# Questions tagged [idempotents]

For questions about elements which satisfy $x\cdot x=x$ where $\cdot$ is a composition law.

366 questions
Filter by
Sorted by
Tagged with
35 views

### $S$ and $T$ are idempotent linear operators in $V$ which is a vector space in $\mathbb{C}$. Prove that if $S+T$ is idempotent then $ST=TS=0$.

If $S+T$ is idempotent, then $$(S+T)^2=S+T$$ $$\implies S^2+ST+TS+T^2=S+T$$ $$\implies S+T+ST+TS=S+T$$ $$\implies ST+TS=0$$ Now, from here how do I show that $ST=TS=0$?
66 views

30 views

### Zero Ideal is Indecomposable as a Finite Intersection of Primary Ideals on an Infinite Ring of Idempotents

Let R be a ring of idempotents, that is, a ring in which holds $x^2=x, \forall x \in R$. Prove that if R is infinite, then (0) is indecomposable as a finite intersection of primary ideals. Any hints ...
43 views

### Idempotents and Direct Sums with Modules

I'm attempting to prove the following statement: "Let $R$ be a ring, let $M$ be an $R$-module, and let $\phi:M \to M$ be an $R$-module homomorphism. Prove that if $\phi$ is an idempotent in the ...
150 views

### Karoubi envelope / idempotent completion of $R-Mod$

I have a question about motivation for building a Karoubi envelope or idempotent completion of a category $C$. A problem in a non-complete category is that it probably contains idempotent elements (...
32 views

### Monoid that is idempotent induces partial ordering

Given a commutative monoid $(M,0,\oplus)$. Then we can define an ordering on $M$ by $a\geq b :\Leftrightarrow \exists c: a=b\oplus c$. The relation is then transitive and reflexive. The claim is now ...
191 views

### When is the composition of two idempotent functions itself idempotent?

If two idempotent functions f and g are individually idempotent, under what conditions is g ∘ f idempotent? I know from this question that it is sufficient for f and g to be commutative: If the ...
Here's what I initially started with: Find a 2x2 non zero matrix $A$, satisfying $A^2=A$, and $A\neq I$. I understand that this is fairly easy, but please keep reading for something ...