TITLE: Question about the necessity of the distributive property axiom used in the definition of a ring
QUESTION [0 upvotes]: Pinter's "A Book of Abstract Algebra" defines a ring $A$ stipulating the following $3$ axioms: 

$A$ with addition alone is an abelian group.
Multiplication is associative.
Multiplication is distributive over addition.

I am a little confused as to why the third axiom is needed. Doesn't axiom $1$ naturally lead to $3$? Consider the following example:
$(a \circ_+ b)^5 = (a\circ_+ b) \circ_+ (a\circ_+ b) \circ_+ (a\circ_+ b) \circ_+ (a\circ_+ b) \circ_+ (a\circ_+ b)$
Because the operation of addition is an abelian group, the above statement is equal to:
$a^5 \circ_+ b^5$
if we simply move the notation around...doesn't this read just like:
$5(a\circ_+ b)=5(a) \circ_+ 5(b)$ ...which could be generalized to any number. 
From this explanation, it seems that multiplication's distributive property over addition is already baked into the system without the need to specify it.
What am I missing / incorrectly concluding? Cheers~

REPLY [2 votes]: I'll (mostly) write "$+$" and "$\times$" for the addition and multiplication, respectively, since I think it's ultimately clearer.

You're conflating exponentiation notation in the context with rings with exponentiation notation in the context of (non-abelian) groups: when we're working in a ring, "$a^n$" (for $a$ an element of the ring and $n\in\mathbb{N}$) denotes the multplicative expression $$a\times...\times a\quad\mbox{ ($n$ times)}.$$ In particular, we have $$(a+b)^5=(a+b)\times (a+b)\times (a+b)\times (a+b)\times (a+b)$$ (and not "$(a+b)+(a+b)+(a+b)+(a+b)+(a+b)$" as in your post). 
The point is that we have two different operations at play on the right hand side. Because of this, we can't really use the commutativity of "$+$" since we have to "get around" the multiplication $\times$ somehow.

Note that this actually does match the notation used in the context of abelian groups: for $(G,*)$ an abelian group, "$a*...*a$" ($n$ times) is generally denoted by "$na$" rather than "$a^n$."


Note that $(3)$ is the only axiom which tells you how $+$ and $\times$ relate; in the absence of $(3)$, we would be able to just "glue together" an abelian group operation and an associative operation on a set to get a ring. For example, look at $\mathbb{Z}$ and - switching to unambiguous notation now - let both $\circ_+$ and $\circ_\times$ be the usual addition $+$. This is clearly not a ring, since it satisfies $(1\circ_+1)\circ_\times 1=3$ but $(1\circ_\times 1)\circ_+(1\circ_\times 1)=4$, but it satisfies axioms $(1)$ and $(2)$ from your list.

REPLY [1 votes]: Note that, in general, $a^5\neq a+a+a+a+a$. For example, consider the ring $\mathbb Z$  and let $a=1$.
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