# Questions tagged [rngs]

A rng is an associative ring without necessarily having a multiplicative identity (rng = ring - $i$dentity).

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### Can $\mathbf{Rng}$ be made into a pre-additive category?

Question: Can $\mathbf{Rng}$ be made into a pre-additive category? Rngs are just rings, without the requirement of an identity. Accordingly, we do not require rng homomorphisms to preserve the ...
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### Formula for two sided principal ideal for a ring without unity

How to express, the two sided principal ideal of a ring without unity(Rng)? How to express, the two sided ideal of a ring $R$ generated by some subset $S\subseteq R$ where $R$ is ring without unity(...
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### Looking for counterexamples (non-domains)

I stumbled upon this question Why doesn't $xa = x$ for all $x \in R$ imply that $a$ is the unit of $R$? and understood the given answers. But it got me thinking about a counterexample, and I was ...
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### Semiring of classes of associates of a ring/rng

Let's call elements that generate the same principal ideal of a ring/rng associates. $[a]$ is the equivalence class of associates of an element $a$ of a ring/rng. Let $A = \{[a],[b], ...\}$ be the ...
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### In a commutative ring $R$ if, for every $a\in R$, the smallest ideal containing $a$ is equal to $Ra$ then $R$ has identity?

Let $R$ be a commutative ring such that, for every $a\in R$, the smallest ideal containing $a$ is equal to $Ra$. Does $R$ have identity? I did this when $R$ has at least one non-zero divisor then it'...
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### Non-unital ring $(2\mathbb{Z})[X]$ is not Noetherian

Let $R = 2\mathbb{Z}$. Then $R[x]$ is not a noetherian ring. I do not understand why this is so, because Hilbert's basis theorem says: If R Noetherian ring, then R[X] a is Noetherian ring (from wiki)....
I'm exploring rings that have the following property: $\rule{10cm}{0.4pt}$ For all $x \in R$, and for all $n \in \mathbb{N}$, there exists $y \in R$ such that $\sum_{i=1}^{n} x = y \cdot x$. $\rule{... 3 votes 1 answer 79 views ###$\mathrm{ann}(R/I) = I$only when$R$is unital? My professor wrote down that$\sqrt{\mathrm{ann}_R(R/I)} = \sqrt{I}$for$I$a proper ideal of$R$. Clearly if$R$is unital ring we have that$\mathrm{ann}_R(R/I) = I$. So the radical relation is ... 3 votes 1 answer 543 views ### Runs-Up/Down RNG test question. There is not much relevant information to be found about Runs-Up/Down test on the great web. All I find is more or less just recycling the info than can be found in Knuth, The Art of Computer ... 0 votes 1 answer 87 views ### Conditions on The Lemma 3.5.1 of Herstein The lemma states that "Let R be a commutative ring with unit element whose only ideals are (0) and R itself. Then R is a field." My question is that what if we don't assume the ring has unit element? ... 4 votes 1 answer 197 views ### Definition: rng ideal versus algebra ideal in a non-unital$C^*$-algebra When I first moved into operator algebras, equipped with the basics of abstract algebra, I encountered the following definition of ideals: Definition (algebra ideals). A left (respectively, right) ... 3 votes 2 answers 332 views ### Every commutative ring of characteristic$p$contains$\mathbb F_p$as a subring? I know that if a commutative ring with unity is of characteristic$p$then it will contain$\mathbb F_p$as a subring, but if the ring is commutative with characteristic$p$and without unity then is ... 4 votes 1 answer 2k views ### Ideal of direct sum of rings. It is known that if$R_1$and$R_2$are rings with unity, then every ideal of$R_1 \oplus R_2$has the form$I_1 \oplus I_2$, where$I_1$and$I_2$are ideals of$R_1$and$R_2$respectively. ... 3 votes 3 answers 856 views ###$R$a ring with 1 and charac.$n>1$(resp. 0)$\implies R$contains a subring$\simeq \Bbb{Z}_n$(resp.$\Bbb{Z}$). What happens if$R$has no 1?! Definition: Let$R$be a ring. We say that$n\in \Bbb{Z}_+$is the characteristic of$R$if it is the least positive integer such that$n r=0$, for all$r\in R$(here$nr$denotes$r+r+\dots+r$, "$n$... 2 votes 1 answer 39 views ### If an ordered rng$A$has a upper or lower bound, then$A = {0}$Let$m$be the upper bound of$A$. Then,$a \leq m$for all$a\in A $. Now I should maybe find a way to show that that inequality is in fact an equality. Or maybe there's an easier way that I don't ... 3 votes 2 answers 384 views ###$R$is a prime right Goldie ring which contains a minimal right ideal. Show that$R$must be a simple Artinian ring.$R$(1 is not assumed to be in$R$) is a prime right Goldie ring (finite dimensional and ACC on right annihilators) which contains a minimal right ideal. Show that$R$must be a simple Artinian ring. ... 2 votes 2 answers 383 views ### Product of Principal Ideals when$R$is commutative, but not necessarily unital When$R$is a ring (not necessarily commutative, and not necessarily with unity), I have a result that tells me that for$x \in R$, the ideal generated by$x$,$(x) $, is$= I_{x} = \langle RxR \... 