Ideal subgroup of the additive group of a ring (R,+, *)

Question

I just started learning ring theory. My professor defined an ideal as follows:

An $\textbf{ideal}$ in a ring R is a nonempty subset $I\subseteq R$ such that if $a\in I$ and $r\in R$, then $ar,ra\in I$ and if $a,b\in I$, then $(a+b)\in I$.

I want to show that by this definition, $I$ is a subgroup of $(R,+)$, the additive group of the ring. I can see that $0\in I$, since $0\in R$, so $0\cdot a = 0 \in I$. Associativity follows from addition, and closure follows from the definition as well. I can't seem to figure out why $I$ contains additive inverses, though.

Answer 1

In any ring, $(-1)\cdot a = -a$, because $a + (-1)\cdot a = (-1+1)\cdot a = 0\cdot a = 0$.

Therefore, if $I$ is an ideal of a ring $R$, then for every $a\in I$, $(-1)\cdot a = -a \in I$.

Answer 2

$-1\in R$ so $(-1)a\in I$ $~~~~~~$