This is an exercise in a book "Rings and Modules- Musili": (in this book, ring may not have unity.)

Give an example of a non-trivial commutative ring in which square of every element is zero.

Here non-trivial means it is not the case that "$xy=0$ for all $x,y$".

This also raises a natural question:

Give an example of a non-trivial non-commutative ring in which square of every element is zero.

I tried the following examples: $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$, with point-wise addition and multiplication like well known "cross product" in vector calculus in $\mathbb{R}^3$. But, this product is not associative.

Another example, $M_2(\mathbb{Z}/2\mathbb{Z})$ with usual addition; but multiplication $*$ defined as $A*B=AB+BA$. But here also, associativity of $*$ fails.

  • 4
    $\begingroup$ One observation: it would have to be a ring without a multiplicative identity, because if it had one, call it $e$, then by definition $e^2 = ee = e$, not $0$. $\endgroup$ – Bungo Dec 11 '14 at 5:41
  • $\begingroup$ yes! I edited in my question about this, first line. $\endgroup$ – Groups Dec 11 '14 at 5:42
  • $\begingroup$ From $(A+B)^2 = 0$ we have $AB + BA = 0$, so multiplication is anticommutative. $\endgroup$ – hardmath Dec 11 '14 at 5:45
  • $\begingroup$ Oh-did you only want a noncommutative example, or is the commutative one I gave also of interest? $\endgroup$ – Kevin Carlson Dec 11 '14 at 5:55
  • $\begingroup$ Book question asks for "commutative" case; I am also asking "non-commutative case", a natural question. $\endgroup$ – Groups Dec 11 '14 at 5:58

Start with $\mathbb{Z}_2$ and its two-variable polynomial ring $\mathbb{Z}_2[x,y]$. Introduce the relations $x^2=y^2=0$ and take the subring $R$ generated by $x$ and $y$. Everything squares to zero, since in characteristic $2$ $(a+b)^2=a^2+b^2$ and every element of $R$ is a sum of terms divisible by either $x$ or $y$. However, $xy\neq 0$.

  • $\begingroup$ $xyxy$ is zero there? $\endgroup$ – Mariano Suárez-Álvarez Dec 11 '14 at 5:51
  • $\begingroup$ Oh, I thought you were taking about non-commutative polyomials. Notice the question wants a noncomutative example! $\endgroup$ – Mariano Suárez-Álvarez Dec 11 '14 at 5:51
  • $\begingroup$ Commutative polynomials, right? Ah, yes. I'm currently wondering whether I can modify this in terms of differential operators or something for a noncommutative example, but I don't see it. $\endgroup$ – Kevin Carlson Dec 11 '14 at 5:51
  • $\begingroup$ @Kevin: Good Example. Thank you! $\endgroup$ – Groups Dec 11 '14 at 6:33
  • $\begingroup$ Also you have to consider the non-unital polynomial ring (so $x+1$ for example is forbidden). Otherwise it is not true that every square is zero. $\endgroup$ – Martin Brandenburg Dec 11 '14 at 9:39

Long comment...

Suppose here exists such a ring $R$. Since it is not commutative, there are two elements $x$ and $y$ in $R$ which do not commute. Consider the ring $F$ which is free as a $\mathbb Z$-module with basis the set of (positive length) words in two letters $a$ and $b$. There is then a ring morphism $f:F\to R$ mapping $a$ and $b$ to $x$ and $y$, repectively. It follows from this that kernel $I=\ker f$ is an ideal of $F$, and it must contain all squares of elements of $F$.

This suggests that we consider precisely the ring $U=F/Q$ with $Q$ the ideal generated by all squares. Indeed, the argument above implies that if there is any example at all of what you are looking for, then $U$ is also an example, so we may just as well consider $U$ as an example! Of course, we have to check that $U$ is an example. In $U$ every square is zero by construction, but wee have to check that it is non-commutative (it could fail to be non-commutative, if for example it turns out that $Q=F$.

...and an example

Now $x^2$ and $y^2$ are in the ideal $Q$, so $Q$ contains the whole ideal $(x^2,y^2)$. This implies that $U$ is a quotient of the algebra $\mathbb\langle x,y\rangle/(x^2,y^2)$. This has as a basis the monomials of positive length which alternate $x$s and $y$s. This means that these monomials also span $U$. But in $U$ $x$ and $y$ anticommute, as hardmath observed. This allows us to see exactly what $U$ is: it is the free abelian group with basis $x$, $y$ and $xy$, with product such that every square is zero and $x$ and $y$ anticommute.

Once we have gotten us this example, we can think how to obtain it all in one swoop. That's easy: let $V=\mathbb Z^2$ be a free abelian group of rank two, let $E=\Lambda V$ be the exterior algebra of $V$, and let $A\subseteq E$ be its augmentation ideal. Then $A$ is isomorphic to the universal example $U$ we constructed above.

  • $\begingroup$ Ah, that's very natural. Is there actually a worry here? $F$ is the noncommutative integer polynomials in $x$ and $y$ without constant term, and you have a degree function there, so $xy-yx$ could only be a square of a degree-1 element, all of which we can write down. Or am I missing something? $\endgroup$ – Kevin Carlson Dec 11 '14 at 5:59
  • 1
    $\begingroup$ It is clearly not a square, but it could possibly be in the ideal generated by squares. $\endgroup$ – Mariano Suárez-Álvarez Dec 11 '14 at 6:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.