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Does there exist a non-commutative or commutative ring or subring $R$ with $x \cdot x = 0$ where $0$ is the zero element of $R$, $\cdot$ is multiplication secondary binary operation, and $x$ is not zero element, and excluding the case where addition (abelian group operation) and multiplication of two numbers always become zero?

Edit: most seem to be focused on non-commutative case. What about commutative case?

Edit 2: It is fine to relax the restriction to the following: there exists countably (and possibly uncountable) infinite number of $x$'s in $R$ that satisfy $x \cdot x = 0$ (so this means that there may be elements of ring that do not satisfy $x \cdot x =0$) excluding the case where addition and multiplication of two numbers always become zero.

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  • $\begingroup$ I forgot to write commutative. So both commutative and non-commutative cases. $\endgroup$ – user69886 Mar 28 '13 at 13:44
  • $\begingroup$ Regarding your edit: If you want arbitrary elements of this type, just take products or tensor products. So take $k[x_1]/(x_1^2) \times k[x_2]/(x_2^2) \times \dotsc$ or $k[x_1,x_2,\dotsc]/(x_1^2,x_2^2,\dotsc)$. $\endgroup$ – Martin Brandenburg Mar 28 '13 at 14:29
  • $\begingroup$ @user69886: I updated my answer with a commutative example. In general and for the future, please do not change your question after answers has been given. In general if you have another question, then just post that as a separate question. $\endgroup$ – Thomas Mar 28 '13 at 14:49
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Example for the non-commutative case: $$ \pmatrix{0 & 1 \\ 0 & 0} \pmatrix{0 & 1 \\ 0 & 0} = \pmatrix{0 & 0 \\ 0 & 0}. $$ Example for the commutative case: Consider the ring $\mathbb{Z} / 4\mathbb{Z}$. What is $2^2$ in this ring?

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  • $\begingroup$ See my answer for a conceptual view of the genesis of this example. $\endgroup$ – Math Gems Mar 28 '13 at 15:54
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Sure, take $k$ any ring and $R=k[x]/(x^2)$. As suggested by Martin Brandenburg, a good example is $k=\mathbb R$, in which case $R$ becomes the ring of dual numbers. The wikipedia site lists a lot of properties, which might be interesting.

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    $\begingroup$ Or almost the same, $\mathbb{Z}/4$. $\endgroup$ – Martin Brandenburg Mar 28 '13 at 13:40
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    $\begingroup$ It's commutative. Though if we take a noncommutative base ring instead of $k$.. $\endgroup$ – Berci Mar 28 '13 at 13:41
  • $\begingroup$ @Berci: Whoops, I misread that. $\endgroup$ – Jesko Hüttenhain Mar 28 '13 at 13:42
  • $\begingroup$ I forgot to write "or commutative.." $\endgroup$ – user69886 Mar 28 '13 at 13:44
  • $\begingroup$ Jesko, you may perhaps add a link to en.wikipedia.org/wiki/Dual_number $\endgroup$ – Martin Brandenburg Mar 28 '13 at 13:53
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It is easy if you know about quotient rings, e.g. $\rm\:\Bbb Z/n^2,\ \Bbb Z[x]/(x^2).\:$ If you don't know about quotient rings, is there still a simple example? Yes! One quickly checks that $\rm\,\Bbb Z^2$ is a ring under pointwise addition and multiplication and, that the set of linear maps on $\rm\,\Bbb Z^2\,$ form a ring under addition and map composition. Now note that the (right) shift map $\rm\: S(a,b) = (0,a)\:$ is linear, and $\rm\,S^2 = 0\,$ since $\rm\: S^2(a,b) = S(0,a) = (0,0).\:$

Remark $ $ If you know matrix rings, note that the matrix of the shift map is that in Thomas's answer. But knowledge of matrices is not required to understand the above simple example.

Where does this example come from? It arises from a general construction. It is, in fact, a linear representation of the generic example $\rm\:R[x]/(x^2),\:$ whose additive group is $\rm\:R^2,\:$ where $\rm\:x\:$ acts as a (right) shift map $\rm\:x(a+bx) = 0+ax,\:$ i.e. $\rm\:(a,b)\mapsto (0,a).\:$ Similarly any ring $\rm\,R\,$ has such a linear representation as a subring of the ring of linear maps on its underling additive group - the so-called (left) regular representation of $\rm\,R.\:$ This is a ring-theoretic analogue of Cayley's theorem, that a group $\rm\,G\,$ may be represented as a subgroup of the group of bijections (permutations) on $\rm\,G.\:$ In both cases the representation arises by sending each element $\rm\:r\:$ into its associated (left) multiplication map $\rm\: x\:\to\:r\:x\:.\:$ See this MO question for further discussion.

As another important example, this view provides a natural way to construct the polynomial ring $\rm\,R[x]\,$ as a subring of the ring of linear maps on $\rm\,R^\Bbb N = (f_0,f_1,f_2,\ldots)\:$ generated by $\rm\,R\,$ and $\rm\,x\,$ i.e. by all constants $\rm\:(r,0,0,\ldots)$ and by the shift map $\rm\,x : (f_0,f_1,f_2,\ldots)\mapsto (0,f_0,f_1,\ldots).$ See also the presentation in this post.

It is worth emphasis that the ring $\rm\:R[x]/(x^2)\:$ is known as the algebra of dual numbers over $\rm\,R,\,$ and that it frequently proves quite handy in various contexts, e.g. when studying (higher) derivatives, or tangent/jet spaces, and when transferring properties of homomorphisms to derivations

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  • $\begingroup$ Actually $S=x$ on $\mathbb{Z}[x]/(x^2)$. So these are the same examples in disguise. $\endgroup$ – Martin Brandenburg Mar 28 '13 at 14:17
  • $\begingroup$ @martin Yes, I meant to segue into that - see the added remark. $\endgroup$ – Math Gems Mar 28 '13 at 14:57
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Yes, for example consider the noncommutative polynomial ring $\Bbb R\langle x,y\rangle$ (so here $xy\ne yx$), and let $R$ be its quotient by the ideal $(x^2)$ generated by $x^2$.

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    $\begingroup$ Non-commutative polynomial rings are often called free (associative) algebras and denoted by other brackets, namely $R\langle x_1,\dotsc,x_n\rangle$. $\endgroup$ – Martin Brandenburg Mar 28 '13 at 13:48

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