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Let $A$ be an associative algebra and consider the poset $I$ of idempotents in $A$, where as usual if $e,f \in A$ we say $e \leq f$ provided that $e = fef$. Clearly $0 \in I$ is minimal, but in general I can't say anything else about this poset.

This leads me to ask: are there any other general properties of $I$ without more assumptions on $A$? In particular, if I am given some poset with a minimal element, is it possible to construct an algebra with the prescribed poset of idempotents?

Edit: I should have said I'm working over a field (or at least a ring with no nontrivial idempotents), since otherwise idempotents in the base ring complicate things. Also: bonus points for producing examples where $A = \cup_{e \in I} eAe$.

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What goes wrong when you take a polynomial ring in a bunch of variables $e$, mod out by the ideal generated by $e^2 = e$ and all the $e = fef$s that you need? Oh, is the problem that you can't prove that some of the idempotents don't coincide in this quotient and/or that there aren't more idempotents in it? –  user29743 Jul 21 '12 at 18:54
    
Do you suspect that this quotient is right but it's hard to prove it does the job, or do you suspect/know that it's wrong? –  user29743 Jul 21 '12 at 18:55
    
@countinghaus: This is a plausible candidate, although $0$ and $1$ cause some trouble. We should leave out the minimal element of the poset from the generators, since $0$ comes along for free. Then pass to the augmentation ideal to avoid $1$. –  Justin Campbell Jul 21 '12 at 19:10

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up vote 5 down vote accepted

The poset of idempotents is equipped with a relative complement operation: $e \le f$ if and only if $ef = fe = e$, and then we compute that $$(f - e)^2 = f - e$$

hence $f - e$ is also an idempotent. This operation, which I"ll write as $e \Rightarrow f$, satisfies the following axioms:

  • $e \Rightarrow f$ is order-preserving in the second variable and order-reversing in the first,
  • $(0 \Rightarrow f) = f$,
  • $((e \Rightarrow f) \Rightarrow f) = e$,
  • $\inf(e, e \Rightarrow f) = 0$.

In particular, for fixed $f$ it defines an order-reversing isomorphism from the interval $[0, f]$ to itself. Consequently, any poset with a minimal element $0$ and an element $e$ such that the interval $[0, e]$ is not isomorphic to its opposite cannot admit a relative complement and so cannot be the poset of idempotents of a ring. A poset with this property of minimal size is the poset $\{ 0 \lt a \lt b, c \lt 1 \}$.

Another minimal counterexample is the total order $3 = \{ 0 \lt 1 \lt 2 \}$, which admits a unique map $e \Rightarrow f$ satisfying the first, second, and third axioms above, and this map does not satisfy the fourth axiom.


Another thing you can say about the idempotents in a rng is that they are equipped with an interesting relation, namely commutativity. If $e, f$ are commuting idempotents, then $ef$ gives their inf and $e + f - ef$ gives their sup. More generally, any commuting subposet of the idempotents is a lattice.

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