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In a ring, a multiple for addition is written as $na$ to stand for $(a + a + ... + a)$.

This is not necessarily the same as $n * a$ (the "multiplication" operation). Is that correct?

Multiple is only the same as multiplication for specific rings such as Integers. Is that right?

I suspect the answer to be the case but I have never seen a proof one way or the other.

Thanks

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  • $\begingroup$ @Pippo I think you mean $a^n$. $\endgroup$
    – kccu
    Nov 30, 2017 at 16:18
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    $\begingroup$ Actually if your ring has a $1$, then $a+...+a$ $n$ times is equal to $(n\cdot 1)×a$, where $n\cdot 1 = 1+...+1$ $n$ times $\endgroup$ Nov 30, 2017 at 16:19
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    $\begingroup$ @Max: That's worth making an Answer. $\endgroup$
    – hardmath
    Nov 30, 2017 at 16:21
  • $\begingroup$ If it would be the same, the word: polynomial would not exist. $\endgroup$
    – LAAE
    Nov 30, 2017 at 21:44
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    $\begingroup$ All rings have multiplicative identify, call it 1. So, multiples always equal multiplication given that 1 + ... + 1 n times is defined as the ring member n. $\endgroup$
    – Jae Noh
    Nov 30, 2017 at 23:15

7 Answers 7

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Let's use two different notations: $n\cdot a$ for repeated addition and $a*b$ for ring multiplication, so that we can discuss when it makes sense to compare them. And when it does make sense to do so, when are they equal.

In a ring $n\cdot a$, for $n$ a natural number, refers to repeated addition: $$n\cdot a=\underbrace{a+\dotsb+a}_{n\text{ times}}.$$ Using additive inverse, we may also define the expression for $n$ any integer: $(-n)\cdot a=-(n\cdot a)$.

The ring also has a multiplication operation $a* b$ which has nothing to do with repeated addition; you can't add $b$ to itself $a$ times if $a$ is not a counting number. Rather multiplication is an operation given as part of the given structure of the ring, separate from addition.

It does not make sense to ask whether $a * b$ and $a\cdot b$ are the same, because the latter expression is undefined for a general ring element. The question is only even defined for expressions of the form $n\cdot b$, where $n$ is an integer.

If the ring has a multiplicative identity element $1$, which most authors do assume, then the numbers $$n=\underbrace{1+1+\cdots+1}_{n\text{ times}}$$ may also be viewed as ring elements. (But beware that in some rings we may have $\underbrace{1+1+\cdots+1}_{n\text{ times}}=0$ for $n\neq 0.$) These elements make a subring of $R$ called the prime ring (see the second definition).

So in rings with identity, multiplication of general ring elements may be viewed as an extension of multiplication of elements of the prime subring. Meaning that when if a ring element $a$ is equal to $n$ for some integer $n$, then $a*b=a\cdot b$. So the answer is that your proposed phrasing "$n*a$ is not necessarily the same as $na$" is not correct; $n*a$ and $n\cdot a$ are necessarily the same. As functions, the operation $\cdot$ is a subset of $*$. Note that multiplication is always commutative on the prime subring $m\cdot n=n\cdot m$, whereas it need not be on the whole ring $a* b\neq b* a.$

In a ring without identity element, (sometimes called rngs. Get it? Remove the 'i' from "rings"), both types of multiplication exist together, but multiplication by integers is not a subset of the ring's multiplication operation. Note that a ring without identity can be canonically embedded in one which does include identity.

So to sum up, comparing $n * b$ and $n\cdot b$ only makes sense when $n$ is an integer, not a general ring element and only if our ring contains a multiplicative identity. In that case, they must agree.

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    $\begingroup$ @Jae Noh - In summary, where $*$ is the ring multiplication: If $R$ has an identity $1$, then you can identify $n \in \Bbb N$ with $n \cdot 1$, and if you do that, then $n \cdot r = n * r$. But if you do not make that identification, or if $R$ doesn't have an identity, then $n * r$ is not even defined. $\endgroup$ Nov 30, 2017 at 17:33
  • $\begingroup$ @PaulSinclair any abelian group is a $Z$-module (and any ring has an additive abelian group). So in my opinion (and this distinction may be purely semantic, notational, or splitting hairs) $n\cdot r$ is always defined, even with no unit. You just cannot view the factorization $2r=r+r=(1+1)r$ as being a statement purely about elements of the ring. $\endgroup$
    – ziggurism
    Nov 30, 2017 at 17:41
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    $\begingroup$ please read my comment again. I never disputed the definition of $n \cdot r$. It is $n * r$ that is undefined unless you can identify $n$ with an element of $R$. I made this addendum to your post because Jae Noh specifically asked about whether $n \cdot r = n * r$, and I wanted to make sure he/she understood the answer. $\endgroup$ Nov 30, 2017 at 17:50
  • $\begingroup$ @PaulSinclair I got it now. You’re using two different symbols for the two multiplications. That’s a good idea to make the distinction clearer. Sorry for the confusion. Indeed if $*$ is the symbol reserved only for the ring operation, $n*r$ is undefined. $\endgroup$
    – ziggurism
    Nov 30, 2017 at 17:53
  • $\begingroup$ @PaulSinclair: Rather than identifying $n$ as an element of $R$, you can identify $R$ as a $\mathbb{Z}$-algebra. $\endgroup$
    – user14972
    Dec 1, 2017 at 3:54
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Addition and multiplication distribute in a ring.

If we start with $$x+x+x+x$$ in a ring $R$, and we assume $1_R$ is an element of your ring (the multiplicative identity, which most definitions of Ring assume is there), we get: $$1_R*x + 1_R*x + 1_R*x + 1_R*x$$ $$(1_R+1_R)*x + (1_R+1_R)*x$$ $$(1_R+1_R+1_R+1_R)*x$$ $$4_R*x$$ where we define $4_R$ as an element of the ring to be $1_R+1_R+1_R+1_R$.

Note that this works equally well with right-multiplication, so $4_R*x = x+x+x+x = x*4_R$.

For elements of your ring that cannot be found through repeated addition of $1_R$, left-multiplication and right-multiplication by them may not be the same. $2\times2$-matrices furnish an example.

Even if multiplication is commutative, the above argument may fail. $R=\mathbb{Z}_3^2$ (2-tuples of integers mod 3) with element-wise $+$ and $*$ has $\{(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)\}$ as elements. Here $1_R$ is $(1,1)$, $0_R$ is $(0,0)$, $3_R$=$0_R$ and $(2,1)*(1,2)=(2,2)$ while no sum of either $(2,1)$ or $(1,2)$ adds to $(2,2)$.

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  • $\begingroup$ wow this is enlightening. $\endgroup$ Nov 30, 2017 at 20:28
  • $\begingroup$ I hope you don't mind my edit to clarify what you meant by "does not hold". At first I thought you meant "if $x$ cannot be found through repeated addition of $1_R$, ..." which of course is not true. Then I realized you must have meant something else. =) $\endgroup$
    – user21820
    Dec 1, 2017 at 4:07
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Let me provide an alternate answer, since I understood your question a little differently than the other answers. Here is what you asked:

In a ring, a multiple for addition is written as $na$ to stand for $(a + a + ... + a)$.

This is not necessarily the same as $n * a$ (the "multiplication" operation). Is that correct?

That's not correct -- they are both the same thing! To clarify, the second notation only makes sense in a ring with unity / multiplicative identity, but as long as we are in such a ring, we must be clear: the definitions of these two notations are equivalent.

In more detail: for $n$ an integer and $r$ a ring element in an arbitrary ring $R$, we define $$ nr =_{\text{def}} \underbrace{r + r + \cdots + r}_{n \text{ times}}. $$ Now, if $r$ has a multiplicative identity $e$, then we also define, for $n$ an integer, $$ n =_{\text{def}} \underbrace{e + e + \cdots + e}_{n \text{ times}}. $$

Now, by distributivity, $(e + e + \cdots + e)* r = r + r + \cdots + r$. In other words, if we do $n * r$ with ring multiplication or with repeated addition, we get the same answer. This is good -- we don't want our two notations to conflict.

As other answers have mentioned, in a general ring we might have $n = 0$ for some positive integer $n$. But that's really irrelevant, because the two notations still coincide even in this case.


P.S. The most important thing is to always keep separate in your head which variables are integers (like $n$, $0$, $1$) and which variables are ring elements (like $r \in R$). It gets confusing since the integers themselves are a ring, and since we have these notations such as $n * r$ for repeated addition, and since $1$ can be used both for a ring element and for an integer, etc. But as long as you are keeping the set of integers separate in your mind from the ring, you should be able to avoid confusion.

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  • $\begingroup$ Thanks. All rings have multiplicative identify, 1, correct? Rings are monoids under the multiplication operation. So, there is a multiplicative identity for all rings. Is that right? $\endgroup$
    – Jae Noh
    Nov 30, 2017 at 23:19
  • $\begingroup$ @JaeNoh There is no one definition of "ring". Some authors define it one way, some another way. In my classes, rings always had a multiplicative identity. Maybe in yours, too. $\endgroup$ Nov 30, 2017 at 23:33
  • $\begingroup$ If the definition you are given includes "monoid under multiplication", then yes, it has a multiplicative identity. $\endgroup$ Nov 30, 2017 at 23:34
  • $\begingroup$ Isn't it worth noting though, that if $\mathbb{Z}\subseteq R$, then for any $n\in\mathbb{N}$, that $n=\sum\limits _{i=1}^{i=n}e$ a priori? $\endgroup$
    – lukeuser
    Oct 30, 2019 at 0:53
  • $\begingroup$ or more to the point, for any $r\in R$, $n\ast r=\sum\limits _{i=1}^{i=n}r$ (which is what one would define $n\cdot r$ as) $\endgroup$
    – lukeuser
    Oct 30, 2019 at 0:59
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For any ring $R$ (not even with identity, necessarily) and $r\in R$ and any positive integer $n$, $n\cdot r$ is defined to be $r+r+\ldots$ $n$ times. This extends to negative numbers and $0$ in the obvious ways.

So yes, it is that way for the product of an element with an integer.

The place where the phrase stops making sense is the product of any two elements of a ring. For instance, in $F[x]$, how do you add $x$ to itself $x$ times?

Under the hood, the notation you're talking about is just the notation for repeated group operations. If you write a group multiplicatively, then taking $n$ copies of $g$ via the operation is written as $g^n$ when the operation is written multiplicatively, and is written $ng$ when the group is written additively.

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Note that in your question the symbol $n$ is used with two different meanings.

In the statement:

  • In a ring, a multiple for addition is written as $\color{blue}{na}$ to stand for $(a + a + ... + a)$.

the symbol $n$ denotes the natural number $n\in\mathbb{N}$ which is used to indicate a multiple of the ring-element $a$. \begin{align*} \underbrace{a+a+\cdots+a}_{\color{blue}{n}\text{ times}}\qquad\qquad n\in\mathbb{N} \end{align*}

It is clever to write it stands for instead of $na=a+a+\cdots+a$, since these elements are not necessarily comparable (the ring $R$ does not necessarily contain natural numbers).

On the other hand, in the statement

  • This is not necessarily the same as $\color{blue}{n \ast a}$ (the "multiplication" operation).

the symbol $n$ denotes a completely different object, namely an element of the ring $R$.

Keeping this in mind the answer to your question is yes, your assumption is correct. We just need to consider a ring which don't contain natural numbers.

Example: We consider the ring $R=\left(M_2(\mathbb{R}),+,\ast\right)$ of all real $(2\times 2)$-matrices with the usual matrix addition and multiplication.

For $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in R$ and $n\in\mathbb{N}$ we obtain \begin{align*} nA=n\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}na&nb\\nc&nd\end{pmatrix}\in R \end{align*} but $n\ast A$ is not defined since $n\not\in R$.

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  • $\begingroup$ Thank you. So, na cannot be compared to n * a if n is not an element of the Ring. The comparison does not make sense since n * A is not defined. $\endgroup$
    – Jae Noh
    Dec 1, 2017 at 18:08
  • $\begingroup$ @JaeNoh: Correct. This is the essence. $\endgroup$ Dec 1, 2017 at 18:16
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Any ring $R$ is set equipped with two binary operations addition and multiplication. Here we required addition to be commutative (additive abelian group). Thus any ring is an $\Bbb{Z}$-module and therefore your intuition is correct.

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None of the other answers clearly stated what I was looking for when I found this question. Namely, if $(R, +, *)$ is a ring (with multiplicative identity $1_R$), then for all $n \in \mathbb{Z}_{\geq 1}$ and $r \in R$, $$\begin{align*} n \cdot r &= \underbrace{r + \cdots + r}_{n\text{ times}} \tag{definition of $n \cdot r$} \\ &= \underbrace{1_R * r + \cdots + 1_R * r}_{n\text{ times}} \\ &= \left( \underbrace{1_R + \cdots + 1_R}_{n\text{ times}} \right) * r \tag{right distributivity} \\ &= \left( n \cdot 1_R \right) * r \\ &= n_R * r \tag{define $n_R := n \cdot 1_R$} \end{align*}$$ If we instead write $\underbrace{r + \cdots + r}_{n\text{ times}} = \underbrace{r * 1_R + \cdots + r * 1_R}_{n\text{ times}}$ and use left distributivity, we get $n \cdot r = r * n_R$.

Additionally, $$\begin{align*} (-n) \cdot r &= -(n \cdot r) \\ &= -(n_R * r) \\ &= -1_R * (n_R * r) \\ &= (-1_R * n_R) * r \\ &= (-(n_R)) * r \\ &= (-n)_R * r \end{align*}$$ since $(-n)_R := (-n) \cdot 1_R = - (n \cdot 1_R) = - (n_R)$. Similarly, $$(-n) \cdot r = n \cdot (-r) = (-r) * n_R = (r * -1_R) * n_R = r * (-(n_R)) = r * (-n)_R$$

With $0$ denoting the additive identity, we define $0_R := 0 \cdot 1_R = 0$. Thus $0 \cdot r = 0 = 0 * r = 0_R * r = r * 0_R$. This brings us to the pleasing conclusion that for all $m \in \mathbb{Z}$ and $r \in R$, $$m \cdot r = m_R * r = r * m_R$$

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