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In A First Course in Abstract Algebra, Fraleigh writes on p. 172 that "a ring in which multiplication is commutative is a commutative ring". Of course, this raises the question: is addition necessarily commutative in a commutative ring? Is it commutative in any ring? Are there examples of rings in which addition is non-commutative?

As far as I could tell, commutativity of the underlying additive operation is not one of the defining properties of a ring. Or is it? I don't think the book has been entirely clear on these issues --- I would be thankful if someone could shed some light on this.

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The definition of a ring $R$ is that it has two binary operations, $+$ which is an abelian group and $\cdot$ which is not necessarily commutative (and need not even be a group, merely associative). Of course, you have your extra distributive properties added as well. – mathematics2x2life Mar 7 '14 at 4:06
Possible duplicate of Why is ring addition commutative? – Najib Idrissi Nov 23 at 18:59

3 Answers 3

up vote 11 down vote accepted

The three basic properties of a ring are;

  • The set under addition makes an abelian group,
  • Multiplication is associative, and
  • Left and right distributive laws hold.

Thus, by definition of "abelian group", the addition must be commutative. Hopes that help.

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Oh. I totally glossed over the fact that the underlying group under addition is abelian. – Newb Mar 7 '14 at 4:11
@Newb A ring with not necessarily commutative addition is sometimes called a near-ring. I don't know a whole lot of study about them (but I'm generally ignorant of that sort of thing). – Mike Miller Mar 7 '14 at 4:25

In order to generalize rings to structures with noncommutative addiiton, one cannot simply delete the axiom that addition is commutative, since, in fact, other (standard) ring axioms force addition to be commutative (Hankel, 1867 [1]). The proof is simple: apply both the left and right distributive law in different order to the term $\rm\:(1\!+\!1)(x\!+\!y),\:$ viz.

$$\rm (1\!+\!1)(x\!+\!y) = \bigg\lbrace \begin{eqnarray} (1\!+\!1)x\!+\!(1\!+\!1)y\, =\, x \,+\, \color{#C00}{x\!+\!y} \,+\, y\\ \rm 1(x\!+\!y)\!+1(x\!+\!y)\, =\, x\, +\, \color{#0A0}{y\!+\!x}\, +\, y\end{eqnarray}\bigg\rbrace\:\Rightarrow\: \color{#C00}{x\!+\!y}\,=\,\color{#0A0}{y\!+\!x}\ \ by\ \ cancel\ \ x,y$$

Thus commutativity of addition, $\rm\:x+y = y+x,\:$ is implied by these axioms:

$(1)\ \ *\,$ distributes over $\rm\,+\!:\ \ x(y+z)\, =\, xy+xz,\ \ (y+z)x\, =\, yx+zx$

$(2)\ \, +\,$ is cancellative: $\rm\ \ x+y\, =\, x+z\:\Rightarrow\: y=z,\ \ y+x\, =\, z+x\:\Rightarrow\: y=z$

$(3)\ \, +\,$ is associative: $\rm\ \ (x+y)+z\, =\, x+(y+z)$

$(4)\ \ *\,$ has a neutral element $\rm\,1\!:\ \ 1x = x$

Said more structurally, recall that a SemiRing is that generalization of a Ring whose additive structure is relaxed from a commutative Group to merely a SemiGroup, i.e. here the only hypothesis on addition is that it be associative (so in SemiRings, unlike Rings, addition need not be commutative, nor need every element $\rm\,x\,$ have an additive inverse $\rm\,-x).\,$ Now the above result may be stated as follows: a semiring with $\,1\,$ and cancellative addition has commutative addition. Such semirings are simply subsemirings of rings (as is $\rm\:\Bbb N \subset \Bbb Z)\,$ because any commutative cancellative semigroup embeds canonically into a commutative group, its group of differences (in precisely the same way $\rm\,\Bbb Z\,$ is constructed from $\rm\,\Bbb N,\,$ i.e. the additive version of the fraction field construction).

Examples of SemiRings include: $\rm\,\Bbb N;\,$ initial segments of cardinals; distributive lattices (e.g. subsets of a powerset with operations $\cup$ and $\cap$; $\rm\,\Bbb R\,$ with + being min or max, and $*$ being addition; semigroup semirings (e.g. formal power series); formal languages with union, concat; etc. For a nice survey of SemiRings and SemiFields see [2]. See also Near-Rings.

[1] Gerhard Betsch. On the beginnings and development of near-ring theory. pp. 1-11 in:
Near-rings and near-fields. Proceedings of the conference held in Fredericton, New Brunswick, July 18-24, 1993. Edited by Yuen Fong, Howard E. Bell, Wen-Fong Ke, Gordon Mason and Gunter Pilz. Mathematics and its Applications, 336. Kluwer Academic Publishers Group, Dordrecht, 1995. x+278 pp. ISBN: 0-7923-3635-6 Zbl review

[2] Hebisch, Udo; Weinert, Hanns Joachim. Semirings and semifields. $\ $ pp. 425-462 in: Handbook of algebra. Vol. 1. Edited by M. Hazewinkel. North-Holland Publishing Co., Amsterdam, 1996. xx+915 pp. ISBN: 0-444-82212-7 Zbl review, AMS review

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I would upvote this answer multiple times if I could. As usual, Bill, your answer is highly informative. – Newb Mar 7 '14 at 7:50

The definition of a ring is that it has two binary operations, $+$ and $\cdot$. The $+$ operation forms an abelian group and $\cdot$ need only be associative. The distributive laws need hold. However, notice that the distributive laws force $+$ to be abelian when $R$ has $1$!



$$ (1+1)(x+y)=(1+1)x+(1+1)y=x+x+y+y $$ so that $x+y+x+y=x+x+y+y$ then adding $-x$ and $-y$ ($+$ forms a group so has inverses $-x,-y$) on the left and right yields $y+x=x+y$.

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A lot of people don't accept a ring without unity as a ring. – Tim Seguine Mar 7 '14 at 11:18
But they do indeed exist and can be useful, thought they do not appear often. So it is important to know they rings are not required to have such properties. Then one can happily continue working with rings with identity. – mathematics2x2life Mar 7 '14 at 21:16

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