Commutative Ring Ideals So I'm having some issues on fully believing my response to the following: 

Let $R$ be a commutative ring. Is $xR$ always an ideal? Prove or show a counterexample. 

So at first glance I believe this is true because if we have a right ideal, then by commutativity it follows that it is also a left ideal and thus a two sided ideal. But I also have this thought that unity has some importance for ideals but I have not found a counterexample to this statement with this key property. 
 A: Yes, it is always an ideal. As for the importance of unity, two things come to my mind that may have something to do with this.


*

*Assuming all rings are required to be unital, a subring would have to contain the unity, but not an ideal. Maybe you're confusing these two definitions with each other?

*Also, there's this property (theorem, lemma) that if an ideal contains the unity, then it's the entire ring.
A: The subset $xR=\{xr:r\in R\}$ is certainly an ideal of $R$:


*

*$0=x0$

*$xr-xs=x(r-s)$

*$(xr)s=x(rs)$


Thus $xR$ is a right ideal, which is automatically also a left ideal in case $R$ is commutative.
What can fail when $R$ is not supposed to have an identity, is that $x\in xR$. For instance in the ring (without identity) $R=2\mathbb{Z}$, if $x=4$, then $x\notin xR$, because $4(2\mathbb{Z})=8\mathbb{Z}$.
In this case, if you want the least ideal containing $x$, you have to consider
$$
xR+\mathbb{Z}x=\{xr+nx:r\in R, n\in\mathbb{Z}\}.
$$
When $R$ has an identity, $xR+\mathbb{Z}x=xR$.
A: If you mean a proper ideal, then $x$ must be non-invertible, but if you just say an ideal, then $xR$ is always an ideal.
