I was studying idempotent ultrafilters when I saw that no principal ultrafilter could ever be idempotent, because $\left\langle n \right\rangle \oplus \left\langle n \right\rangle = \left\langle 2n \right\rangle$. A Ramsey ultrafilter is, by definition, never principal, so the question that arose to me was: do Ramsey idempotent ultrafilters exist? First some explanation: One of the […]

Here’s a variation of a question I was given during a research internship. Some Definitions: Definition 1: Let $S$ be a semigroup. For any $a, b\in S$, define Green’s $\mathcal{L}$-relation by $a\mathcal{L}b$ if and only if $S^1a=S^1b$ and define Green’s $\mathcal{R}$-relation by $a\mathcal{R}b$ if and only if $aS^1=bS^1$, where $S^1$ is $S$ with a one […]

Let $f:R\rightarrow S$ be a ring homomorphism of commutative rings with identities such that $f(1_R)=1_S$. And let $J$ be an ideal of $S$ such that for every prime ideal $P$ of $R$ and every $n\in \mathbb{N}$, $J\subseteq {\langle f(P)+J \rangle}^{n}$, where $\langle f(P)+J \rangle$ is the ideal generated by $f(P)+J$ in the ring $f(R)+J$, as […]

Let $A$ be a ring which may not have a unity. Suppose every element $a$ of $A$ is an idempotent. i.e. $a^2 = a$. It is easily proved that $A$ is commutative. Suppose every ideal of $A$ is finitely generated. Can we determine the structure of $A$?

It is a well-known result that if a ring $R$ satisfies $a^2=a$ for each $a\in R$, then $R$ must be commutative. See here for proof. I am wondering whether the same result holds for finite rings if we only assume sufficiently many (but not necessarily all) elements of the ring are idempotents. Recall that an […]

Let $\cal A$ be the (noncommutative) unitary $\mathbb Z$-algebra defined by three generators $a,b,c$ and four relations $a^2=a,b^2=b,c^2=c,(a+b+c)^2=a+b+c$. Is it true that $ab\neq 0$ in $A$ ? This question is natural in the context of an older question here on MSE. My thoughts : The following two relations follow easily from the axioms : $$ […]

Let $R \neq 0$ be a ring which may not be commutative and may not have an identity. Suppose $R$ satisfies the following conditions. 1) $a^2 = a$ for every element $a$ of $R$. 2) $ab \neq 0$ whenever $a \neq 0$ and $b\neq 0$. Is $R$ isomorphic to the field $\mathbb{Z}/2\mathbb{Z}$?

Given a ring $A$, an element $e \in A$ is called an idempotent if one has $e^2 = e$. If $e$ is an idempotent, then so is $1 – e$, since$$(1 – e)^2 = 1 – 2e + e^2 = 1 – 2e + e = 1 – e.$$Also, we have $e(1 – e) = […]

Let $A$ be a square matrix of order $n$. Prove that if $A^2=A$ then $\mathrm{rank}(A)+\mathrm{rank}(I-A)=n$. I tried to bring the $A$ over to the left hand side and factorise it out, but do not know how to proceed. please help.

An element $a$ of the ring $(P,+,\cdot)$ is called idempotent if $a^2=a$. An idempotent $a$ is called nontrivial if $a \neq 0$ and $a \neq 1$. My question concerns idempotens in rings $\mathbb Z_n$, with addition and multiplication modulo $n$, where $n$ is natural number. Obviously when $n$ is a prime number then there is […]

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