$aR$ and $Ra$ ideals but $aR\ne Ra$ It's easy to prove the following:

Let $R$ be a ring and $a\in R$. Then we have the following sequence of implication:
$$
a\text{ is central}\implies aR = Ra\implies aR, Ra \text{ are ideals}
$$

Now I began contemplating on the converse directions. I was able to come up with the following counterexamples:

*

*$aR = Ra$ but $a$ not central: Easy. Take $R := \mathbb R^{2\times 2}$ and $a$ to be any invertible non-central matrix.


*$Ra$ an ideal although $aR\ne Ra$: A bit artificial. Take $R := \bigl\{ \left[\begin{smallmatrix}x & 0\\
y & 0\end{smallmatrix}\right] : x, y\in\mathbb R \bigr\}$ (this is a left ideal of $\mathbb R^{2\times 2}$) and $a := \left[ \begin{smallmatrix}1 & 0\\
0 & 0\end{smallmatrix} \right]$.
Now I am looking for an example where $aR$, $Ra$ are both ideals, but $aR\ne Ra$ (or prove that this doesn't happen). And I am stuck. (In example 2 above, $aR = R$ which is trivially an ideal, hence I am not done.)
I however make the following observations:

*

*$R$ has to be non-commutative.

*$a$ has to be non-invertible. $R$ can't have an identity. (As Jeremy's answer shows below.)

*$RaR\subseteq aR\cap Ra$.

Any lead/hint from here?

Unless I say left or right, I mean two-sided.
I allow rings to be non-commutative and not have identity.
 A: If $R$ is a unital ring and $aR$ and $Ra$ are both two sided ideals, then $aR=Ra$.
$aR$ is the smallest right ideal containing $a$, so if it is a two-sided ideal then it must be the smallest two-sided ideal containing $a$.
Similarly, $Ra$ is the smallest left ideal containing $a$, so if it is a two-sided ideal then it must be the smallest two-sided ideal containing $a$.
So if both $aR$ and $Ra$ are two-sided ideals, they are both the smallest two-sided ideal containing $a$, and so they are equal.
However, if R is non-unital then there are counterexamples.
For example, let $R$ be the three-dimensional nonunital algebra over a field spanned by $a$, $b$, $ba$, with $a^2=ab=b^2=0$. Then $aR=0$, a two-sided ideal, and $Ra$ is the one-dimensional two-sided ideal spanned by $ba$.
This example can be realised as a ring of matrices. $R$ is the ring of matrices of the form $\begin{pmatrix}0&0&0\\x&0&y\\z&0&0\end{pmatrix}$, where $x$, $y$ and $z$ are real numbers, and $a=\begin{pmatrix}0&0&0\\0&0&0\\1&0&0\end{pmatrix}$.
Then $aR=\{0\}$, and $Ra$ is the set of matrices of the form $\begin{pmatrix}0&0&0\\y&0&0\\0&0&0\end{pmatrix}$, both of which are two-sided ideals.
