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If $R$ is a ring, is it abuse of notation to write $2ab$ for general elements $a,b\in R$ for $ab+ab$?

I have a question regarding this answer. I am trying to prove that any Boolean ring is commutative. The answer at that link is incredibly intriguing, but there's something that bugs me slightly:

The answer starts off with $2ab$. But what is the meaning of $2ab$ exactly? I understand that rings have an underlying commutative group structure under the operation $+$, but is it safe to write the $2$ as a coefficient to denote $ab + ab$? It makes it seem as if $2$ is an element of the ring, but we never defined what the elements of $R$ are.

There is some precedence in my textbook of using $-$ to denote the opposite of $+$, i.e., $(-a)(-b) = ab$, analogous to how we pretend there's a $-1$ out in front of the $a$ and $b$ in elementary algebra, but it honestly seems as if it's "abuse of notation" to write $2ab$ in this case where $a,b$ are elements of a ring.

EDIT: As an addendum, my textbook does not require that rings be unital. However, the problem at hand appears in a section that says "assume $R$ is a ring with identity $1$."

Bill Dubuque
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Decaf-Math
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  • Would you also argue against writing $2a$ for $a+a$? Do you require your rings to be unital? – Improve Apr 10 '17 at 08:31
  • for any ring with multiplicative unity 1, you can define 2:=1+1 – Bananach Apr 10 '17 at 08:32
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    Notice that $(R,+)$ is an abelian group and each abelian group is a $\mathbb Z$-module, thus making sense of expressions like $ka\in R$ for $k\in \mathbb Z$ and $a\in R$, although $k$ need not lie in $R$. – Claudius Apr 10 '17 at 08:44
  • @Improve, I would very much like to answer your question. However, the more I read here and in my textbook, the more I realize I do not understand the properties of a ring. I'm assuming that "unital" means that $R$ has an element $x\in R$ such that $rx = x = xr$ for all $r\in R$, yes? In other words, has an identity $1$? My textbook states that "$R$ is said to have an identity if there exists an element $1\in R$ such that $1\times a = a = a\times 1$ for all $a\in R$." Does this means that not all rings have a multiplicative identity? – Decaf-Math Apr 10 '17 at 08:51
  • Yes, that is correct. Consider for example the even integers. They behave like a ring in the sense that if you add two even integers, you get another even integer, and if you multiply two even integers, you get another even integer. Can this ring have an identity element? – Improve Apr 10 '17 at 09:00
  • @Improve, definitely not. Thanks so much. – Decaf-Math Apr 10 '17 at 09:02
  • I believe there is typo in question. You are asking whether we can write $ab + ba$ as $2ab$. This has a problem if the ring is not commutative. The answers given here are related to the question: Can we write $ab + ab$ as $2ab$? – Paramanand Singh Apr 10 '17 at 09:45
  • @ParamanandSingh, sorry. This has been fixed. – Decaf-Math Apr 10 '17 at 10:02

3 Answers3

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If a ring $R$ is unital, we adopt for natural numbers $n$ the notation $$ n = \underbrace{1 + \cdots + 1}_{\text{$n$ times}}. $$ If $n$ is a negative integer, we similarly define it as the sum of $-n$ copies of $-1$. Then $$ ab + ab = 1\cdot ab + 1\cdot ab = (1 + 1)\cdot ab = 2ab. $$ Note that the mapping defined above gives a ring homomorphism $\mathbb Z \to R$, which tells you that with these "integers in the ring" you can do a lot of ordinary arithmetic.

Edit: If the ring is not unital, it does not make sense to talk about $2$ as an element of the ring. However, the notation $2ab = ab + ab$ still does. One way of thinking about this is as follows: even if $R$ is not unital, the endomorphism ring $\mathrm{End}(R)$ of the underlying additive group of $R$ still will be, with the identity as unit. Hence, we can interpret natural numbers not as elements of $R$, but nonetheless as a subring of the endomorphisms of the additive group of $R$. Then the above sequence of equations becomes $$ ab + ab = \mathrm{id}_{R}(ab) + \mathrm{id}_{R}(ab) = (\mathrm{id}_{R} + \mathrm{id}_{R})(ab) = (1_{\mathrm{End}(R)} + 1_{\mathrm{End}(R)})(ab) = 2_{\mathrm{End}(R)}(ab), $$ and we denote that last element as $2ab$.

Mees de Vries
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No, it's not abuse.

If $M$ is a (multiplicative) monoid, you can recursively define, for $x\in M$, $$ x^0=\mathbf{1},\qquad x^{n+1}=x^nx \tag{1} $$ and, if $x$ is invertible, $x^n=(x^{-1})^{-n}$ when $n$ is a negative integer.

When the operation is denoted additively, the exponent is usually set besides the element: $$ 0x=\mathbf{0},\qquad (n+1)x=nx+x \tag{2} $$ and, as before, $nx=(-n)(-x)$ if $x$ is invertible. Note that I use $\mathbf{1}$ and $\mathbf{0}$ for the neutral element, in order to avoid ambiguities.

A ring is a group with respect to addition, so the notation with multiples (2) can be freely used, hence $$ ab+ab=2ab $$ is fine. Here it should be interpreted as $2(ab)$. On the other hand, interpreting it as $(2a)b$ would yield the same result: $$ (2a)b=(a+a)b=ab+ab=2(ab) $$ and this can be generalized to any integer in place of $2$. Therefore it's not necessary to add parentheses. Also $a(2b)=2(ab)$ can be easily proved and generalized.

Note that it's not really necessary to distinguish $\mathbf{1}$ (the identity in the ring) with $1$ (the integer) and $\mathbf{0}$ (the zero in the ring) with $0$ (the integer), because $$ 0a=\mathbf{0}=\mathbf{0}a,\quad 1a=a=\mathbf{1}a $$ It's not necessary that the ring has an identity: if it has it, denoting it by $1$ will not raise ambiguities.

If the ring $R$ has an identity (let me still denote it by $\mathbf{1}$), then there exists a unique ring homomorphism $\chi\colon\mathbb{Z}\to R$, defined by $n\mapsto n\mathbf{1}$. It's easy to prove that, for $x\in R$, $$ nx=\chi(n)x $$ (the left-hand side has the notation defined in (2), in the right-hand side there's the ring multiplication). So there's no ambiguity even if $2$ (or any other integer) is thought to be $2\mathbf{1}=\mathbf{1}+\mathbf{1}$. Hence usually the distinction between the two possible interpretations of $1$ is dropped. Thus it's common to see that $10=0$ in a ring of characteristic $5$ or similar identities (this is indeed a slight abuse of notation).

Note also that $-a$ can be interpreted as $(-1)a$, because both yield the same result; but it's better to think to $-a$ just as the opposite element of $a$, that is, the unique element such that $a+(-a)=\mathbf{0}=(-a)+a$.

By the way, (1) explains why, in unital rings, $\mathbf{0}^0=\mathbf{1}$ (usually written $0^0=1$).

egreg
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  • It might perhaps be a good idea to mention that the notation $2ab$ is unambigious in the sense that $2(ab) = (2a)b = a(2b)$. – Improve Apr 10 '17 at 08:56
  • @Improve Thanks for the comment: it's really helpful and I incorporated it in my answer. – egreg Apr 10 '17 at 09:05
  • @egreg I had some connection problems for half an hour and when I was able to finally post my answer, I realised it was very close to yours. My apologies. In your answer, $M$ should be a monoid rather than a semigroup. – J.-E. Pin Apr 10 '17 at 09:09
  • @J.-E.Pin I corrected (although I don't like “monoid” at all). – egreg Apr 10 '17 at 09:17
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You are probably more familiar with the notation $a^n$ to denote the $n$-th power of an element $a$. But how do you define it formally? Well, you do it by induction. First set $a^0 = 1$ and then, for all $n \geqslant 0$, $a^{n+1} = a^na$.

Now, just do the same thing for addition: $0a = 0$ and for all $n \geqslant 0$, $(n+1)a = na + a$. Thus $na = \underbrace{a + {}\dotsm {} + a}_{n \text{ times}}$, which is exactly what you need.

J.-E. Pin
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