I'm experimenting around with ring ideals (perhaps ideals is always for rings, so when speaking of ideals we always refer to these ring subsets?), and my book gives me the definition that an ideal $I$ is a subset of a ring $R$, for which the elements are closed under addition and $ra\in I$ for all $r\in R$ and $a\in I$.
Wikipedia say that an ideal is an additive subgroup which I clearly see, but I suspect that $I$ is also a subring to $R$. I cannot find any info on this in my book, but I tried to apply the "subring criterion" theorem (let $S$ be a subset of a ring $R$): (i) additive and multiplicative closures, (ii) if $a\in S \implies -a \in S$ and (iii) $S$ contains the identity.
This is what I did:
(i) addition follows from the definition, but for multiplication let $a,b \in I$ then for any $r_1,r_2 \in R$ we get $r_1ar_2b = (r_1ar_2)b \in I$ because $r_1ar_2 \in R$. Same argument for $a$, thus closure for multiplication.
(ii) This holds from definition of a ring.
(iii) $I$ is an additive subgroup hence $0\in I$.
So, does my "proof" hold? Is it really true that an ideal is always a subring? If it is not true, ca anyone illustrate some counter example?