Is $Z$ generated by $\pm 1$ as a ring

Question

I know that $Z$ = <$\pm 1$> as a group,$Z$ generated by $\pm 1$ as a ring. If yes, how do you show that?

Answer 1

The ring generated by a subset contains the additive group generated by this subset.

Answer 2

$(R,..+)$ is a ring if satisfying the following three sets of axioms, called the '''ring axioms:

1. ''R'' is an [[abelian group]] under addition. In your case $\mathbb{Z}$ is.
2. ''R'' is a [[monoid]] under multiplication, meaning that: $$(a · b) · c = a · (b · c)$$ for all $a,b,c\in R$.

There is an element 1 in R such that $a · 1 = a$ and $1 · a = a$ for all $a\in R$ (1 is the [[multiplicative identity]]).

$\color{red}{Note~ that~ the~ existence~ of~ 1~ is~ not~ assumed~ by~ some~ authors.}$

3. Multiplication is [[distributive]] with respect to addition: for all $a,b,c\in R$
   (left distributivity) $$a.(b+c)=(a.b)+(a.c)$$ (right distributivity) $$(b+c).a=(b.a)+(c.a)$$

all the above hold in $\mathbb{Z}$. In order to show it is generated by $\pm1$ it is enough to show every elements is the sum of finitely many $\pm1$.