Rings in which every non-unit is a zero divisor 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?
 A: 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).
A: I don't know the name for this class of commutative rings.  Two quick examples:

*

*Any finite ring: then for all $x$ there exist $0 < k < l$ 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.


*Any Boolean ring, i.e., each element is an idempotent: if $x^2 = x$, then $x(1-x) = 0$.
Added: 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


*Any local Artinian ring: each nonunit is a member of the unique maximal ideal, which is nilpotent (e.g. Theorem 82 of http://alpha.math.uga.edu/~pete/integral.pdf).

Also


*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 $xy = 0$.

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 http://alpha.math.uga.edu/~pete/integral.pdf), so every Artinian ring is in the OP's class.
A: How about calling these things balanced commutative rings?
Recall that a category satisfying "monic + epic --> isomorphism" is said to be balanced. Therefore:


*

*a monoid is balanced iff every element that is both left-cancellative and right-cancellative is a unit.

*a commutative monoid is balanced iff every element that is cancellative is a unit.
It makes sense to apply this terminology to rings, too:


*

*a ring is balanced iff every element that is neither a left zero-divisor nor a right-zero divisor is a unit.

*a commutative ring is balanced iff every non-zero divisor is a unit.
A: I have studied a noncommutative version of this. There is such a thing called a right cohopfian ring 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.
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.
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:
"Varadarajan, K. Hopfian and co-Hopfian objects. Publ. Mat. 36 (1992), no. 1, 293–317."
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.
Too bad I didn't see this a year ago :)
A: 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".
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) Noetherian rings of this form have been investigated in J. T. Stafford: Noetherian Full Quotient Rings (Proc. London Math. Soc. (3) 44 (1982) pp. 385-404).  Quote:

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.

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.
Ad: I have asked if this property is Morita invariant in MO 124856.
A: 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".
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.
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?
A: 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).  
Edit: Sorry, I meant a connected graded algebra (where the degree 0 part is the field of coefficients). 
A: Any (commutative unitary) ring of Krull dimension 0 has this property. This includes the class of Artinian rings. 
[Edit] 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. 
Artining rings are zero-dimensional and semi-local. Boolean rings are reduced and zero-dimensional. 
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. 
[Edit] 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. 
A: 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.
At first let me give you some special information about these examples.
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: $Z(f)=${$x\in X$: $f(x)=0$}  
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 $x\in int_X Z(f)$.
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.
Theorem: In the ring $C(X)$, every non-unit is a zero divisor iff the topological space $X$ is almost $P$-space.
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.
A: Pace Chris Leary, the standard terminology is that a module all monomorphisms of which are automorphisms is said to be cohopfian (or co-Hopfian, if you're checking MathSciNet).   A Dedekind-finite (a.k.a. directly finite) 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.
T. Y. Lam, in his book Lectures on Modules and Rings, pp. 320–322, calls a noncommutative ring in which every regular element (i.e. neither right nor left zero-divisor) is a unit a classical ring, and he provides various examples.   One that has not already been mentioned here is that any right (or left) self-injective ring is classical.   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, 
“On automorphisms of an injective module,”
Proc. Japan Acad. 44 (1968), 120–124, and G. F. Birkenmeier,
“On the cancellation of quasi-injective modules,”
Comm. Algebra 4 (1976), no. 2, 101–109).
