3
$\begingroup$

Any idempotent element (other than 0 or 1) must be a zero-divisor, so in particular there are no nontrivial idempotents in domains.

Are there examples of rings with zero divisors without nontrivial idempotents? (If so, is there anything interesting to say about such rings, e.g., characterizations?)

$\endgroup$
0
4
$\begingroup$

$\newcommand{\Z}{\mathbb{Z}}$Quotient rings $\Z/p^{n}\Z$, with $p$ a prime, have zero divisors for $n \ge 2$, but no idempotents $\ne 0, 1$. As noted in another answer, more generally local rings have no non-trivial idempotents.

$\endgroup$
4
$\begingroup$

You may find this interesting. To each ring $A$ we may attach a geometric space $\text{Spec } A$. For instance, if the ring is of the form $\mathbb{C}[x,y]/f(x,y)$ you may picture this space as the graph of the equation $f(x,y) = 0$ (although you have to pretend that $\mathbb{C}$ looks like $\mathbb{R}$ to draw this graph on paper).

A ring has a non-trivial idempotent if and only if this space is disconnected. It has a zero-divisor which is not nilpotent if and only if this space is "reducible" (google it for the formal definition, but the idea is that it looks like a union of two spaces meeting in a space of smaller dimension).

From this we can get many examples. For instance, $\mathbb{C}[x, y]/(xy)$. This looks like the union of two lines meeting at the origin. Contrast a ring with idempotents, e.g. $\mathbb{C}[x, y]/(x^2 - x)$ which looks like the disjoint union of two vertical lines.

$\endgroup$
2
  • 1
    $\begingroup$ I swear I just saw your answer! Upvoting you because great minds think alike... :) $\endgroup$ – Bruno Joyal Dec 28 '13 at 22:23
  • $\begingroup$ @BrunoJoyal Do you mean "great and modest"? just kidding... $\endgroup$ – Matemáticos Chibchas Dec 28 '13 at 23:20
3
$\begingroup$

Idempotents of $R$ correspond to clopen subsets of $X = \text{Spec } R$. When $R$ is noetherian, a clopen subset is necessarily a finite disjoint union of connected components.

On the other hand, zero divisors of $R$ correspond to closed subsets of $X$ whose complement is contained in a proper closed subset of $R$. When $R$ is noetherian, such a closed subset is a finite, nonempty union of irreducible components.

Thus, we can obtain a wide variety of examples from geometry. Whenever $X$ is connected but not irreducible, $R$ has nontrivial zero divisors but no nontrival idempotents. For instance, the variety consisting of two intersecting lines is connected but not irreducible, so the ring $\mathbf C[x,y]/(xy)$ has nontrivial zero divisors (eg. $x$ and $y$!) but no nontrivial idempotents.

$\endgroup$
0
$\begingroup$

One example: $\mathbb{Z}_2$, or more generally, $\mathbb{Z}_{p^n}$ for prime $p$ and integer $n$.

In this instance, all zero divisors are nilpotent (of order at most $n$).

$\endgroup$
0

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.