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Questions tagged [idempotents]

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

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The residue class of a complete set of primitive orthogonal idempotents

I was studying the book Elements of the Representation Theory of Associative Algebras: Volume 1 and this question occurred to me: In page 29, it says that Because $\{e_1,…,e_n\}$ is a complete set of ...
Chestnuto's user avatar
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Set of primitive idempotents of group algebras

Let $G$ be a finite group and $K$ be a field with characteristic zero. Can we construct the set of primitive orthogonal idempotents of $KG$? By this set, I mean the set of idempotents such that $...
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On idempotents of group rings and group algebras [closed]

I want to know the name and characterization of idempotent elements in group algebras. Let $G$ be any finite group and $K$ be any field such that its characteristic is either zero or does not divide $|...
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Primitive non-central idempotents of a group algebra

Let $W_i;\ 1 \le i \le m$ be irreducible $\mathbb{C}$-representations of a finite group $G$ with $\mathbb{C}$-characters $\chi_i$. Let $(V,\rho$) be a $\mathbb{C}$-representation of $G$ with isotypic ...
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irreducible representations of Group generate left ideal in Groupring CG.

Let W be the vectorspace of an irreducible representation of a group G.$char_W(g)$ denotes the character of the representation W. With $$e:=dim W \cdot \frac{1}{|G|}\sum\limits_{g \in G} \overline {...
VeryGenericUsername's user avatar
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Decomposition of primitive central idempotents in group algebras

Let $W$ be an irreducible $\mathbb{C}$-representation of a finite group $G$ with character $\chi_W$. A primitive central idempotents of the group algebra $CG$ is: $$e=\frac{\dim_{\mathbb{C}}(W)}{|G|}\...
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When are Idempotents elements of a semisimple algebra primitive

Let $A=KG$ be a $K$-algebra such that $|G| \in K^{\times}$. Here $A$ is a semisimple algebra. Consider the decomposition of $A$ into simple components:$$A=A_1 \times A_2 \times \cdots \times A_k.$$ ...
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Primitive idempotent and bilateral ideals

I'm trying to show for my algebra class that in a semisimple ring with unity $R$ (not necessarily commutative), every primitive idempotent element must belong to a minimal two-sided ideal. Here, by ...
F. Salviati's user avatar
5 votes
1 answer
116 views

Dimension of the center of a block

Let $G$ be a finite group and let $F$ be an algebraically closed field of characteristic $p$. I've been studying some modular character theory from Navarro's "Characters and blocks of finite ...
Gauss's user avatar
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primitive idempotent

I’ve tried to prove: $R$ local, Artin ring $\phi:End(P) \rightarrow End(Q) $ ,$e\in End(P) $ is primivite idempotent and $P$ and $Q$ free module so $P= (\oplus R_i)$ and $Q=(\oplus R_j) $then how ...
wanheda's user avatar
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Von Neumann regular rings and its ideals

Let R be a strongly regular Von Neumann ring, this is, that for every $r \in R$ there exists $x \in R$ such that $r^2x=r$. From here, how to prove that $R$ is strongly Von Neumann regular if and only ...
Apopip's user avatar
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1 answer
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Existence of non-trivial idempotent $e$ in $R/x$

Let $R ≠ 0$ be a principal ideal domain that is not a field. Let $0 ≠ x \in R.$ Show that if $x = yz$ for $y, z \in R \backslash R^\times$ with $\text{gcd}(y, z) = 1$, then there exists an idempotent $...
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Show that two summands of a direct sum are invariant under a linear operator

Given two linear operators $T,S\in L(V)$ such that $T$ and $S$ commute ($T\circ S=S\circ T$) and $T^2=T$. Show that: (a) $V=ker(T)\bigoplus im(T)$. (b) each summand is invariant under $S$. So far I ...
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Can the equality $r = e + u$ hold for all elements $r \in R[x]$, where $R$ - a commutative ring?

Can the equality $r = e + u$ hold for all elements $r \in R[x]$, where $R$ - a commutative ring, $e$ - an idempotent ($e^2 = e$) and $u$ has a multiplicative inverse $u^{-1}$? I think not. My ...
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Complete set of local orthogonal idempotents in $\mathbb{Z}_n$

I want to find a set of local orthogonal idempotents in $\mathbb{Z}_n$, i. e. such idempotents $e_1, \ldots e_n \in \mathbb{Z}_n$ that $1 = e_1 + \ldots + e_n$, $e_i e_j = e_j e_i = 0, (i \ne j)$, and ...
Nickeil's user avatar
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Proof that if idempotent commutes with all nilpotent that it belong to center of ring $Z(R)$

Let $R$ - a associative ring with 1. Let $e \in R$ - an idempotent ($e^2 = e$). I want to proof fact, that if $e$ commutes with all nilpotents of $R$ then $e \in Z(R)$. My ideas. Let $r$ - an ...
Irene's user avatar
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If A is idempotent then A+AB−ABA isidempotent for any square matrix B with the same dimension as A. [closed]

If A is idempotent then A+AB−ABA is idempotent for any square matrix B with the same dimension as A. I have this question to solve and I tried squaring the entire expression and then simplifying it ...
Valentina Tanguy's user avatar
1 vote
2 answers
100 views

If the trace of a matrix equals its rank, is it idempotent?

It is well-known and can easily be proven that if a matrix $A$ is idempotent, then its trace equals its rank: $$ A^2 = A \Rightarrow \mathrm{tr}(A) = \mathrm{rk}(A) $$ Does the inverse also hold? If ...
Joram Soch's user avatar
4 votes
1 answer
121 views

Is a finite semigroup with a unique idempotent $e$ always a group with unit $e$?

Given a finite semigroup $S$ containing a unique idempotent $e$, I can show that every element $s\in S$ has an ''inverse" $s^{-1}\in S$ in the sense that $s s^{-1} = s^{-1}s = e$: Since $S$ is ...
Lemma 5's user avatar
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Is there an idempotent e in a ring $R$ such that the functor $T_e$, induced by $e$, is not full?

Let $R$ be a ring and let $e$ be a non-zero idempotent in $R$. For each $R$-module $M$, define $T_e: M \rightarrow eM$, where $eM$ is the left $eRe$-module. For each pair of $R$-modules $M, N$ and ...
Liang Chen's user avatar
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1 answer
69 views

If R is a ring with idempotent then eR is clean ring if and only if eRe is clean ring

I am reading this paper on clean rings: Clean general rings, by W.K. Nicholson, Y. Zhou, and I am puzzled by the proof of Corollary 11: Corollary 11. If $R$ is a ring and $e^2=e \in R$, then $eR$ is ...
thewizardofmath's user avatar
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1 answer
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How to determine the eigenvalues of a $k$-order idempotent (symmetric) matrix $A$?

I have a following question: Let $A$ satisfying $A^{\prime} = A$, $A^k = A$ and $A^i \neq A$ for $i=2, \cdots, k-1$. In this case, $A$ is referred to as a $k$-order idempotent matrix. How to ...
John Stone's user avatar
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1 answer
49 views

Subgroups in semigroups vs monoids

Let $S$ be a semigroup (i.e. $S$ is endowed with an associative operation). With some work, one can prove that the idempotents of $S$ are in one-to-one correspondence with maximal subgroups of $S$: ...
mathfan24's user avatar
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What is the history of clean rings?

We recall that a ring is called a clean ring if each element is the sum of an idempotent and a unit. Is there someone who can explain to me the history of the emergence of the definition of a clean ...
Belajar Matematika's user avatar
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Does an idempotent function has to be 1-Lipschitz?

A function $f$ satisfying $f(f(x)) = f(x)$ for all $x$ is called an Idempotent function. I want to know if it has to be a Lipschitz function. Actually, I am training a neural network which has to be ...
beluga's user avatar
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0 votes
1 answer
123 views

$\text{rank}(A)+\text{rank}(A-I)=n \iff A^2=A$. I don't know my mistake in my proof

Let $A$ be an $n\times n$ matrix over a field $F$. Show that $A^2=A$ if and only if $\text{rank}(A)+\text{rank}(A-I)=n$. My attempt: $A^2=A \implies \text{rank}(A)+\text{rank}(A-I)=n$ Using the ...
user avatar
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0 answers
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Idempotents in centralizers of non-nilpotent zero divisors

Let $R$ be a (say left-right-Artinian) noncommutative unital ring and $r \in R$ a non-nilpotent zero divisor. Does there necessarily exist a nontrivial idempotent $e$ that commutes with $r$? This ...
Amateur_Algebraist's user avatar
1 vote
1 answer
70 views

Noetherian commutative ring is a product of connected rings

Let $R$ be a noetherian commutative ring. Why is $R\cong R_1 \times \cdots \times R_n$ with $R_i$ connected? I know that one can prove this result using topological arguments on $\mathrm{Spec}(R)$, ...
BillyJohny's user avatar
0 votes
1 answer
48 views

Incorrect proof for $e$ idempotent $\Rightarrow$ $eA$ projective

I have seen several proofs for $e$ idempotent $\Rightarrow$ $eA$ projective where $A$ is an algebra. I tried something different and produced a proof without using the fact that $e$ is idempotent (so ...
kubo's user avatar
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1 vote
1 answer
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Two comaximal radical ideals, with product the nilradical, are radicals of principal ideals generated by complementary idempotents

Let $\mathfrak a,\mathfrak b\subset R$ be two comaximal radical ideals of the ring $R$, with $\mathfrak{ab}=0$. Then $\mathfrak a=Ra$ and $\mathfrak b=Rb$ for complementary idempotents $a,b\in R$. The ...
Jerry Scott's user avatar
2 votes
1 answer
50 views

Is an idempotent logical expression considered to be in conjunctive normal form?

Is $$A \lor B \lor A$$ technically in conjunctive normal form? Or must we apply the idempotent law for it to be in CNF? $$A \lor B \lor A \equiv A \lor B$$?
justanotheruser's user avatar
2 votes
1 answer
70 views

Number of idempotent functions

I'm supposed to show that the number of functions $\ $$\mathrm{f}$: $[n]$ $\to$ $[n]$ $\ $ such that $\mathrm{f\circ f=f}$ is $$1+\sum _{k=1}^{n}{n \choose k}k^{n-k}$$ But I guess that this result ...
J P's user avatar
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2 votes
1 answer
154 views

about $\operatorname{tr}(A) = \operatorname{rank}(A)$ for idempotent matrix $A$

I'm trying to prove the above statement, and I had a look at this site. This ends the proof with the following statement the rank is the number of non-zero eigenvalues But, this is what makes me ...
jason 1's user avatar
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1 vote
1 answer
109 views

Prove $\lVert A \rVert _{p} \geq 1$ for any idempotent matrix $A \neq 0$

As the title states, I'm facing a problem to prove: for any idempotent matrix $A \in \mathbb{C}^{n \times n}$ and $A \neq 0$, $\lVert A \rVert_{p} \geq 1$. Here the p-norm $\lVert A \rVert_{p}$ ...
Yuchun's user avatar
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3 votes
0 answers
65 views

Category of algebraic vector bundles $VB(X)$ over a scheme $X$, which is not idempotent complete

I have recently come across the category of algebraic vector bundles over a scheme $X$. In short, it is the category of locally free $\mathcal{O}_X$-modules of finite rank. An additive category $\...
Divya's user avatar
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0 answers
401 views

Show that a division ring contains exactly two idempotent elements.

My proof: Suppose that $R$ is a division ring. Since $0a=a0=0$, then $0^2 =0$. If $a \neq 0$ and $a^2 = a$, then the inverse $a^{-1}$ of $a$ exists. So, $a^{-1} (a^2)=a^{-1} (a)$. This implies that $a=...
user avatar
-1 votes
1 answer
75 views

Are idempotents always central in a von Neumann regular ring? [closed]

A ring $R$ is called von Neumann regular if for every $a \in R$; $a=axa$ for some $x \in R$. A ring $R$ called Abelian if every idempotent in $R$ is central. Are von Neumann regular rings abelian?
Cary's user avatar
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2 votes
1 answer
70 views

Term for $eSe$ where $S$ is a semigroup and $e \in S$ is an idempotent

For a (possibly non-unital) ring $R$ and an idempotent $e \in R$, $eRe$ is a unital ring with identity $e$ and is known as a corner ring. Now, given any semigroup $S$ and any idempotent $e \in S$, $...
Geoffrey Trang's user avatar
1 vote
0 answers
43 views

Directly irreducible ring and central idempotents

My question is mainly concerned with the characterisation of directly irreducible rings as defined on Wikipedia. We say that a (unital, not necessarily commutative) ring $R$ is directly irreducible, ...
Gargantuar's user avatar
1 vote
1 answer
65 views

The natural partial order on the set of idempotents in the semigroup of Boolean relation matrices.

Let $\mathcal{E}_n$ be the set of idempotents in the semigroup $\mathcal{B}_n$ of $n \times n$ Boolean relation matrices. The relation $E \leq F$ iff $EF=FE=E$ is called the natural partial order on $...
geoffrey's user avatar
1 vote
1 answer
128 views

If $R$ is commutative and $I$ is a finitely generated ideal with $I^2=I$, then there exists an idempotent $e\in I$ with $I=Re$

Question Let $R$ be a commutative ring. Let $I$ be a finitely generated ideal. Assume that $I^2=I$. Show that $I$ is a direct summand of $R$. Answer I know that $I=Re$ for some idempotent $e\in I$ and ...
confused's user avatar
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6 votes
3 answers
316 views

How to prove $\mathbf{1}^\top\mathbf{Q}^+\mathbf{Q}=\mathbf{1}^\top$, where $\mathbf{Q}$ is any element-wise squared correlation matrix?

Let $(X_1,…,X_n)$ be a random vector with $0<\prod_{j=1}^n\text{Var}(X_j)<∞$. Let $\mathbf{Q}=(\mathbf{q}_{1},…,\mathbf{q}_{n})=(ρ_{jk}^2)_{n×n}$, where $ρ_{jk}$ is the Pearson correlation ...
woody's user avatar
  • 87
2 votes
1 answer
80 views

Does there exist an idempotent, pseudo-constant $p$-adic function with an uncountable image?

$f$ is defined to be a pseudo-constant function if $f'(x)=0$. The question simply comes from idly wondering, "What are p-adic idempotent functions like?" Differentiable idempotent functions ...
Merosity's user avatar
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1 vote
0 answers
49 views

Orthogonal idempotents in rings without unity?

Some context In this answer and comments, an argument is given for why an ordered ring with unity can't have nontrivial idempotents. I'm trying to extend the argument for an ordered ring $R$ without ...
Nick F's user avatar
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5 votes
1 answer
113 views

Find all idempotent matrices such that $(A-B)^2 = 0$

Find all idempotent matrices such that $(A-B)^2 = 0$ We can see that the hypotheses imply that $A+B=AB+BA$, and if we multiply by $AB$ on the right, we get $AB+BAB=(AB)^2+BAB$, which also implies ...
Noname's user avatar
  • 585
1 vote
1 answer
251 views

About idempotent matrices

It seems to me that there do not exist two distinct idempotents matrix such that $(A-B)^2=0$. I have not found a counter-example with $2\times 2$ matrix We can see that the hypotheses imply that $A+B=...
Noname's user avatar
  • 585
15 votes
3 answers
619 views

Does every commutative ring have $2^n$ idempotents?

I've spent a lot of time looking for examples, and I can't find any commutative rings which have a finite number of idempotents other than a power of $2$. Intuitively, adjoining an extra idempotent $a$...
Zoe Allen's user avatar
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6 votes
0 answers
34 views

An element commuting with all non-trivial idempotents is central in a simple ring

Suppose that $R$ is a simple ring that has non-trivial idempotents. I try to prove that if an element $a$ commutes with all the idempotents, then $a$ is in the center of the ring. If we define $[x, y] ...
eeerase's user avatar
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0 votes
0 answers
178 views

Prove that for every idempotent element a belonging to $A$ (ie such that $a^2 = a$) and for every $b$ belonging to $A$ then $ab = ba$.

Let $A$ be a ring without nonzero nilpotent elements. Prove that for every idempotent element a belonging to $A$ (ie such that $a^2 = a$) and for every $b$ belonging to $A$ then $ab = ba$. I know that ...
Misimiausis's user avatar
3 votes
0 answers
38 views

A nice characterization of commuting idempotent endomaps?

Let $X$ be a set, and let $f\colon X\to X$ be a map. It is idempotent when for each $x\in X$, $f(f(x))=f(x)$. It is equivalent to the requirement that the restriction of $f$ on its image $f(X)$ is the ...
Amaru's user avatar
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