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Is there an easy example (one that an undergraduate math education student who learned the definition of a ring 5 minutes ago can understand) of a set who satisfies all the conditions of a ring except for associativity of multiplication?

epsilon
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3 Answers3

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Let $R$ be a ring. The commutator $$[·~·]\colon R × R → R,~(x,y) ↦ xy - yx$$ then defines an operation on $R$ which is trivial if and only if $R$ is commutative.

It is easy to check that this operation satisfies distributive laws. However, it has no unit in general.

If $R$ is non-commutative, this operation tends to be non-associative. For $x, y, z ∈ R$, $$[[x~y]~z] = (xy - yx)z - z(xy - yx) ≠ x(yz - zy) - (yz - zy) x = [x~[y~z]]$$ if and only if $$yxz + zxy ≠ xzy + yzx.$$

For example, if $x$ and $y$ are a symmetric pair of zero divisors, that is $xy = yx = 0$ and $z$ isolates them, the operation becomes non-associative. To specify further, take $x = [\begin{smallmatrix}1 & 0 \\ 0 & 0\end{smallmatrix}]$, $y = [\begin{smallmatrix}0 & 0 \\ 0 & 1\end{smallmatrix}]$ and $z = [\begin{smallmatrix}0 & 1 \\ 1 &0\end{smallmatrix}]$. Then $yxz + zyx = 0$, but $xzy + yzx ≠ 0$. (This assumes the student knows how to multiply matrices and recognizes matrix rings as rings.)

k.stm
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  • (Well, that operation has a unit in the zero ring.) ​ ​ –  Jan 15 '18 at 15:03
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    On the other hand, every x satisfies ​ [x,x] = 0 , ​ so R being a zero ring is the $\hspace{1.87 in}$ only way for the bracket to have an identity. ​ ​ ​ ​ –  Jan 15 '18 at 15:09
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You can get the ring of vectors with $\times$ because for all $a$, $b$ and $c$ we obtain: $$(a\times b)\times c+(b\times c)\times a+(c\times a)\times b=0.$$

Michael Rozenberg
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  • That gives a rng, not a ring. ​ ​ –  Jan 15 '18 at 14:26
  • See here: https://en.wikipedia.org/wiki/Ring_(mathematics) This structure apply all axioms. – Michael Rozenberg Jan 15 '18 at 14:32
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    @MichaelRozenberg In the "Definitions" section of your link, item two, the second bullet is not fulfilled. There is no $1$ in this ring. So no, this structure doesn't "apply all axioms". – Arthur Jan 15 '18 at 14:36
  • I think it's not necessary for ring. At least my teacher (professor Dorofeev) said it. OK. If now it's not a ring I'll delete my post. Or maybe Wikipedia wrong? – Michael Rozenberg Jan 15 '18 at 14:38
  • @RickyDemer Some authors allow rings to be non-unital. The questioner has so far not specified if he or she wishes the ring to be unital or not. – k.stm Jan 15 '18 at 14:52
  • @Arthur you might find this article interesting. – P. Siehr Jan 15 '18 at 14:58
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    @P.Siehr I know that there are varying conventions. See, for instance, my comment to the original question post. However, when Rozenberg claims that this structure fulfills all ring axioms and links to the Wikipedia article where the unit requirement is spelled out explicitly, then that is wrong without further explanation and contextualization. – Arthur Jan 15 '18 at 15:18
  • I think, now you understood all points. If you wish, you can fix my post. I don't want to do it. Also, I need to say to you that my name is Michael. – Michael Rozenberg Jan 15 '18 at 15:25
  • Non unital ring is fine. Thank you – epsilon Jan 15 '18 at 20:08
  • You are welcome! – Michael Rozenberg Jan 15 '18 at 20:23
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Here's an example of a non-associative ring.

Let's call it $R$.

Start with the free Abelian group on two generators $x$ and $y$.

Define an operation $*$ in the following way.

  • $x*x = y$
  • $x*y = 0$
  • $y*x = 0$
  • $y*y = x$

Additionally, note that the following two equations hold, which can be used to define $*$ on terms headed by $-$, noting that $a + b$ is equivalent to $a - (0 - b)$.

  • $(a - b) * c = (a * c) - (b * c)$
  • $a * (b - c) = (a * b) - (a * c)$

We can use the triple $(x, y, y)$ to witness the non-associativity of $R$.

$$ x * (y * y) = x * x = y $$ $$ (x * y) * y = 0 * y = 0 $$

Greg Nisbet
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