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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.

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    $\begingroup$ 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
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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, $xy\neq 0$.

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  • $\begingroup$ $xyxy$ 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
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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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

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