# Why is ring addition commutative?

What is the motivation behind axiomatically forcing the underpinning group of a ring to be abelian? Noncommutative rings are vastly more complex than commutative ones, so I am assuming that allowing the additive operation to be noncommuting would just make matters worse. Is there something deeper here, or is this a restriction for the sake of convenience and simplicity?

• I believe I've heard that in any ring with unity, commutativity follows from the other axioms. Fact-checking now... :) Here's a source: math.stackexchange.com/questions/346375/… . Though since it's on MSE maybe this is a duplicate :( Dec 16, 2013 at 17:54
• Possibly useful: near-rings. Dec 16, 2013 at 17:56
• If ring addition weren't commutative, then the distributive property wouldn't hold in general. Dec 16, 2013 at 17:58
• A definition may come about by induction from examples. Rings of polynomials, rings of matrices, etc. Dec 16, 2013 at 17:58
• All useful definitions . Dec 16, 2013 at 18:01

Yes, there is a deeper reason, at least in my opinion. Most abstract axiomatic constructions in mathematics are inspired by or even equivalent to concrete examples. One of the most basic algebraic objects one can think of is the set $S^S$ of mappings from a set $S$ into itself, which has a natural associative multiplication (composition) and a unit (the identity map). Taking these properties as abstract axioms, one obtains the definition of a unital monoid. However, an element $m\in M$ of an abstractly defined unital monoid $M$ can be viewed as the map $[l_m:M\rightarrow M]\in M^M$ given by $l_m(x)=mx$. Mapping $m\in M$ to $l_m\in M^M$ induces an injective homomorphism of monoids so the abstract definition is really just language (albeit quite helpful) since $M$ is isomorphic to a submonoid of a mapping monoid. One can also consider submonoids of $S^S$ which contain only bijective maps. Again, taking this additional property (i.e. existence of inverses) as another axiom, one obtains the definition of a group - but every group can be viewed as a subgroup of a group of bijective maps on a set, so the abstract definition is really just language since it does not provide you with anything that truly generalizes the guiding example.

Now for unital rings the story is essentially the same, but whereas for unital monoids the basic model is $S^S$ and its submonoids, the basic model for rings is $\operatorname{End}(G)$ and its subrings, where $G$ is a abelian group. $\operatorname{End}(G)$ is the submonoid of the mapping monoid $G^G$ which contains only those maps which are group homomorphisms. The restriction to abelian groups is necessary since the pointwise product $(\varphi,\psi)\mapsto [x\mapsto \varphi(x)\psi(x)]$ (which is the addition in the ring $\operatorname{End}(G)$) of endomorphisms is another endomorphism only if $G$ is assumed to be abelian, in general. As with unital monoids and groups, an abstract unital ring $R$ embeds homomorphically and injectively into its own endomorphism ring $\operatorname{End}(R)$ by multiplication operators, and this is a homomorphism of unital rings (not just of unital monoids).

Thus, generalizing the definition of a ring in a manner which is suggested by the question is tantamount to generalizing the example $\operatorname{End}(G)$ to nonabelian $G$. As for how this could be done, there are two possibilities (actually more... see below in the edit). First, one could attempt to define a group structure on $\operatorname{End}(G)$ other than the pointwise product. This seems somewhat unnatural since the whole point of using $\operatorname{End}(G)$ is to exploit the presence of the group product in the first place. Still, one might speculate that for certain nonabelian groups there are structures on $\operatorname{End}(G)$ which are somehow derived from the group product and interact well with the composition - such a structure would be quite bizarre since at the very least, the distributive property will be lost, as explained in the other answers. The other avenue of generalization is to continue using the pointwise product, which as you may have noticed makes the entire mapping monoid $G^G$ into a group. Thus, it is possible to consider subsets of $G^G$ which are closed under both operations: pointwise product and composition. This is the guiding example of a "near ring" as described on Wikipedia. In general, a submonoid $M\subset G^G$ such that $M\subset \operatorname{End}(G)\subset G^G$ can be one of these only when $G$ is abelian, and these examples are the traditionally defined rings.

EDIT 12/23/2013 Earlier today I realized that there is an easy way to create an example of a "ring with nonabelian underlying group": just take your favorite nonabelian group $G$ written with $+$ (which is horrible, I know) and impose a second law of composition $\ast$ given by $(x,y)\mapsto x\ast y=e_G$. It is easily verified that $\ast$ is associative and distributes over $+$ so $(G,+,\ast)$ satisfies all ring axioms with the exception of the commutativity of $(G,+)$.

Now there is no conflict with the other answers here because the weird ring I've just described has no multiplicative identity. Thus, I have tried to insert the word "unital" = "possessing a two-sided identity" in various appropriate places in the above text in order to emphasize the assumption that an identity exists. Now if one does not assume the existence of an identity things can get complicated rather quickly. For instance, an arbitrary set $M$ can be made into a non-unital monoid by choosing a point $x\in M$ and imposing the law of composition $(m,n)\mapsto x$ for every pair $(m,n)\in M\times M$. This is clearly an associative law of composition, but there is no identity and, perhaps more importantly, the canonical homomorphism $m\in M\mapsto l_m\in M^M$ as defined above has only one point in its image (the point map onto $x$).

The point is that if $M$ does not contain an identity then $m\mapsto l_m$ is not necessarily injective so one does not necessarily see an exact copy of $M$ inside of $M^M$ - only a quotient. This means that the abstract definition of a (not necessarily unital) monoid can produce examples which are not canonically equivalent to the guiding example of $S^S$ and its submonoids, so in this case the definition is not "just language", as I wrote above.

There is a sufficient condition for the injectivity of $m\mapsto l_m$ which is more general than the existence of an identity element: if the right anti-representation $x\mapsto r_x\in M^M$ of $M$ is defined as usual ($r_x(m)=mx$) then $m\mapsto l_m$ will be injective provided that there is at least one $x\in M$ such that $r_x\in M^M$ is an injective map, for then $l_m=l_n$ implies $$r_x(m)=l_m(x)=l_n(x)=r_x(n)$$ and therefore $m=n$ by the injectivity of $r_x$. I have a feeling that this is a manifestation of some more general phenomenon involving monomorphisms in the category of sets but I don't know enough about category theory to discuss this in any detail (maybe @Martin Brandenburg would like to leave a comment). In particular, if $M$ contains a two-sided identity $1$ then $r_1\in M^M$ is injective so $m\mapsto l_m$ injective.

At the level of rings, this means that in an abstractly defined ring, not necessarily with identity, the presence of at least one $x\in R$ such that $r_x\in \operatorname{End}(R)$ is injective forces the canonical representation $m\mapsto l_m$ to be injective and therefore to faithfully reproduce $R$ inside of $\operatorname{End}(R)$.

Now alot has been said on this thread about how the distributive property in a ring forces the underlying group to be abelian and the computation presented by Bill Dubuque and drhab uses the existence of a ring identity to show this. In fact, this can be proved assuming only that the canonical representation is injective:

Proposition. If $(R,+,\ast)$ satisfies all the ring axioms except the commutativity of the group $(R,+)$, then the canonical representation $x\mapsto [z\mapsto l_x(z)=x\ast z]$ is a homomorphism of both products into $R^R$ which takes values in $\operatorname{End}(R)$ and if this homomorphism is injective then $(R,+)$ is abelian.

Remark. Since the proposition requires two-sided distributivity of $\ast$ over $+$, the hypotheses are somewhat stronger than simply stating that $(R,+,\ast)$ is a near ring.

Proof.

1."$x\mapsto l_x$ takes values in $\operatorname{End}(G)$" uses distributivity from the left: $$l_x(y+ z)=x\ast (y+ z)=(x\ast y) + (x\ast z)=l_x(y)+ l_x(z).$$

2."$x\mapsto l_x$ is a homomorphism $\ast\rightarrow \circ$" uses associativity of $\ast$: $$l_{x\ast y}(z)=(x\ast y)\ast z=x\ast (y\ast z)=l_x\circ l_y(z).$$

3."$x\mapsto l_x$ is a homomorphism $+\rightarrow +$" uses distributivity from the right: $$l_{x+ y}(z)=(x+ y)\ast z=(x\ast z)+ (y\ast z)=l_x(z)+ l_y(z)=(l_x+l_y)(z).$$

4.(R,+) is abelian if $x\mapsto l_x$ is injective: $$l_{x+y}(z+z)=l_x(z+z)+l_y(z+z)=l_x(z)+l_x(z)+l_y(z)+l_y(z) =l_x(z)+l_{x+y}(z)+l_y(z)$$ but also $$l_{x+y}(z+z)=l_{x+y}(z)+l_{x+y}(z)=l_x(z)+l_y(z)+l_x(z)+l_y(z)=l_x(z)+l_{y+x}(z)+l_y(z).$$ Canceling the outer terms (which is valid since $(R,+)$ is assumed to be a group), we have $l_{x+y}(z)=l_{y+x}(z)$. This being true for all $z$, we conclude that $x+y=y+x$ provided that $x\mapsto l_x$ is injective. The proposition is proved.

What if we drop the assumption that the canonical representation is injective? Then we can produce examples of "rings with nonabelian underlying groups" as I did at the beginning of the edit - to reiterate: just take your favorite nonabelian group $G$ written with $+$ (which is atrocious, I know) and impose a second law of composition $\ast$ given by $(x,y)\mapsto x\ast y=e_G$. It is easily verified that $\ast$ is associative and distributes over $+$ so $(G,+,\ast)$ satisfies all ring axioms with the exception of the commutativity of $(G,+)$. The proposition shows that the canonical representation cannot be injective and sure enough, it takes values only on a single point: the trivial endomorphism $[x\mapsto e_G]\in \operatorname{End}(G)$.

To sum things up I would say that

1. Yes, the ring axioms can be relaxed to produce "rings with nonabelian underlying groups", however some nice property will have to be sacrificed: either

• multiplication will not distribute from the left, which can give you a near-ring but then the image of the canonical representation will not lie in $\operatorname{End}(R)$, in general; or

• the canonical representation will not be injective in which case you cannot adjoin an identity without severely disrupting the given algebraic structure (I assume that you will have to pass to some abelian quotient of the underlying group).

2. In either case (and especially in the second case) these objects will not interact well with $\mathbb{Z}$, and in my estimation that is why they are mostly curiosities rather than the object of intense study.

• +1 Welcome to MSE. Dec 17, 2013 at 2:31
• The basic point of this answer is not to generalize purely for the sake of generalization. All new math is an extension of slightly older math, which is to say, all generalizations are guided by example lest they be vacuous. As an example: the legend of the PhD candidate and the anti-metric space (singular!) Dec 23, 2013 at 15:55
• @Ryan Reich: Could you elaborate which "legend" you mean? Dec 23, 2013 at 16:03
• Found it by a quick googling, e.g., here and here (page 50). An anti-metric space satisfies the axioms of a metric space, except the triangle inequality is replaced with $d(x, y)+d(y,z) \le d(x,z)$. In fact, there aren't very many anti-metric spaces. Dec 23, 2013 at 16:18
• @Martin It's in the urban legends question on MO, with slightly different details. Basically, student defines said spaces without giving examples and is hoist on his own petard at defense when famous mathematician observes there is only the one-point space. Dec 23, 2013 at 16:44

In order to generalize rings to structures with noncommutative addition, 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.

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

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

• In fact, (4) can even be weakened to the existence of a non-zero-divisor (either of left or right). Jun 16, 2016 at 1:45
• This answer is weirdly similar (including spelling errors) to this one: math.stackexchange.com/a/346682/101420. Are the authors the same person under different names? Jun 16, 2016 at 8:12
• @Vincent They are both excerpted from our lecture notes and/or sci.math posts, etc. Jun 16, 2016 at 12:35
• This means that a ring without a commutative identity doesn't necessarily have to have commutative addition right? Feb 23, 2019 at 21:23
• @grenmester It means you have to drop at least one of the listed $4$ laws. Feb 23, 2019 at 21:38

$\left(1+1\right)\left(a+b\right)=1\left(a+b\right)+1\left(a+b\right)=a+b+a+b$

$\left(1+1\right)\left(a+b\right)=\left(1+1\right)a+\left(1+1\right)b=a+a+b+b$

So distributivity (on both sides) demands that $a+b=b+a$

• (This is precisely the argument given in the link I mentioned; there a detailed explanation of what axioms are used is also given) Dec 16, 2013 at 18:02

Most rings known to man have their addition operation commutative.

The definition of ring tries to capture that.

• (The fact that this follows form the other axioms is very minor and unimportant) Dec 16, 2013 at 17:55
• (See here) Dec 16, 2013 at 17:57
• Dear Mariano, Or here! Cheers, Dec 17, 2013 at 2:23
• @Mariano: This is really the best possible answer. It seems to me that nowadays abstract notions are not introduced via a couple of important and motivating examples, which is very regrettable. Instead, students learn them and play around them, instead of really DOING something with them. Dec 23, 2013 at 15:48
• @MartinBrandenburg, I couldnot agree more. The periodic downvotes on this answer reflect that situation :-) Mar 9, 2014 at 20:55

There is another answer from a category-theoretic perspective. Recall the notion of a monoid object in a monoidal category. It is natural to study them, they appear in many situations, and many "theorems" about monoids, rings, topological rings or algebra in general etc. are actually special cases of abstract nonsense with monoid objects in (nice) monoidal categories.

• Monoids = monoid objects in $(\mathsf{Set},\times)$
• Monoids with zero = monoid objects in ($\mathsf{Set}_*,\wedge)$
• $H$-spaces = monoid objects in $(\mathsf{hTop}_*,\wedge)$
• Semirings = monoid objects in $(\mathsf{CMon},\otimes)$
• Rings = monoid objects in $(\mathsf{Ab},\otimes)$
• Rings with noncommutative addition = monoid objects in ... ???

Well, it is tempting to take $(\mathsf{Grp},\otimes)$ here, but what should $\otimes$ be here? Although there are various variants of tensor products of groups, none of them makes $\mathsf{Grp}$ a monoidal category.

Therefore, rings with a noncommutative addition fall out of the general picture. This doesn't necessarily imply that they are uninteresting, but rather that their theory is more exotic.

As already mentioned in the other answers, a near-ring is a "ring" with a noncommutative addition and only the one-sided distributive law $(x+y)z=xz+yz$. These may be interpreted as groups $G$ equipped with an associative map $G \otimes G \to G$, where $G \otimes G$ is defined to be the free group generated by symbols $x \otimes y$ subject to the (additively written) relations $(x+y) \otimes z = x \otimes z + y \otimes z$. This seems to be just the coproduct (aka free product) of $|G|$ copies of $G$, where $x \otimes z$ belongs to the copy indexed by $z$. A similar definition should work for arbitrary algebraic categories.

Its not just for convenience and simplicity; its actually totally fundamental. The problem with replacing abelian groups with general groups (or general monoids, for that matter) is that neither of these are the models of a commutative algebraic theory. Compared to this, the fact that commutativity follows from the other ring axioms pales to insignificance.

Let me explain.

First, a problem.

Exercise 1. Let $X$ and $Y$ denote commutative semigroups written additively, and let $f,g : X \rightarrow Y$ denote homomorphisms. Show that $f+g$ is a homomorphism.

I highly recommend you solve the above (fairly easy) exercise before going on. In doing so, you'll have noticed that both associativity and commutativity were used in the proof. However, with a bit of thought, we see that they weren't really fundamental to the proof. Basically, we have the expression

$$(f(x)+f(x'))+(g(x)+g(x'))$$

and we use associativity to drop the brackets, obtaining

$$f(x)+f(x')+g(x)+g(x')$$

then we switch the two inner terms, obtaining

$$f(x)+g(x)+f(x')+g(x')$$

and then we put the brackets back in:

$$(f(x)+g(x))+(f(x')+g(x'))$$

Thus, our proof ultimately comes down to the fact that $Y$ satisfies the following identity.

$$(a+b)+(c+d) \equiv (a+c)+(b+d)$$

So we get an idea: call magmas satisfying the above identity medial, prove the result for medial magmas rather than commutative semigroups, and then simply observe that every commutative semigroup is in fact a medial magma. Of course there might be a niggling doubt in the back of our minds; is the mediality condition actually a "natural" thing to consider? It looks just like a warped form of commutativity. The answer is: "Yes, it is a natural property to consider," but the justification for this claim will have to wait for later. So, lets temporarily adopt the position that mediality is interesting and natural, and lets take some time to consider its consequences.

Exercise 2. Verify that every commutative semigroup is indeed a medial magma (I suggest using additive notation). Conversely, prove that every medial magma with a two-sided identity element is both associative and commutative.

The above exercise begs the question; do there exist magmas that are medial, but which aren't commutative semigroups? The answer is a resounding "yes!"

Example 1. If $(X,+)$ is an abelian group, then $(X,-)$ is a medial magma that is neither commutative, nor associative.

Example 2. If $I \subseteq \mathbb{R}$ and $f : I^2 \rightarrow I$ is defined by $f(x,y) = (x+y)/2$, then $(I,f)$ is a medial magma that is commutative, but not associative.

The best way to verify Example 1 is with the result of the following exercise; Example 1 is then obtained as the special case in which $*$ is taken as addition, $f$ denotes the identity function on $X,$ and $g$ represents unary negation.

Exercise 3. Let $(X,*)$ denote a medial magma, and suppose $f,g : X \rightarrow X$ are commutative endomorphisms, in the sense that $f \circ g = g \circ f$. Defining $x \diamond y = f(x) * g(y)$, show that $(X,\diamond)$ is also medial.

Finally, since we've come this far, lets check for a moment that we're not fooling ourselves, by verifying the result that motivated our interest in mediality in the first place:

Exercise 4. Let $X$ and $Y$ denote medial magmas written with an asterisk, and let $f,g : X \rightarrow Y$ denote homomorphisms. Verify that $f*g$ is a homomorphism.

So to recap, if $f$ and $g$ are parallel homomorphisms in the category of medial magmas, then so too is $f*g$. But, is this really a big deal? I think this is a huge deal; here's one reason why.

Let $(X,*)$ denote a medial magma and suppose $A$ is the set of all endomorphisms of $X$. Then we observe that $A$ has the following properties.

1. $(A,*)$ is a medial magma, where $*$ on $A$ is defined pointwise from $*$ on $X$.
2. $(A,\circ)$ is a monoid
3. $\circ$ distributes on both sides over $*$

Thus, $A$ is kind of like a ring. This motivates the following definition: a ringlike-entity appropriate for medial magmas is a triple $(A,*,\circ)$ satisfying the aforementioned axioms. Now suppose $A$ is a ringlike-entity appropriate for medial magmas and that $X$ is a medial magma. I leave it as an exercise for the reader to define the notion of "multiplication of a scalar in $A$ with a vector in $X$"; that is, a function $f : A \times X \rightarrow X$ that satisfies all the reasonable compatibility conditions. As inspiration; let $X$ denote an arbitrary medial magma, suppose $A$ is the set of all endomorphisms of $X$, and think of $f : A \times X \rightarrow X$ as the evaluation function.

Therefore, since we have notion of scalar multiplication, for any ringlike-entity appropriate for medial magmas $A$, we have a notion of $A$-module.

Lets go picture for a moment. Commutative algebra (think: abelian groups, rings, modules and algebras) is built on the theory of "abelian groups." However, what I'm trying to say here, is that we can replace this theory with others (e.g. the theory of commutative semigroups, or medial magmas) in order to obtain different versions of commutative algebra, each with their own version of "ring", "module" etc. Most generally, I think, is that we can replace the theory of "abelian groups" with any commutative algebraic theory. Examples of such theories include:

• The theory of abelian groups, with addition and unary negation.
• As above, except with with addition and binary subtraction this time.
• The theory of medial magmas.
• The theory of commutative semigroups.

Note that a theory with a single binary operation $*$ is a commutative algebraic theory if and only if it proves that $*$ is medial; that is why I argued earlier that mediality is a natural condition to consider. Furthermore, not every group is medial, and not every monoid is medial, and not every semigroup is medial, so it wouldn't make sense to try to build commutative algebra over these structures.

Anyway, I have much, MUCH more to say about this stuff, some of it quite abstract, but I don't want to give away too much of what I've been toying with, since it might be publishable someday.

In conclusion though, its not just for convenience and simplicity. For everything to work, we have to build commutative algebra over some base commutative algebraic theory.

• Hey why the downvote? Dec 23, 2013 at 17:20