Are there different left and right ideals in a ring without identity?

Question

For a non commutative ring without identity, is it possible that there will be right and left ideals which are different?

Answer 1

To elaborate on Orat's suggestion:

If $R=2\Bbb Z$, $\begin{bmatrix}R&0\\R&0\end{bmatrix}$ is a left ideal of $M_2(R)$, and $\begin{bmatrix}R&R\\0&0\end{bmatrix}$ is a right ideal of $M_2(R)$.

You could let $R$ be any ring, really. It doesn't have anything to do with identity.

To give a different construction entirely, take a field homomorphism $\sigma: F\to F$ such that $[F:\sigma F]=n$ where $n$ is a natural number greater than $1$.

Form the twisted polynomial ring $F[x;\sigma]$ where for each $a\in F$, $xa:=\sigma(a) x$ and the rest of the ring multiplication is defined linearly. Then let $R=F[x;\sigma]/(x^2)$. The set $Fx$ is a simple left ideal, and in fact the ring has exactly three left ideals. But as a right ideal $Fx$ is a direct sum of $n>1$ right ideals. Therefore there are many right ideals of $R$ which are not left ideals.