localised ring $\mathbb{Z}_{(2)}$ is an integral domain

Question

Given the localised ring $\mathbb{Z}_{(2)}=\{\frac{a}{b}:a,b \in \mathbb{Z}, 2 \nmid b \}$, I want to show that this is an integral domain.

We choose some fraction $ \frac{a}{b}\in \mathbb{Z}_{(2)}$,where $a \in \mathbb{Z}$ and $b \in \mathbb{Z}$, such that $2 \nmid b$ and pick up another fraction $ \frac{a'}{b'}\neq 0\in \mathbb{Z}_{(2)}$ with $a' \in \mathbb{Z}$ and $b' \in \mathbb{Z}$ such that $2 \nmid b'$. We look at the term $ \frac{a}{b}*\frac{a'}{b'}=0$. Can we just conclude that $a$ and $b$ have to be zero because the only zero divisors in $\mathbb{Z}$ are the zeroes? How could I argue alternatively with the prime ideal $(2)$ ?

Answer 1

Let $A$ be a commutative ring with unit and $S$ multiplicative subset of $A$ (contains $1$ by definition), and $S^{-1}A$ the localization. Then $\frac{a}{s} \frac{a'}{s'} = 0$ means that there is an $s'' \in S$ such that $s''(aa' \times 1 - 0 \times ss') = 0$ in $A$. (As $\frac{0}{1}$ is the localization's zero.) With $A = \mathbf{Z}$ and $S = A \backslash \mathfrak{p}$ where $\mathfrak{p} = (2)$ which is the setup you are dealing with, one sees that $a$ or $a'$ must be zero.

Remark. If $\mathbf{Z}_{(2)}=\{\frac{a}{b} \in\mathbf{Q}\;|\;a,b \in \mathbf{Z}, 2 \nmid b \}$ then $0 = \frac{a}{b} \frac{c}{d} = \frac{ac}{bd}$ implieds that $a$ or $c$ is zero, same conclusion.

Answer 2

I think that the slickest proof of this is that $\mathbb{Z}_{(2)}$ embeds into $\mathbb{Q}$. Then since $\mathbb{Q}$ is a field, all subrings must be domains.