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# ring

A *ring* is a set $R$ together with two binary operations, denoted $+:R\times R\longrightarrow R$ and $\cdot:R\times R\longrightarrow R$, such that

1. $(a+b)+c=a+(b+c)$ and $(a\cdot b)\cdot c=a\cdot(b\cdot c)$ for all $a,b,c\in R$ (associative law)

2. $a+b=b+a$ for all $a,b\in R$ (commutative law)

3. 4. For all $a\in R$, there exists $b\in R$ such that $a+b=0$ (additive inverse)

5. $a\cdot(b+c)=(a\cdot b)+(a\cdot c)$ and $(a+b)\cdot c=(a\cdot c)+(b\cdot c)$ for all $a,b,c\in R$ (distributive law)

Equivalently, a ring is an abelian group $(R,+)$ together with a second binary operation $\cdot$ such that $\cdot$ is associative and distributes over $+$. Additive inverses are unique, and one can define *subtraction* in any ring using the formula $a-b:=a+(-b)$ where $-b$ is the additive inverse of $b$.

We say $R$ has a *multiplicative identity* if there exists an element $1\in R$ such that $a\cdot 1=1\cdot a=a$ for all $a\in R$. Alternatively, one may say that $R$ is a *ring with unity*, a *unital ring*, or a *unitary ring*. Oftentimes an author will adopt the convention that all rings have a multiplicative identity. If $R$ does have a multiplicative identity, then a *multiplicative inverse* of an element $a\in R$ is an element $b\in R$ such that $a\cdot b=b\cdot a=1$. An element of $R$ that has a multiplicative inverse is called a *unit* of $R$.

A ring $R$ is *commutative* if $a\cdot b=b\cdot a$ for all $a,b\in R$.

## Mathematics Subject Classification

16-00*no label found*20-00

*no label found*13-00

*no label found*81P10

*no label found*81P05

*no label found*81P99

*no label found*

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## Attached Articles

examples of rings by matte

Klein 4-ring by pahio

left and right unity of ring by rspuzio

additive inverse of one element times another element is the additive inverse of their product by cvalente

additive inverse of a sum in a ring by rspuzio

additive inverse of an inverse element by Mathprof

ring hierarchy by Algeboy

inverses in rings by Wkbj79

hollow matrix rings by Algeboy

## Comments

## commutative

if it were me I'd get rid of the sentence about commutative rings. I'd probably also express it in terms of additive abelian groups and multiplicative semigroups or monoids, so maybe you shouldn't listen to me.

## commutative

I think he's planning on still having a separate "commutative ring" entry...

-apk

## group?

Can't a ring (R,+,*) be described as a commutative group (R,+) and a * (closed) operation?

## Re: group?

yes, if you also ask

for * being associative

and add a distributive law that links + with *

f

G -----> H G

p \ /_ ----- ~ f(G)

\ / f ker f

G/ker f

## Re: group?

This definition is not adequate because we are missing the very important property of distribution. Distribution ensures other structural features of the ring, specifically with the zero.

## Ring definition ( by djao )

In Dummit & Foote it is proved that the following definition of a ring "A ring (R,+, Ã‚Â·) is a group (R, +) with an additional associative multiplication Ã‚Â· with identity, which distributes over the addition." implies commutativity in the additive group. Yet authors persist to include commutativity as an axiom in Ring Theory.

Any comments?

## Re: Ring definition ( by djao )

The requirement that a ring have a multiplicative identity is not, in general, a standard ring axiom. I suppose the reason that authors explicitly list the requirement that $(R,+)$ be abelian is that it may not be immediately obvious that commutativity of $+$ is implied by the left and right distributive laws. It wasn't to me anyway :)

Keenan

## Re: Ring definition ( by djao )

It's really just a matter of taste. Commutativity of addition isn't just a nice property that unital rings happen to have, it's part of what being a ring is all about. We want it even for non-unital rings (where it isn't automatic). So it's natural to include it in the definition.

Note that djao's definition doesn't include a multiplicative identity, so he has no choice but to specify commutativity of addition explicitly. So your question appears to have no relation to the title of your post.

## Re: Ring definition ( by djao )

You are right about the relation with djao's post. ( I am a bit tired. ) - Maybe it's all caused by the 1, do we include a 1 or not? I noticed there are a lot of competing definitions. A matter of taste, as you say.

Thanks for your reply.

## Re: Ring definition ( by djao )

Could you add two entries:

1. A proof that multiplicative identity and distributive

law imply commutativity of addition.

2. An non-trivial example of an algebraic system where

multiplication distributes over addition but addition is not

commutative. (Any noncommutive group is a trivial example

if we define the product of any two elements to be zero.)

## Re: Ring definition ( by djao )

Ok, after work.

## Re: Ring definition ( by djao )

rspuzio wrote:

> 1. A proof that multiplicative identity and distributive

> law imply commutativity of addition.

By the way, this is exercise 2.1.3 in Jacobson's

_Basic Algebra I_.

> 2. An non-trivial example of an algebraic system where

> multiplication distributes over addition but addition is not

> commutative. (Any noncommutive group is a trivial example

> if we define the product of any two elements to be zero.)

Jacobson gives a construction in section 2.17 to embed a

``rng'' (what I would call a ring without unity) into a ring

with unity. Following his idea, let (R, +, .) be a tuple

for which (R, +) is a group, (R, .) is a semigroup, and .

distributes over + on the left *and* on the right. Define a

tuple (S, +, .) by S = Z \times R, where (S, +) is the

direct product of (Z, +) and (R, +), and the semigroup

structure (S, .) is given by the multiplication

(m, a)(n, b) = (mn, mb + na + ab).

It is not difficult to verify that (S, .) is a monoid with

unit (1, 0) and that . distributes over +. By exercise

2.1.3, it follows that (S, +) is abelian. Moreover, there

is an *injective* homomorphism f : R -> S defined by

f(a) = (0, a), so we can pull the abelian property back to R.

(I would *really* appreciate it if anyone could detect and

point out errors in the above reasoning.)

This really goes back to Taussky-Todd's 1936 paper ``Rings

with non-commutative addition''. Taussky-Todd proves that

in any distributive near-ring (N, +, .), elements of

NN = { ab : a, b in N } commute additively with each other.

As a consequence, if N is a distributive near-ring and

NN = N, then N is actually a ring.

To get a nontrivial example where commutativity fails, you

may have to be satisfied with permitting distributivity only

on *one* side. You might look into endomorphism near-rings.

I found the books _Near-Rings_ (by Pilz) and _Near-rings and

their links with groups_ (by Meldrum) helpful in composing

this post, but I don't know enough about this area to give

you a concrete example I know is correct.

The person who contributed PM's entry on near-rings,

J\"urgen Ecker, is a current researcher in the field and is

one of the author's of GAP's SONATA package for working with

near-rings, but he no longer logs in to PM.

## Re: Ring definition ( by djao )

> author's

This is why PM needs post-editing capabilities.

## Re: Ring definition ( by djao )

>

> It is not difficult to verify that (S, .) is a monoid with

> unit (1, 0) and that . distributes over +. By exercise

> 2.1.3, it follows that (S, +) is abelian. Moreover, there

> is an *injective* homomorphism f : R -> S defined by

> f(a) = (0, a), so we can pull the abelian property back to

> R.

>

> (I would *really* appreciate it if anyone could detect and

> point out errors in the above reasoning.)

>

What errors are you speaking of?

## Re: Ring definition ( by djao )

> What errors are you speaking of?

I meant ``Which claim did I make that was incorrect?''. (We

know that there must be an error somewhere, since it directly

contradict's rspuzio's example: *any* group (G, +), even a

nonabelian one, satisfies the assumptions I gave if (G, .)

has the trivial product.)

As it turns out, the incorrect claim is ``(S, .) is a monoid''.

It is easy to show that with the defined multiplication,

(S, .) is a unital magma. However, proving that . is

associative requires that (S, +) be abelian. In my

attempted proof (not posted) I implicitly assumed (S, +)

was abelian even though it wasn't supposed to be.

A lesson is that one should not rely too much on notation.

## Re: Ring definition ( by djao )

Or, maybe that ab=0 for all a, b will render S=0, if S contains a multiplicative unity 1, since a=1.a=0.