Let's say I have a ring $R=(\mathbb{R}, +, *)$. Is


($a \in \mathbb{R}$)

always true for any ring¹? (So that when this equation is not true then I can definitively infer that $R$ is not a ring.)

1 ) With "any" I mean any other ring which is also using $\mathbb{R}$ as underlying set, thank you for your comment nick.

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    $\begingroup$ If $R = \Bbb{R}$ and the usual number $1 \in \Bbb{R}$ is the identity, then yes. The distributive law in a ring says that $a+a = a(1+1)$. However, when you say "any ring", this makes it confusing, because the underlying set does not always have to be $\Bbb{R}$. $\endgroup$ – Nick Nov 29 '20 at 23:06
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    $\begingroup$ In particular, the answer depends on what the definition of "2" is in the ring. If it always means "1+1", then the equation is true. $\endgroup$ – Greg Martin Nov 29 '20 at 23:07
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    $\begingroup$ In any ring, for $n\in\mathbb{Z}$, we define $0a = 0_R$, $(n+1)a=na+a$, and $(-n)a = -(na)$ for $n\gt 0$. This is not “really” multiplication, but rather abbreviation for some addition operations. When the ring has a unity, then one can show that $na = (n1_R)a$, and so we can turn this “abbreviated addition” into actual multiplication in the ring. But one does not normally think of “$2a$” as being a product in an arbitrary ring; instead, it is addition and stands for $a+a$. $\endgroup$ – Arturo Magidin Nov 29 '20 at 23:39

If your ring has a unit, i.e. a multiplicative identity, (and the definition that almost everybody uses these days requires that a ring does have an identity), then yes.

As the commenters point out, $2$ is defined* to be $1 +1$, where $1$ is the multiplicative identity, and so it follows from the distributive law and the fact that $1$ is the multiplicative identity.

The only thing to be careful about is that it is possible that $ 2 = 0$ (e.g. in $\mathbb Z_2$), or perhaps $2 = -1$ (e.g. in $\mathbb Z_3$), so these "integers" inside your ring might not behave the way you expect integers to behave.

BTW, if you are dealing with an algebraic structure that doesn't have a $1$, people will often define an "action" of $\mathbb Z$ on your elements, and use multiplication to denote it, where

$$ n \cdot a = a + a + .... + a \text{ (n times)}$$

Edit: Okay, you added "With 'any' I mean any other ring which is also using $\mathbb{R}$ as underlying set", and this needs to be addressed: You can take the underlying set $\mathbb R$, and define a wacky new addition and multiplication on it. The simplest is $a \oplus b = a + b -1$ and $a \otimes b = ab - a -b + 2$.

Let's use the symbol $S$ to denote this new ring $\langle \mathbb R, \oplus, \otimes\rangle$. Then the number 1 in $\mathbb R$ (which I'm going to write as $1_{\mathbb R}$) is not the multiplicative identity for the ring $S$. $1_S$, which is the standard notation for the multiplicative identity in a ring named $S$, is, in fact $2$, by which I mean good old 2 in good old $\mathbb R$, which we might want to write as $2_{\mathbb R}$, and yes $2_{\mathbb R} = 1_{\mathbb R} + 1_{\mathbb R}$.

But what your question asks is still true in $S$, i.e. $a \oplus a =2_{S} \otimes a$; however notice that you have to be sure to use the ring operations of $S$, and remind yourself that you're using $2_{S}$, which is defined to be $1_{S} \oplus 1_{S}$. (And corresponds to the underlying real number $3_{\mathbb R}$!)

The ring $S$ is of course extremely confusing to work with, and I've never seen it used seriously, only to break the brains of undergraduate math majors, to show them how we can define groups, rings, fields, etc. that behave very differently from what they are used to. I.e. $\langle \mathbb R, \oplus, \otimes\rangle$ is a cautionary tale, not a commonly used mathematical tool, but the only requirement you put was that $\mathbb R$ was the underlying set, and so you left it open to me to define really weird addition and multiplication. I wouldn't spend too much time agonizing over it, but it can be a fun example to contemplate and sharpen your wits.

*If someone uses the symbol "$2$" and says it is not equal to $1+1$, you get to look at them funny, ask just what the heck they think they are doing, and demand that they explain why they are using that symbol.

  • $\begingroup$ thank you for your detailed explanation! $\endgroup$ – anion Nov 30 '20 at 11:29

This is basically true by definition, although there are some things you should be aware of.

Some people require that any ring $(R,+_R,\cdot_R)$ contains a multiplicative identity $1_R,$ and that ring homomorphisms $f : (R,+_R,\cdot_R)\to (S,+_S,\cdot_S)$ satisfy $f(1_R) = 1_S.$ If you require this condition, then for any ring $(R,+_R,\cdot_R)$ there is a unique ring homomorphism $i_R : (\Bbb{Z},+,\cdot)\to(R,+_R,\cdot_R).$ In this case, even if the set $R$ does not literally contain $2,$ you may think of $i_R(2)\in R$ as being $2$ (you might even write $i_R(2) = 2_R$). It is true then that for any $r\in R,$ $$ 2_R\cdot_R r = i_R(2)\cdot_R r = r +_R r, $$ because $$ \begin{align*} i_R(2)\cdot_R r &= i_R(1 + 1)\cdot_R r\\ &= (i_R(1) + i_R(1))\cdot_R r\\ &= (1_R + 1_R)\cdot_R r\\ &= r + r. \end{align*} $$ As JonathanZ supports MonicaC notes, it might be the case that $i_R(2)$ behaves differently than you'd expect, or looks different than you'd expect. It could be that $i_R(2) = -1_R$ or even $i_R(2) = 0_R$! See the last paragraph for a particularly outrageous example of this.

If you do not require that your rings have multiplicative identitites and/or that ring homomorphisms need not send multiplicative identities to multiplicative identities, then this is still true to some extent, although we should be careful about what we mean.

Let $(R,+_R,\cdot_R)$ be our possibly non-unital ring. In this case, we can't use the unique homomorphism $i_R :(\Bbb{Z},+,\cdot)\to(R,+_R,\cdot_R)$ from before -- there might be more than one ring homomorphism now! Additionally, the set $R$ might not contain $2.$

So, what do we do? Well, remember that any ring has an underlying abelian group $(R,+_R).$ An abelian group is the same as a $\Bbb{Z}$-module (see here for the definition of a module over a ring if you're not familiar). This means explicitly that we have an action of $\Bbb{Z}$ on $R$ which interacts nicely with addition. We define this action by setting $$ n\cdot r :=\begin{cases} \underbrace{r + \dots + r}_{n\textrm{ times}},&n > 0,\\ 0,&n=0,\\ \underbrace{-r + \dots + -r}_{-n\textrm{ times}}, &n <0. \end{cases} $$ Notice that I'm not writing $n\cdot_R r$ -- that's because there's not necessarily an element $n\in R$ which behaves like $n.$ However, it's still sensible to think of adding the element $r$ to itself $n$ times, which is what $n\cdot r$ means by definition. The $\cdot$ refers to the action of $\Bbb{Z}$ on the underlying abelian group of $(R,+_R,\cdot_R),$ not multiplication in the ring itself. In this sense, the equality $$ 2\cdot r = r+r $$ always holds, and this is basically by definition!

One last remark. You asked if this is true of any ring which has $\Bbb{R}$ as its underlying set. You should be slightly careful here. Consider the following ring structure on $\Bbb{R}$: $$ \begin{align*} +' : \Bbb{R}\times\Bbb{R}&\to\Bbb{R}\\ (r,s)&\mapsto r+'s:=\sqrt[3]{r^3 + s^3},\\ \cdot' : \Bbb{R}\times\Bbb{R}&\to\Bbb{R}\\ (r,s)&\mapsto r\cdot's := rs. \end{align*} $$ This is not the standard ring structure on $\Bbb{R}$ -- the multiplication is the same, but the addition is "twisted." In this case, $2\in \Bbb{R}$, but it is not true that $2\cdot' r = r +' r.$ Suppose $r = 2.$ Then: $$ \begin{align*} 2 +' 2 &= \sqrt[3]{2^3 + 2^3}\\ &= \sqrt[3]{16}\\ &= 2\sqrt[3]{2}. \end{align*} $$ On the other hand, $$ 2\cdot'2 = 4. $$ What happened? I'll let you think about this for yourself before revealing the answer below!

What happened here is that $2\in\Bbb{R}$ is no longer playing the same role it was before. Our ring $(\Bbb{R},+',\cdot')$ still has a multiplicative identity, but our ring homomorphism $i_{(\Bbb{R},+',\cdot')} : (\Bbb{Z},+,\cdot)\to(\Bbb{R},+',\cdot')$ now sends $$i_{(\Bbb{R},+',\cdot')}(2) = i_{(\Bbb{R},+',\cdot')}(1) +' i_{(\Bbb{R},+',\cdot')}(1) = 1 +' 1 = \sqrt[3]{2}.$$So there is an element of $(\Bbb{R},+',\cdot')$ which behaves like $2$ should -- it's $\sqrt[3]{2}$. We thus have$$\sqrt[3]{2}\cdot' r = r +' r$$for any $r\in\Bbb{R}.$ This is very confusing, because we already have $2\in\Bbb{R}$! In this case, it would be very important to distinguish between $2\cdot r$ (which is $2\in\Bbb{Z}$ acting on $r,$ giving $r +'r$) and $2\cdot' r$ (which as we computed, is not $r +' r$ in general). In the notation of the first paragraph, $2_{(\Bbb{R},+',\cdot')} = \sqrt[3]{2}$ and $2\neq 2_{(\Bbb{R},+',\cdot')}$.

To be even more explicit about what happened, given any set $X,$ any ring $(R,+_R,\cdot_R),$ and any bijection $f : X\to R,$ we can give $X$ the structure of a ring by defining addition on $X$ by $x +_X y := f^{-1}(f(x)+_R f(y))$ and $x\cdot_X y := f^{-1}(f(x)\cdot_R f(y)).$ We're taking the ring structure on $R$ and transporting it to $X$ via the bijection $f$: first, take your elements $x$ and $y$ in $X,$ send them over to $R$ where you add or multiply them, and then bring them back to $X.$ In my example above, I'm using the bijection $\Bbb{R}\to\Bbb{R}$ which sends $x$ to $x^3.$

  • $\begingroup$ Stahl, we wrote very similar answers! In a fair world our points would be pooled and split evenly. I think I like your "weird $\mathbb R$" better -- mine is just the standard conjugating by $\phi(a) = a-1$, but I'd never seen yours before. Having multiplication stay the same while getting strange with addition is a really nice touch. $\endgroup$ – JonathanZ supports MonicaC Nov 30 '20 at 0:10
  • $\begingroup$ @JonathanZsupportsMonicaC Thanks! I think your answer is very nice -- it's simple and to the point, while I've rambled on a bit. You posted yours while I was in the middle of writing mine, and I thought I would keep going and post in case the extra details helped someone. $\endgroup$ – Stahl Nov 30 '20 at 0:20
  • $\begingroup$ Yeah, I figured that it was an overlap. These days if I find myself writing an essay, I've taken to posting a few times as I complete sections (I'm still trying to get to 10k rep!). And I ended up rambling too when addressing the OP's edit. But I'm glad you finished and posted yours; I'm going to have to go and have a think about your twisted addition. I think my answer targets the average Fraleigh reader, and yours is good for the Hungerford crowd. $\endgroup$ – JonathanZ supports MonicaC Nov 30 '20 at 0:28
  • $\begingroup$ thank you for your detailed explanation! $\endgroup$ – anion Nov 30 '20 at 11:29

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