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Prove that the ring $(\Bbb Z_n, +_n, \cdot_n)$ is a commutative ring with unity.

I know how to prove this for a particular integer $n=5, 6 ,7$ etc but I don't know how to prove it for the general case $n$.

How do I draw the composition table for $n$ values?enter image description here

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  • $\begingroup$ If you can do it for $n=5$, I really doubt that it should be too difficult to adapt your proof with $n$ in place of $5$. $\endgroup$ Commented Sep 20, 2016 at 17:59
  • $\begingroup$ The reason I ask is because I would have to draw the composition table for n as a prime number and also n as a composite number ... Which didn't seem correct .... Also how would I do the multiplication modulo table.... As n can be any number ... $\endgroup$
    – VCA
    Commented Sep 20, 2016 at 23:40
  • $\begingroup$ Like in the picture you see.... $\endgroup$
    – VCA
    Commented Sep 20, 2016 at 23:50
  • $\begingroup$ See this answer for an overview. $\endgroup$ Commented Oct 6, 2022 at 14:38

3 Answers 3

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Since you want to avoid thinking about $\mathbb{Z}_n$ as a quotient ring, write $\mathbb{Z}_n=\{[a]\mid a\in\mathbb{Z}\}$, where $$[a]=\{b\in\mathbb{Z}\mid a\equiv b \mbox{ (mod n) }\}$$ is the congruence class of $a$. Then, addition and multiplication are defined by $$[a]+[b]=[a+b]\;\;\mbox{ and }\;\;[a][b]=[ab].$$ We now have enough to prove that $\mathbb{Z}_n$ is a commutative ring (the ring structure being inherited from $\mathbb{Z}$). For example, associativity of addition can be proved as follows: $$ \begin{align} ([a]+[b])+[c]&=[a+b]+[c]\\&=[(a+b)+c]\\&=[a+(b+c)]\\&=[a]+[b+c]\\&=[a]+([b]+[c]). \end{align} $$ All the other axioms can be proved in the same manner.

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Note that $R=\Bbb Z/n\Bbb Z$ is a quotient of the ring $\Bbb Z$, so if you let

$$\phi_n:\Bbb Z\to R$$

be the usual "mod n" map, then each element of $R$ is of the form $\phi_n(k)$ for some integer, $k$. But then

$$\phi(k_1)+\phi_n(k_2)=\phi_n(k_1+k_2)=\phi_n(k_2+k_1)=\phi_n(k_2)+\phi_n(k_1)$$

showing R is commutative because $\phi_n$ is a ring homomorphism. Similarly we can see that $phi_n(1)$ multiplies all elements of $R$ and preserves them as

$$\phi_n(1)\cdot\phi_n(k)=\phi_n(1\cdot k) = \phi_n(k)$$

The basic idea is that $R$ inherits almost all of its structure from $\Bbb Z$ which you already know a lot about, and homomorphisms preserve ring operations.

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  • $\begingroup$ Someone edited my question .. in a wrong manner.. I never wanted the proof for a quotient ring $\endgroup$
    – VCA
    Commented Sep 20, 2016 at 14:19
  • $\begingroup$ @VCA assuming your question is as you intend it, this proof should still work. $\Bbb Z_n$ is a quotient ring (by definition) so the key to the proof is exploiting that fact to get your desired result. $\endgroup$ Commented Sep 20, 2016 at 14:58
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I feel like there is no way around looking at the integers modulo $n$ as a quotient ring of $\mathbb Z$. That is, it is a quotient ring of the form $\mathbb Z/I$ where $I=(n)=\{nz\mid z\in \mathbb Z\}$.

In fact, I don't think there's any reason not to prove that $R/I$ has a ring structure for any ideal $I$ in the ring $R$. The main requirement is to prove that the set of cosets $R/I=\{r+I\mid r\in R\}$ has two well-defined operations that make it a ring, namely:

$$ r+I+s+I=(r+s)+I $$

and

$$ (r+I)(s+I)=rs+I $$

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  • $\begingroup$ Someone edited my question .. in a wrong manner before... I never wanted the proof for a quotient ring $\endgroup$
    – VCA
    Commented Sep 20, 2016 at 14:19
  • $\begingroup$ @VCA The thing you are talking about is defined as a quotient ring, so I don't really see the sense in avoiding thinking about it as such. It's fine to write up everything in terms of $R=\mathbb Z$ without thinking about general rings. $\endgroup$
    – rschwieb
    Commented Sep 20, 2016 at 14:32

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