Why is $\mathbb{Z}_m$ a commutative ring?

Question

I understand for it to be a ring it has to be closed under the addition and product, have inverses for the additive group, distributive and identity in the product.

I can see that it is closed under addition and product, but I can't get the inverses in the additive group.

If $\mathbb{Z}_m$ is composed of equivalence classes ${\underline{0},\underline{1},\underline{2},...,\underline{m-1}}$, I think I have to find an equivalence class $\underline{y}$ for every class $\underline{x}$ such that $\underline{x} - \underline{y} = 0$. Am I right?

And how do I do that?

Answer 1

Let us choose a specific example: $m = 10$. Then

• $1 + 9 = 0$ so $9 = -1$
• $2 + 8 = 0$ so $8 = -2$
• $3 + 7 = 0$ so $7 = -3$
• $4 + 6 = 0$ so $6 = -4$

and so on. This is because $10 = 0$ in $\mathbf{Z}_{10}$.

In general $x + (m - x) = m = 0$ in $\mathbf{Z}_m$.

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You can also think of this as a clock. For example:

• 9 o'clock is 3 hours to midnight. Thus $9 = -3$.
• 11 o'clock is 1 hour to midnight. Thus $11 = -1$.