0
$\begingroup$

Pinter's "A Book of Abstract Algebra" defines a ring $A$ stipulating the following $3$ axioms:

  1. $A$ with addition alone is an abelian group.
  2. Multiplication is associative.
  3. 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~

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ I was using the exponent notation to reflect how one thinks about "repeated composition" in groups. $\endgroup$ – S.Cramer Jan 11 '20 at 17:19
  • 2
    $\begingroup$ @S.Cramer But the point is that you're using addition sometimes when you should be using multiplication: the right hand side of your expression shouldn't just use "$\circ_+$," regardless of how you want to notate it. $\endgroup$ – Noah Schweber Jan 11 '20 at 17:21
  • 1
    $\begingroup$ @S.Cramer Serendipity - see the bottom of my edit! $\endgroup$ – Noah Schweber Jan 11 '20 at 17:26
  • 1
    $\begingroup$ @S.Cramer No, that's not right. The "iterated addition" version of multiplication is not the ring operation: rather, it's a multiplication between a natural number and a ring element (just as in an abelian group). Distributivity governs how ring addition interacts with ring multiplication. $\endgroup$ – Noah Schweber Jan 11 '20 at 17:42
  • 1
    $\begingroup$ @S.Cramer Exactly. (Incidentally, if you've seen vector spaces already, there's an analogy between them, abelian groups, and rings. An abelian group is (or rather, can be canonically thought of as) a "vector space over $\mathbb{Z}$" (or more precisely, a $\mathbb{Z}$-module since $\mathbb{Z}$ doesn't allow division). A ring is more complicated since it allows multiplication of "vectors;" the analogous fact is that a ring "is" a $\mathbb{Z}$-algebra.) $\endgroup$ – Noah Schweber Jan 11 '20 at 17:45
1
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ I guess I am familiar with what exponents mean in the context of single operators. For example, in groups, $a^5$ would just mean $a \circ a \circ a \circ a \circ a$ I can edit the post to clarify. $\endgroup$ – S.Cramer Jan 11 '20 at 17:21
  • $\begingroup$ My point is that $(a \circ_+ b)^5 = (a\circ_+ b) \circ(a\circ_+ b) \circ (a\circ_+ b) \circ (a\circ_+ b) \circ (a\circ_+ b)$ and not what you wrote in the post. I presume that $\circ_+$ is the addition and $\circ$ is the multiplication. $\endgroup$ – Shivering Soldier Jan 11 '20 at 17:26
  • $\begingroup$ Ahhh, so the definition of exponents changes when we are dealing with two operators, such as in a ring. $\endgroup$ – S.Cramer Jan 11 '20 at 17:28
  • 2
    $\begingroup$ @S.Cramer Actually, it's already changed in the switch from general groups to abelian groups - see my edit. $\endgroup$ – Noah Schweber Jan 11 '20 at 17:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.