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I am looking to prove the following properties of $\mathbb{Z}_n$

  1. Commutativity and associativity of addition
  2. $[0]$ is identity element for addition and existence of additive inverses
  3. Commutativity and associativity of multiplication and $[1]$ as the multiplicative identity.

I'm mainly looking for confirmation that my solutions are correct. I'm wondering primarily because they seem so simplistic, almost trivial, leveraging mainly the facts that $[a] + [b] = [a +b]$ and $[a][b] = [ab]$

$1\alpha.$ $[a] + [b] = [a+b] = [b + a] = [b] + [a]$. Since regular addition is commutative.

$1\beta.$ $([a] + [b]) + [c] = [a+b] + [c] = [a+b+c] = [a] + [b+c] = [a] + ([b]+[c])$

$2.\alpha$ $[a] + [0] = [a+0] = [a]$

$2.\beta$ $[a] + [-a] = [a + (-a)] = [a-a] = [0]$

$3.\alpha$ $[a][b] = [ab] = [ba] = [b][a]$. Since regular multiplication is commutative.

$3.\beta$ $([a][b])[c] = [ab][c] = [abc] = [a][bc] = [a]([b][c])$

$3.\gamma$ $[a][1] = [a \cdot 1] = [a]$

jjagmath
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Numerical Disintegration
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    I think these "simplistic" verifications really are all that's needed. Indeed these are all properties that quotient rings inherit from the original rings, so there's certainly nothing special that's needed about $\Bbb Z_n$ to prove them. – Greg Martin Dec 15 '21 at 02:40
  • Thank you for the verification Greg. I am always a bit paranoid when my answers are so short and simplistic so I just needed to be sure. I appreciate the insight, makes sense why having regular mult. and add. being commutative is all that'd really necessary to prove them. – Numerical Disintegration Dec 15 '21 at 02:55
  • Avoid the use of $*$ to denote multiplication. That's a common practice in programming, not in Mathematics – jjagmath Dec 15 '21 at 02:57
  • Noted. I usually use a circle but wasn't sure how to format that in. I'll keep that in mind for further questions I post. Thanks.

    I see in your edit is \cdot. Okay thanks again.

     – Numerical Disintegration Dec 15 '21 at 03:04
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    When writing a proof at this level, would you not need to show $[a]+[b] = [a+b]$? – Eric Towers Dec 15 '21 at 03:09
  • Yes, that's the correct idea, and the proofs are indeed quite simple when done that way. But you should give a bit more detail, e.g. as in this answer in the linked dupe. Once you know how to do one of these proofs the others follow all essentially the same way. – Bill Dubuque Dec 15 '21 at 08:28
  • @Eric That is the definition of addition in the quotient ring, so no proof is needed (normally that has already been shown well-defined before questions like this are posed). And even if one is not working with quotient rings, that still is usually assumed known before such exercises are posed. – Bill Dubuque Dec 15 '21 at 08:30
  • See here for elaboration on the prior comment. – Bill Dubuque Dec 15 '21 at 11:01
  • @BillDubuque : I see the "elementary-number-theory" tag, which says to me that this discussion of modular arithmetic occurs in a context where this is the first encounter with a ring of positive characteristic. If so, facts about quotient rings are not yet available -- one is still showing that the operations are well-defined. My dim recollection of the first version of this question I ever saw required passing through "$(a+k_1 n) + (b+k_2 n)$" to show commutativity and similarly for the other properties. – Eric Towers Dec 15 '21 at 13:08

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