Integers form Totally Ordered Ring

Theorem

The structure $\left({\Z, +, \times, \le}\right)$ is a totally ordered ring.

Proof

From Integers form Commutative Ring, $\left({\Z, +, \times}\right)$ is a commutative ring.

From Integer Addition forms Totally Ordered Group, $\left({\Z, +, \le}\right)$ is a totally ordered group.

We need to show that $\le$ is a compatible ordering on $\Z$.

That is, that:

$(1): \quad \le$ is compatible with $+$

$(2): \quad \forall x, y \in \Z: 0 \le x, 0 \le y \implies 0 \le x \times y$

The first one follows from the fact that $\left({\Z, +, \le}\right)$ is a totally ordered group.

Now, from Multiplicative Ordering on Integers, we have:

$\forall z, x, y \in \Z, 0 < y: z \le x \iff y \times z \le y \times x$

Let $z = 0, 0 \le x$.

Then:

$0 = x \times z \le x \times y = y \times x$

So if $y \ne 0$, the condition $0 \le x, 0 \le y \implies 0 \le x \times y$ holds.

Now if $y = 0$, we have $x \times y = 0$ and thus $0 \le x \times y$.

Thus $\le$ is compatible with $\Z$, and the result follows.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 23$