Does an ideal of a ring contain the zero element of the additive group of the ring? If yes does this apply to the multiplicative group? Well, i was wondering about this, because this is the only thing i can deduct right now in a proof in reading. If it turns out the this is not the case i will post the detail in the proof that is confusing me..
 A: Yes. By definition of an ideal, if $r\in I$ and $s\in R$ then $sr\in I$. Take $s=0$ to get that $0r=0\in I$.
A: An ideal always contains the additive identity $0$, as by definition it is an additive subgroup of the additive group structure in the ring. 
The multiplicative structure of the ring (which need not be a group structure!) may or may not have an identity element, namely the multiplicative identity $1$. If the ring does have $1$ and the ideal contains $1$, then necessarily the ideal is the entire ring. Indeed, if $I$ is the ideal and $R$ is the ring, and $r\in R$ is any element, then by definition $r\cdot 1 \in I$ if $1\in I$. So, $R\subseteq I$ and $I\subseteq R$, hence $I=R$. Usually, the more interesting ideals end up being ones that do not contain $1$. 
A: An ideal of a (commutative with $1$) ring $R$ is defined to be 


*

*A subgroup $I$ of the additive group of $R$ 

*for which $rx\in I$ for every $r\in R,\ x\in I$.


Since $I$ is a subgroup of the additive group of $R$, it must contain the zero element.
However, if $I$ contains $1$, the identity of the multiplicative group, then by part $2$ of the definition, for every $r\in R$, $r\cdot 1 \in I$. So $I = R$. Hence, if $I$ is a proper ideal, then $1\notin I$.
