This is the first time I hear this term. Specifically the assertion is that $\mathbb{Z}$ has no zero divisors. So, from my understanding this is because there are not two non-zero numbers $a,b\in \mathbb{Z}$ such that $ab=0$.

Also I can see that this definition is related to the one I learnt in high school that $a$ divides $0$ if $\exists b\in\mathbb{Z}\ (ba=0)$, the difference being that we need to consider the restriction $a,b\neq 0$ when dealing with zero divisors.

What is the motivation of zero divisors? what are they used for?

  • 8
    $\begingroup$ A small part answer: It is not so much what they are good for, it is more that they are things to avoid. The familiar algebra can often be extended to the elements of a ring that are not zero divisors. $\endgroup$ Commented Jul 12, 2013 at 4:22
  • 11
    $\begingroup$ This seems like asking "what is the motivation for asteroids?" We didn't invent them so much as we just found them upon looking outwards beyond our comfortable home turf. $\endgroup$
    – anon
    Commented Jul 12, 2013 at 4:43
  • 4
    $\begingroup$ @anon, I'm not sure I agree. While it may not be exactly phrased this way, it seems to me that it is asking "What is the motivation FOR STUDYING zero divisors?" Just because we observe that something foreign to us exists does not justify why it is interesting, fun or useful to study - and that understanding is what I think is sought here. Of course, I may be liberally adding my own meaning here... $\endgroup$ Commented Jul 12, 2013 at 5:09

3 Answers 3


Trying to shed additional light by means of an example involving a most typical deduction where the presence/absence of zero divisors plays a role. Let's imagine that we need to solve the equation $$x^2-x=0.$$ Before we can make any progress on this, we need to know a bit more. What kind of an object is $x$? Is it a number (real/complex/rational)? Is it a matrix? Is it an element of some other universe, where the equation makes sense? A couple things are certain. Whatever $x$ is, we need the ability to multiply it with itself (otherwise $x^2$ does not exist) as well as subtract it from its square. Furthermore, this universe must have an object called zero. Let's assume that the universe is at least a ring, because then we have all of the above operations available. In a ring we have a unit (a neutral element for multiplication, "$1$") and the distributive law. This allows us to rewrite the equation as $$ 0=x^2-x=x^2-1\cdot x=(x-1)x. $$ Well, does that help? Depends! If $x$ is an element of any of the school level number systems, we know that the system has no zero divisor. Meaning that if $ab=0$ in such a system, we are allowed to deduce that either $a=0$ or $b=0$. If our $x$ comes from such a system, we can make swift progress with our equation and conclude that either $$ x-1=0\qquad\text{or}\qquad x=0. $$ This tells us that either $x=1$ or $x=0$ (standard applications of ring axioms).

So we have completely solved the equation PROVIDED that $x$ resides in a ring that does not have zero divisors. This is just an abstraction of the techniques we learned in school, but there no attention to the zero divisor concept is usually given, even though it did play a key role. That is all fine, because drawing attention to the zero divisor concept begs the question: "Are there meaningful systems with zero divisors?"

Let us next assume that $x$ is an element of the residue class ring $\mathbb{Z}_6$. Students might nevertheless want to use the above technique. But some wiseguy will notice that $x=\overline{3}$ is also a solution, as $x^2=\overline{9}=\overline{3}$. What was wrong with the earlier approach? Let's check: $$ (x-1)x=\overline{(3-1)}\cdot\overline{3}=\overline{2}\cdot\overline{3}=\overline{6}=0. $$ Lo and behold, in this ring a prodct can be zero without either factor being so. But the need to solve our equation still is there. If we are not allowed to use the familiar trick, then what can we do? Quadratic formula? That requires an ability to calculate square roots, and an ability to divide by two. Without going into details I will just state that both of these are suspect in more general rings. Also the absence of zero divisors is hidden in the derivation of the quadratic formula. So what? Thankfully this ring is finite, so we can get away by simply trying all the possible values of $x$. Such checking reveals that the residue class $x=\overline{4}$ is also a solution.

But it gets worse! If $x$ is a $2\times2$ matrix with real entries, then in addition to the "obvious" solutions $$ x=\left(\begin{array}{rr}1&0\\0&1\end{array}\right),\qquad\text{and}\qquad x=\left(\begin{array}{rr}0&0\\0&0\end{array}\right) $$ we also have solutions like $$ x=\left(\begin{array}{rr}1&0\\0&0\end{array}\right),\qquad\text{and}\qquad x=\left(\begin{array}{rr}0&0\\0&1\end{array}\right).$$ And that's not all. I invite you to check that the matrix $$ x=\left(\begin{array}{cc}\cos^2\alpha&-\cos\alpha\sin\alpha\\-\cos\alpha\sin\alpha&\sin^2\alpha\end{array}\right) $$ is a solution of the equation $x^2-x=0$ for all values of the angle $\alpha$.

So the main "motivation" to study the concept of zero divisors is, as André pointed out, to know when they are absent. A simple task may become quite harrowing, if we don't know about the possibility of zero divisors.

Above we saw that a quadratic equation may have four or infinitely many solutions, if the ring has zero divisors. With a little bit of extra work we can prove the familiar result that a degree $n$ equation in a commutative ring without zero divisors has at most $n$ solutions. The reason for adding the commutativity assumption is a bit subtle, and I won't get into that. If I managed to awake your curiosity about that, I advice you to take a peek at this question for an example of a quadratic equation with infinitely many solutions in a non-commutative ring without zero divisors. I also recommend the answer by Arturo Magidin, but it does require some familiarity with ring concepts.

  • 1
    $\begingroup$ If this was all clear to you, and you were looking for something deeper, then I apologize for wasting your time. I couldn't quite ascertain the level you desired, and gambled on the level of "first course in abstract algebra". $\endgroup$ Commented Jul 12, 2013 at 6:39
  • 2
    $\begingroup$ Thank you so much, Jyrki, for your great explanation. Actually, as I said this is the first time I see this concept so, although it's just curiosity because I haven't study abstract algebra yet, it's very important for me to hear opinions of people that work on these grounds and understand the importance of the concept. Your explanation is totally clear and I can see now that everything is motivated for the solution of the equation $ab=0$ because we always assume that $a=0$ and $b=0$, which might not always be possible if the ring has zero devisors, just as AWertheim said in his comment. $\endgroup$ Commented Jul 12, 2013 at 7:10
  • 3
    $\begingroup$ A related viewpoint is that cancellation is only possible in rings without zero divisors. We all know that if $xy=xz$ and $x$ is nonzero, then $y=z$. But this requires absence of zero divisors (to see this, rearrange to get $x(y-z)=0$). $\endgroup$
    – Mark Grant
    Commented Jul 12, 2013 at 10:05
  • $\begingroup$ @MarkGrant I think this does not differ from any ring with zero that much. Anyway you have to check whether the denominator is valid. You have more elements by which you cannot divide, but they are usually the minority anyway. $\endgroup$
    – Anixx
    Commented Sep 25, 2021 at 12:42

Just to add something new to what is already said in the previous answers: in some cases you have "useful" zero-divisors. For example, let $V$ be a vector space and let $W$ be a subspace. Consider $A=\mathrm{End}(V)$, the ring of linear endomorphisms of $V$, endowed with the usual pointwise sum and the composition product. You can define a nontrivial endomorphism $\pi : V \to V$ in $A$ such that $\pi(V) = W$ and $\pi_{|W} = 1_W$. This is what you call a "projection", and it is idempotent, that is $\pi \circ \pi = \pi$. It is also a zerodivisor: if it weren't, then you would have $\pi \circ(1-\pi)=0$ and then $\pi = 0$ or $\pi = 1_V$. It should be pointed out that the existence of nontrivial idempotents on a ring is useful to obtain decomposition results for modules over the given ring. Read, for example, this article:



The motivation for the definition of zero-divisors is to identify a phenomenon. Some rings may have zero divisors, some may not. It's a very different definition than, say, that of a group or a ring. When you define groups or rings you, in some sense, create new structures that did not exist before. It's like creating a new machine that turns vodka into flowers. You may wonder what the motivation is to turn vodka into flowers. Similarly, you may wonder what the motivation is for defining groups or rings. On the other hand, the definition of zero-divisors does not create anything new (other than the ability to talk about a phenomenon). Its like looking around you and seeing these huge rocks on the horizon, and naming them mountains. You can't really ask what the motivation for mountains is. Similarly, you can't ask what the motivation for zero-divisors is.

Additional note: You can create a group, but you can't create zero-divisors. A ring either has zero-divisors or it does not. It's a phenomenon.

  • $\begingroup$ oh, I see. but then what are the consequences of having zero divisors in a structure? I mean in the sense of André's answer above I guess it causes some limitations to something... $\endgroup$ Commented Jul 12, 2013 at 4:51
  • 4
    $\begingroup$ @DanielaDiaz, here are two immediate consequences: (1) In the reals (or any integral domain), whenever we have $ab = 0$, we can conclude at least one of $a$ or $b$ is $0$. We cannot make this conclusion in a ring with zero-divisors. (2) Suppose we have a ring with zero divisors, with $ab = 0$ and $a \neq 0, b \neq 0$. What is $\frac{1}{a}$? Does it even exist? Depending on your point of view, these may be banal examples, but it might be fun to explore the kinds of difficulties (referenced by André Nicolas) you can run into in a ring with zero divisors. $\endgroup$ Commented Jul 12, 2013 at 5:15
  • $\begingroup$ @AWertheim I don't know but I have the idea that something specific is behind this concept, I mean something more natural just like the other properties like conmutativity, etc. For example looking on google I found this paper about the relation between zero divisors and the cancelation law msme.us/2008-1-1.pdf $\endgroup$ Commented Jul 12, 2013 at 5:37
  • $\begingroup$ @AWertheim I finally understood your point after reading Jyrki's answer. It's like I was just complicating my life trying no to see the obvious things, sorry for that and thank you so much for your comments. $\endgroup$ Commented Jul 12, 2013 at 7:18
  • 3
    $\begingroup$ @DanielaDiaz the ring of dual numbers has 'useful' zero divisors. $\endgroup$ Commented Jul 12, 2013 at 8:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .