In a ring, result of multiple (of "addition" operation) is not the same as result of multiplication, correct?

Solution 1:

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.

Solution 2:

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 $(2,2)$.