Proving that $\mathbb{Z}(p)$ is a ring

Question

Let $p$ be prime. Prove that $\mathbb{Z}(p)= \{a/b\ |\ a,b \text{ are elements of $\mathbb{Z}$ and $\gcd(b,p)=1$}\}$ is a ring. (This is called the ring of integers localized at $p$.)

What should be the first step that I should do? Should I show that it is closed under addition, multiplication and inverse since $\mathbb{Z}$ must contain $0$.

Any hint..??

Answer 1

Any time you want to prove that a set $(R,\cdot, +)$ is a ring, you must prove the same things:

1. $(R,+)$ is an abelian group
2. $\cdot$ is an associative operation
3. For all $a,b,c$ in $R$, the equalities $a\cdot (b+c) = a\cdot b + a\cdot c $ and $(a+b)\cdot c = a\cdot c + b\cdot c$ hold.

Answer 2

I am going to use the notation $\mathbb{Z}_{(p)}$ for $\mathbb{Z}(p)$.

Your definition of $\mathbb{Z}_{(p)}$ suggest that you view it as subset of $\mathbb{Q}$ with the multiplication and addition inherited. This means that you actually should show that $\mathbb{Z}_{(p)}$ is a subring of $\mathbb{Q}$.

This boils down to:

1. Show that $1 \in \mathbb{Z}_{(p)}$.
2. Show that for all $a,b \in \mathbb{Z}_{(p)}$, also $a - b \in \mathbb{Z}_{(p)}$.
3. Show that for all $a,b \in \mathbb{Z}_{(p)}$, also $a \cdot b \in \mathbb{Z}_{(p)}$.

Alternatively, instead of (2.), you can show

• for all $a,b \in \mathbb{Z}_{(p)}$, also $a + b \in \mathbb{Z}_{(p)}$, and $-a \in \mathbb{Z}_{(p)}$.