1
$\begingroup$

We know that when a ring is an integral domain, we have that:

$$x^2=x \implies x^2-x = 0 \implies x(x-1) = 0$$

Since this is an integral domain, a product giving $0$ forces one of the the terms in the product to be $0$, therefore the solutions are:

$$x=0, x -1 = 0$$

However, what about when we have a ring but that's not an integral domain, can we find solutions to this equations such that neither of the two terms are $0$?

$\endgroup$
2
$\begingroup$

Yep. In $\mathbb{Z}_6$ let $x=3$. Then $x^2 - x = 9-3 = 6 = 0$. Of course $3\neq 0\wedge (3-1)\neq 0$.

It is possible in $\mathbb{Z}_n$, if $n = k(k-1)$ for some $k\in\mathbb{N}$.

$\endgroup$
5
$\begingroup$

This means the ring has a non-trivial idempotent. It corresponds to a decomposition of the ring, let's call it $R$ as a product of rings. If $e$ is such an idempotent, we have $$R\simeq Re\times R(1-e)$$ $e$ and $1-e$ are the units of the rings $Re$ and $R(1-e)$ respectively. W.r.t. the ring $R$, they're orthogonal idempotents. A simple example is the product of two fields.

Topologically, $\DeclareMathOperator{\spec}{Spec}\spec R=\spec(Re)\cup\spec R(1-e)$ is a partition of $\spec R$ in two open sets. We see $\;\spec R\;$ is connected if and only if $R$ has non non-trivial idempotents. bg)

In the domain of non-commutative algebra, the results are analog, but correspond to a decomposition of $R$ as a module. We obtain a ring decomposition if we have central idempotents.

$\endgroup$
  • $\begingroup$ In case anyone interprets the question in noncommutative context, it should be said that idempotents correspond to module decompositions of $R$, while the ring decompositions correspond to central idempotents $\endgroup$ – rschwieb Sep 23 '15 at 16:47
  • $\begingroup$ @rschwieb: I'll add that, thanks for explaining. I don't know well non commutative algebra, and I seldom venture on these lands… $\endgroup$ – Bernard Sep 23 '15 at 16:58
1
$\begingroup$

Take $A=A_1\times A_2$ a product of two rings. Then $(1,0)^2=(1,0)$ and $(0,1)^2=(0,1)$.

$\endgroup$
0
$\begingroup$

A Boolean ring is, by definition, a ring in which every element is idempotent. Since any Boolean algebra defines a Boolean rings (and conversely), there are plenty of examples.

$\endgroup$

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.