Justifying $\mathbb{Z}$ is a commutative ring with 1.

Question

I am having trouble justifying $\mathbb{Z}$ is a commutative ring with 1 under multiplication and addition.

Book defines a "ring with 1" as a ring $R$ that has the identity if there is an element $1a=a1=a$.

The additive inverse of the ring element $a\in R$ is denoted as $-a$. Is it enough to state,

Do I just show that $a+(-a)=0$ for any $a\in\mathbb{Z}$ for addition? How can I justify it for multiplication?

Answer 1

$\mathbb{Z}$ is a ring and not a field, so you don't have to find multiplicative inverse if that's your concern.

what is $\mathbb{Z}$ ?

Define the following equivalence relation $\operatorname{E}$ on $\mathbb{N} \times \mathbb{N}$ : $$(a,b) \operatorname{E} (c,d) :\Longleftrightarrow a+d = c+b$$

• Setwise, $\mathbb{Z}$ is $\mathbb{N} \times \mathbb{N} /\operatorname{E}$. The idea behind the scene is that the class $\overline{(a,b)}^{\operatorname{E}}$ represents $a-b$, eg $-5$ is $0-5$ or $1-6$ or $2-7$ etc.
• Addition is defined componentwise : the sum of $\overline{(a,b)}^{\operatorname{E}}$ and $\overline{(c,d)}^{\operatorname{E}}$ is $$\overline{(a,b)}^{\operatorname{E}} + \overline{(c,d)}^{\operatorname{E}} := \overline{(a+c,b+d)}^{\operatorname{E}}$$.
• Multiplication is a bit trickier. The product of $\overline{(a,b)}^{\operatorname{E}}$ and $\overline{(c,d)}^{\operatorname{E}}$ is $$\overline{(a,b)}^{\operatorname{E}} \times \overline{(c,d)}^{\operatorname{E}} := \overline{(ac + bd,ad + bc)}^{\operatorname{E}}$$ Can you check that these operations are well defined (this is often the trickiest part)?

What do one need to check?

You need to check that

• $+$ and $\times$ are associative.
• There is a neutral element for addition (try $0:=\overline{(0,0)}^{\operatorname{E}}$).
• $+$ is commutative.
• Every element has an additive inverse (hint : $-\overline{(a,b)}^{\operatorname{E}} = \overline{(b,a)}^{\operatorname{E}}$).
• There is a neutral element for multiplication (try $1 :=\overline{(1,0)}^{\operatorname{E}}$).
• $\times$ is distributive over $+$.
• $\times$ is commutative.

If you edit your question to include the proofs of each of these points, I would be glad to help you.

Answer 2

You can't justify it for multiplication. You have stumbled onto the fact that $\Bbb Z$ is a ring but not a field (it lacks multiplicative inverses).