x$ is a unit. If $k \geq 1$, $x (x^k y - x^{k-1})=0$ and $x^{k-1} \neq y x^k$, i.e. $x$ is a zero divisor.</p> <p>The class of total rings of fractions is closed under (infinite) products and directed unions. Is it the smallest such class containing the artinian rings?</p> http://mathoverflow.net/questions/42647/rings-in-which-every-non-unit-is-a-zero-divisor/42690#42690 Answer by Qing Liu for Rings in which every non-unit is a zero divisor Qing Liu 2010-10-18T20:00:25Z 2010-10-20T19:00:55Z <p>Any (commutative unitary) ring of Krull dimension 0 has this property. This includes the class of Artinian rings. </p> <p>[<b>Edit</b>] Proof: If $A$ has Krull dimension $0$, then any maximal $m$ ideal of $A$ is also a minimal prime ideal by the definition of Krull dimension. Applying Krull's theorem on the intersection of prime ideals to the localization $A_m$, we find that $mA_m$ is the nilradical of $A_m$. So for any $f\in m$, there exists a positive integer $n$ such that $f^n=0$ in $A_m$. So there exists $s\in A\setminus m$ such that $sf^n=0$ in $A$. We can choose $n$ smallest with respect to this property so that $sf^{n-1}\ne 0$. Therefore $f$ is a zero divisor. Now any non-unit element $f$ belong to some maximal ideal, it is a zero divisor. </p> <p>Artining rings are zero-dimensional and semi-local. Boolean rings are reduced and zero-dimensional. </p> <p>If $(A, m)$ is a local ring, then it has this property if and only if $\mathrm{depth}(A)=0$ (e.g. the example in Daniel Erman's comment). If $A$ is not necessarily local but all localizations $A_m$ at maximal ideals of $A$ have depth $0$, then $A$ has your property. But I don't think this is a necessary condition. </p> <p>[<b>Edit</b>] Similarly one can construct local rings of depth 0 of any (even infinite) dimension. But I don't know whether there exists a ring of positive dimension with infinitely many maximal ideals and such that all its localizations at maximal ideals have depth 0. </p> http://mathoverflow.net/questions/42647/rings-in-which-every-non-unit-is-a-zero-divisor/43246#43246 Answer by Greg Marks for Rings in which every non-unit is a zero divisor Greg Marks 2010-10-22T23:09:09Z 2010-10-22T23:09:09Z <p><i>Pace</i> Chris Leary, the standard terminology is that a module all monomorphisms of which are automorphisms is said to be <i>cohopfian</i> (or <i>co-Hopfian</i>, if you're checking MathSciNet).&#160; A <i>Dedekind-finite</i> (a.k.a. <i>directly finite</i>) module usually means a module whose left invertible endomorphisms are also right invertible, equivalently, a module that is not isomorphic to any proper direct summand of itself.</p> <p>T. Y. Lam, in his book <i>Lectures on Modules and Rings</i>, pp. 320&#8211;322, calls a noncommutative ring in which every regular element (i.e. neither right nor left zero-divisor) is a unit a <i>classical</i> ring, and he provides various examples.&#160; One that has not already been mentioned here is that any right (or left) self-injective ring is classical.&#160; Right self-injective rings need not have the property that every element that is merely not a left zero-divisor is a unit; interestingly, for right self-injective rings the latter condition is equivalent to the ring being Dedekind-finite (in the sense of the preceding paragraph), and also equivalent to the ring having stable range 1 (see Y. Suzuki, &#8220;On automorphisms of an injective module,&#8221; <i>Proc. Japan Acad.</i> <b>44</b> (1968), 120&#8211;124, and G. F. Birkenmeier, &#8220;On the cancellation of quasi-injective modules,&#8221; <i>Comm. Algebra</i> <b>4</b> (1976), no. 2, 101&#8211;109).</p> http://mathoverflow.net/questions/42647/rings-in-which-every-non-unit-is-a-zero-divisor/83454#83454 Answer by rschwieb for Rings in which every non-unit is a zero divisor rschwieb 2011-12-14T18:53:46Z 2011-12-14T18:53:46Z <p>I have studied a noncommutative version of this. There is such a thing called a <em>right cohopfian ring</em> in the sense that if the right annihilator of r is zero, then r is a unit. If you add commutativity and look at the contrapositive, you get that nonunits are zero divisors.</p> <p>I don't think this terminology has caught on, but here is the rationale. A "cohopfian object" is one for which injections are surjections. Looking on elements of the ring as maps sending x-->rx, we are saying that if such a map is injective, it is surjective.</p> <p>Right Artinian, right perfect and strongly-pi regular rings (commutative VNR rings are strongly pi regular) are all right and left cohopfian. Finding a one-sided cohopfian ring seems tough, but Varadarajan did it here:</p> <p>"Varadarajan, K. Hopfian and co-Hopfian objects. Publ. Mat. 36 (1992), no. 1, 293–317."</p> <p>I think someone has noted above that right cohopfian rings have ot be Dedekind finite, and it is interesting that Dedekind finite=right Hopfian=left Hopfian.</p> <p>Too bad I didn't see this a year ago :)</p> http://mathoverflow.net/questions/42647/rings-in-which-every-non-unit-is-a-zero-divisor/97253#97253 Answer by AliReza Olfati for Rings in which every non-unit is a zero divisor AliReza Olfati 2012-05-17T19:15:19Z 2012-05-17T19:15:19Z <p>I think if you have some information about rings of continuous functions$(C(X))$, you can construct a wide class of rings with this property.</p> <p>At first let me give you some special information about these examples.</p> <p>Def 1. For topological space $X$ we denote the ring of all continuous functions on $X$ by $C(X)$. for $f\in C(X)$ the zero-set of $f$ is defined as: $Z(f)=${$x\in X$: $f(x)=0$} </p> <p>Def 2. A completely regular topological space $X$ is called an almost $P$-space, if for every $f\in C(X)$ with nonempty zero-set ,i.e.$Z(f)$, this set has nonempty interior, i.e.there exist $x$ that $x\in int_X Z(f)$.</p> <p>With the above definition, I can introduce a theorem which classifies all Rings of continuous functions $C(X)$ with the property that every non-unit is a zero-divisor.</p> <p>Theorem: In the ring $C(X)$, every non-unit is a zero divisor iff the topological space $X$ is almost $P$-space.</p> <p>The simplest examples of almost $P$-spaces are discrete spaces. for example if $X$ is a discrete space, then $C(X)$ is equal to the usual cartesian product $\mathbb{R}^X$. So for arbitrary set $X$ you can construct $\mathbb{R}^X$ to have the property of your question.</p> http://mathoverflow.net/questions/42647/rings-in-which-every-non-unit-is-a-zero-divisor/124990#124990 Answer by Torsten Schoeneberg for Rings in which every non-unit is a zero divisor Torsten Schoeneberg 2013-03-19T17:33:50Z 2013-03-19T17:33:50Z <p>As Greg Marks says, there is a natural generalisation of this property to a non-commutative ring $R$: "Every regular element (= neither left nor right zero-divisor) in $R$ is a unit", which is equivalent to "The set of regular elements in $R$ is (left and right) Ore, and the natural localisation morphism is an isomorphism". This has been called "full quotient ring" or the like, now Lam calls it "classical".</p> <p>Since I have not found it in any comment or answer here as yet (and neither in Lam's book), let me add that (two-sided) <em>Noetherian</em> rings of this form have been investigated in J. T. Stafford: <em><a href="http://plms.oxfordjournals.org/content/s3-44/3/385.full.pdf" rel="nofollow">Noetherian Full Quotient Rings</a></em> (Proc. London Math. Soc. (3) 44 (1982) pp. 385-404). Quote:</p> <blockquote> <p>Let $A(R)$ be the largest Artinian ideal and $J(R)$ the Jacobson radical of a Noetherian ring $R$. Then $R$ equals its own full quotient ring if and only if $$l\text{-}ann (A(R)) \cap r\text{-}ann (A(R)) \subseteq J(R).$$ It follows that a Noetherian full quotient ring is semilocal [= $R/Jac(R)$ is Artinian semisimple] and has a non-zero Artinian ideal.</p> </blockquote> <p>In the commutative case, the criterion is simply $ann (A(R)) \subseteq J(R)$, and "semilocal" equals "finitely many maximal ideals". Without the "Noetherian" assumption, the criterion does not make sense as $A(R)$ may not exist, and its corollary is false in general.</p> <p>Ad: I have asked if this property is Morita invariant in <a href="http://mathoverflow.net/questions/124856/is-being-a-full-ring-of-quotients-a-morita-invariant-property" rel="nofollow">MO 124856</a>.</p>