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Questions tagged [rngs]

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

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4 votes
2 answers
121 views

Why do not we exclude the zero ring from ring?

In the definition of field, we exclude $\{0\}$ by assuming $1\ne0$ (i.e. there must be at least two elements in the field). Therefore, it is not a field. However, in the case of ring definition, even ...
Prown's user avatar
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7 votes
1 answer
126 views

Show a ring is commutative if $r^2 = r + r$

The claim is that any ring $R$ in which for all $r \in R$ we have that $rr = r + r$, must be commutative. No assumptions are made about $R$ having multiplicative identity or being commutative. I was ...
Andrew E's user avatar
3 votes
0 answers
66 views

About the notion of invertibility in non unitary rings (or semigroups).

Sorry if this one was already asked, I did not find it. Here is the story: Let us take a unitary ring $R$, the notion of unit element is very clear: $$r \in R^{\times} \iff \exists r' \in R \mid rr' = ...
julio_es_sui_glace's user avatar
1 vote
1 answer
30 views

Matrix construction of Dorroh extension

Let $R$ be a commutative unital ring and $S$ an associative rng (nonunital) that is an $R$-module (= $(R,R)$-bimodule with $\forall s \in S, r \in R$: $rs = sr$). The Dorroh extension is a well-known ...
Amateur_Algebraist's user avatar
0 votes
1 answer
19 views

Generated Ideal Set Representation

Let $R$ be a rng (general ring not requiring existence of identity), and $U\subseteq R$ be a subset. Let $I$ be ideal generated by $U$, which is defined as the intersection of all ideals containing $U$...
spicychicken's user avatar
0 votes
0 answers
31 views

What is the chance of getting same decimals of percentage 4 times in a row with RNG?

I was rolling some random percentage with RNG between 0 and 1; if you multiply the result to 100, you're getting the percentage between 0 and 100, decimal amount was 4. And then I got 0.7171% the ...
ʈɦɘ ʙɑɕʞ's user avatar
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
  • 1,249
2 votes
1 answer
62 views

Finitely generated module over non-unital ring

Let $R$ be a non-unital ring, that is, a ring without a multiplicative identity (or a rng if you prefer). If I want to talk about finitely generated modules over $R$, one factible definition is that a ...
Albert's user avatar
  • 3,052
1 vote
1 answer
216 views

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 ...
stoic-santiago's user avatar
0 votes
0 answers
60 views

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(...
Akash Patalwanshi's user avatar
1 vote
1 answer
52 views

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 ...
guacamowlie's user avatar
0 votes
0 answers
30 views

If ring $R\neq \{0\}$ then subring $R_0 \neq \{0\}$ (proof verification)

I want to check my proof: Given: Let $R$ be a ring with $1$ and $H$ an additive subgroup of $R$. We define $R_0=\{x\in R : \forall h \in H \text{ we have } xh \in H\} \subset R$. To prove: $R\neq \{0\}...
guacamowlie's user avatar
1 vote
2 answers
81 views

$R$ a non-unitary ring such that $R^+ \cong \mathbb{Q}/\mathbb{Z}$

We are given $R$ a non-unitary ring such that $R^+ \cong \mathbb{Q}/\mathbb{Z}$. Prove: $ab=0$ for all $a,b \in R$. Here is what I have attempted so far: We take two arbitrary elements of $\mathbb{Q}/\...
guacamowlie's user avatar
1 vote
1 answer
100 views

How can you prove Nakayama's lemma over nonunital rings using the unitization?

For a nonunital commutative ring $A$, an $A$-module $M$ is called finitely generated over $A$ if there is a finite set of elements $x_1,...,x_n\in M$ such that every element of $M$ is of the form $...
William Sun's user avatar
  • 2,503
2 votes
0 answers
127 views

Can the union of an ascending sequence of ideals of $R$ be $R$?

I am confused about the equivalence of the different definitions of Noetherian Rings. The confusion essentially stems from the fact that a Noetherian ring need not be finitely generated. Essentially ...
Daniel's user avatar
  • 5,589
0 votes
1 answer
92 views

Structural differences between $2 \mathbb{Z}$, $3 \mathbb{Z}$ as rings. [duplicate]

Is usual to find in abstract algebra books this exercise: Show that $2 \mathbb{Z}$ and $3 \mathbb{Z}$ are isomorphic as groups (with usual sum) but they aren't as rings (with usual sum and addition). ...
Samuel Díaz's user avatar
6 votes
2 answers
2k views

In a finite commutative ring , every prime ideal is maximal?

I am stuck in a true/false question. It is In a finite commutative ring, every prime ideal is maximal. The answer says it's false. Well what I can say is (Supposing the answer is right) $(1)$ The ring ...
user-492177's user avatar
  • 2,589
0 votes
2 answers
117 views

Nontrivial subring of $\mathbb{R}$ not containing $1$

Are there examples of nontrivial subrings of $\mathbb{R}$ that do not contain $1$? If not, how can we prove they don't exist? The definition of "ring" here is really "rng"; rings ...
jskattt797's user avatar
  • 1,751
1 vote
1 answer
98 views

Simple example for a rng with an inverse semi-group as the multiplicative group

I’m looking for a ring without an multiplicative identity, and in which every element $x$ has a weak inverse $y$ such that $xyx=x,yxy=y$ preferably simple to construct and or of finite size. If it has ...
razivo's user avatar
  • 2,225
1 vote
1 answer
115 views

Characterization of Injective rings homomorphism from $\mathbb{Z}_m$ to $\mathbb{Z}_n$

Show that there is an injective ring homomorphism $f:\mathbb{Z}_m \rightarrow \mathbb{Z}_n$ if and only if $m\mid n$ and $\frac{n}{m}$ is relatively prime with $m$. In one direction, was not ...
Juan P Ocampo's user avatar
1 vote
2 answers
127 views

Is the ring $3\mathbb Z$ a ring homomorphic image of the ring $2\mathbb Z$.

Is the ring $2\mathbb Z$ isomorphic to the ring $3\mathbb Z$ ? Solution: Let if possible,$\phi:\mathbb {2Z\to 3Z}$ be a ring isomorphism. Then $\phi$ is a group isomorphism between the additive ...
Kishalay Sarkar's user avatar
1 vote
2 answers
101 views

Given prime ideal $P$ of ring $R$, does $RrR\subset P$ imply $r\in P$ for $r\in R$? (Hungerford)

I am trying to understand a problem in Hungerford (p. 134, Exercise 14). Specifically, it says that if $P$ is a prime ideal in a not necessarily commutative ring $R$, and $r,s\in R$ such that $rRs\...
Bart's user avatar
  • 1,148
3 votes
1 answer
1k views

Sum of principal ideals in a commutative rng

Let's define the principal ideal of an element $a$ of a commutative ring $R$ with or without identity as $\langle a \rangle = R \cdot a + \mathbb Za$. It looks like with this definition $\langle a \...
Alex C's user avatar
  • 1,120
1 vote
1 answer
391 views

Maximal non-unit ideal in a ring with or without identity

Every maximal ideal is prime in a commutative ring with identity. There were several posts on the site about analogues of the claim for rngs (rings with or without identity): A maximal ideal is ...
Alex C's user avatar
  • 1,120
0 votes
0 answers
41 views

Product of classes of associates in a commutative ring/rng

Let $[a]$ be the equivalence class of associates of an element $a$ of a ring/rng. Two elements of a ring/rng are associates ($\sim$) if they generate the same principal ideal. $[a] \cdot [b] \sim [...
Alex C's user avatar
  • 1,120
1 vote
0 answers
33 views

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 ...
Alex C's user avatar
  • 1,120
3 votes
1 answer
289 views

In a non-unitary commutative ring, every maximal ideal is primary?

Let $R$ be a commutative ring without identity. My question: is it true or false that every maximal ideal of $R$ is primary? (An ideal I of R is said primary if is proper and $\forall a,b\in R, ab\...
Barry Allen's user avatar
1 vote
0 answers
117 views

Factorization in a principal ideal ring/rng

It is known that every PID is a UFD. Is it true that every element of a commutative principal ideal ring (PIR) or rng that is not zero and not a unit is a product of a finite number of irreducible ...
Alex C's user avatar
  • 1,120
4 votes
1 answer
213 views

Element of a ring acting as a permutation on an ideal

I am investigating cases when $r \cdot I = I$ for some element $r$ and an ideal $I$ of a commutative ring or rng R. Clearly, $r \cdot \langle 0 \rangle = \langle 0 \rangle$ for any element $r$ of $R$,...
Alex C's user avatar
  • 1,120
1 vote
0 answers
934 views

In a commutative rng with comaximal ideals, product equals intersection as well?

In a commutative ring $R$ with unity, the product of every two comaximal ideals equals their intersection, that is, if $I + J = R$, then $I\cap J = IJ$. The proof I know involves the unity of $R$, so ...
horned-sphere's user avatar
1 vote
1 answer
200 views

Example of two subrings with unity of a ring with unity whose intersection is non trivial and has no unity.

I was just thinking about the intersection of rings and this question popped up, I tried giving an example and proving that the intersection had to have an unity, but was unsuccessful in both.
HelloDarkness's user avatar
7 votes
5 answers
2k views

Can a ring have no zero divisors while being non-commutative and having no unity?

I was wondering if, in a ring, the property of having no zero-divisors (except for zero itself) is independent from the ring being commutative or from having a unity (i.e.multiplicative identity) so I ...
UndefinedBehavior's user avatar
2 votes
0 answers
37 views

Prime property in noncommutative rings without identity

Let $R$ be a ring (without assuming identity or commutativity), and $P$ a proper ideal of $R$. Show that the following are equivalent: (a) For ideals $A,B$: $AB\subseteq P$ implies $A\subseteq P$ or $...
George's user avatar
  • 2,586
1 vote
1 answer
71 views

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'...
Geol's user avatar
  • 23
5 votes
1 answer
652 views

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)....
Pennywise's user avatar
  • 461
2 votes
1 answer
232 views

A non-Boolean ring without unity with this property

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{...
Jonathan Hebert's user avatar
3 votes
1 answer
103 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 ...
green frog's user avatar
  • 3,434
3 votes
1 answer
683 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 ...
NonStandardModel's user avatar
0 votes
1 answer
99 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? ...
Santanu Datta's user avatar
4 votes
1 answer
290 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) ...
Josse van Dobben de Bruyn's user avatar
3 votes
2 answers
433 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 ...
Mini_me's user avatar
  • 2,185
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. ...
omega-stable's user avatar
  • 1,235
3 votes
3 answers
1k 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 $r+r+\dots+r$, "$n$ ...
Derso's user avatar
  • 2,773
2 votes
1 answer
42 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 ...
dumb_undergrad's user avatar
3 votes
2 answers
440 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. ...
Henry's user avatar
  • 315
2 votes
2 answers
441 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 \...
user avatar
2 votes
1 answer
2k views

Smallest subring of $\mathbb Q$ containing $3/10$

Let $R$ be the smallest subring of $\mathbb Q$ (the field of rational numbers) that contains $3/10$ ($R$ doesn't have to be a unital ring). Does $1 \in R$? Is the desired smallest subring this one: $...
John's user avatar
  • 327
0 votes
1 answer
95 views

Does $\operatorname{End}_R(R)\cong R^{\text{op}}$ still hold when $R$ doesn't have an identity?

When I try to prove that $\operatorname{End}_R(R):=\text{Hom}_R(R,R)\text{(left)}\cong R^{\text{op}}$, I used the map $f \mapsto f(1)$, which requires $R$ to have an identity. Is it really necessary ...
No One's user avatar
  • 8,039
5 votes
1 answer
753 views

Peirce decomposition of a (not necessarily commutative) ring and the cartesian product

Let $R$ be a (not necessarily) commutative ring and $e \in R$ some idempotent. Then the Peirce decomposition writes $$ R=eRe \oplus (1-e)Re \oplus eR(1-e) \oplus (1-e)R(1-e). $$ I tried to construct ...
StefanH's user avatar
  • 18.2k
0 votes
2 answers
1k views

In the ring $6\mathbb{Z}$ is $12\mathbb{Z}$ maximal ideal but not prime ideal?

I need to prove that in the ring $6\mathbb{Z} = \left\{x \in \mathbb{Z} \mid x = 6q, q \in \mathbb{Z}\right\}$ the subset $12\mathbb{Z}$ is a maximal ideal but not a prime ideal. I first wanted to ...
Kamil's user avatar
  • 5,149