Ideal of a ring with element $-1$

Question

Consider the ring $(\Bbb{Z}, +, \cdot)$. Let $I$ be an ideal of the ring that contains the element $-1$. Show that $I = \Bbb{Z}$

Could some one please explain how to solve this.

Answer 1

$I$ is in particular the additive subgroup of $\Bbb{Z}$ and $-1 \in I$ implies $1 \in I$ since it is the additive inverse of $-1$ and so $I=\Bbb{Z}$

Answer 2

This works in pretty general rings. If $R$ is a (not-necessarily) commutative ring with unit $1$, and $I \subset R$ is a (left, right, or two-sided) ideal, then

$-1 \in I \Longrightarrow I = R; \tag 1$

for $I$ being an ideal, it is a subgroup under the "$+$" operation of $R$; thus $-i \in I$ for $i \in I$; but then $1 = -(-1) \in I$, so for $r \in R$,

$r = r1 \in I \Longrightarrow R \subset I \Longrightarrow R = I. \tag 2$

So basically, I solve it by applying the facts that $I$ is an additive subgroup (to get $1 \in I$), and that $I$ is closed under multiplication by $r \in R$, that is $rI \subset I$, to get $r \in I$.