Articles of rngs

Equality in rng with no zero divisors.

I’m working on this problem, but I’m missing some manipulation. Suppose $R$ is a rng without zero divisors and has elements $a$ and $b\neq 0$ such that $ab+kb=0$ for some $k\in\mathbb{N}$ (that is, $ab+\underbrace{b+b+\cdots+b}_{k\ \textrm{times}}=0$.) I’m trying to show that $ca+kc=0=ac+kc$ for all $c\in R$. I observe that $$ (ca+kc)b=cab+kcb=c(ab+kb)=c\cdot 0=0 $$ so $ca+kc=0$ since […]

The structure of a Noetherian ring in which every element is an idempotent.

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 $A$?

What is an element of a rng called which is not the product of any elements?

Let $R$ be a non-unital ring. Let $F:R\times R\longrightarrow R$ be a function given by the formula $F(x,y)=xy.$ Let $r\not\in\operatorname{im}(F).$ Such elements can exists, for example $2\in 2\mathbb Z$ isn’t a product. It seems to be a major difference between unital and non-unital rings. I’m only starting to study non-unital rings and I thought it […]

if $S$ is a ring (possibly without identity) with no proper left ideals, then either $S^2=0$ or $S$ is a division ring.

I’m having trouble with this homework problem (from Algebra by Hungerford). If $S$ is a ring (possibly without identity) with no proper left ideals, then either $S^2=0$ or $S$ is a division ring. We’ve just proven in the previous part that if $S$ is a ring with identity and with no proper left ideals, then […]

Is there any non-monoid ring which has no maximal ideal?

Is there any non-monoid ring which has no maximal ideal? We know that every commutative ring has at least one maximal ideals -from Advanced Algebra 1 when we are study Modules that makes it as a very easy Theorem there. We say a ring $R$ is monoid if it has an multiplicative identity element, that […]

$I+J=R$, where $R$ is a commutative rng, prove that $IJ=I\cap J$.

So I basically have to prove what is on the title. Given $R$ a commutative rng (a ring that might not contain a $1$), with the property that $I+J=R$, (where $I$ and $J$ are ideals) we have to prove that $$IJ=I\cap J$$One inclusion is easy. If $x\in IJ$, then $x=\sum a_ib_i$ where $a_i\in I$ and […]

Does a finite commutative ring necessarily have a unity?

Does a finite commutative ring necessarily have a unity? I ask because of the following theorem given in my lecture notes: Theorem. In a finite commutative ring every non-zero-divisor is a unit. If it had said “finite commutative ring with unity…” there would be no question to ask, I understand that part. What I’m asking […]

Pathologies in “rng”

There is no general consensus regarding whether a ring should have a unity element or not. Many authors work with unital rings , and other does not essentially require unity. If we do not assume unity to be a necessary part of ring, lets call that structure , a “rng” (which may or may not […]

Non-commutative rings without identity

I’m looking for examples (if there are such) of non-commutative rings without multiplicative identity which have the following properties: 1) finite with zero divisors 2) infinite with zero divisors 3) finite without zero divisors 4) infinite without zero divisors I’ll be grateful for any examples and hints. Thanks in advance.

A finite commutative ring with the property that every element can be written as product of two elements is unital

I was struggling for days with this nice problem: Let $A$ be a finite commutative ring such that every element of $A$ can be written as product of two elements of $A$. Show that $A$ has a multiplicative unit element. I need a hint for this problem, thank you very much.