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

  • 1
    $\begingroup$ How about matrix ring $M_2(2\mathbb{Z})$? $\endgroup$
    – Orat
    Commented Jul 23, 2015 at 9:57
  • $\begingroup$ What exactly do you mean? Do you want a right ideal which is not a left ideal and a left ideal which is not a right ideal? Or do you want such a ring $R$ such that the set of its right ideals and the set of its left ideals intersect trivially (i.e., only at $0$ or $R$ can be both right and left ideals simultaneously)? $\endgroup$ Commented Jul 23, 2015 at 9:59
  • $\begingroup$ Thanks.Can you help me for regular elements not for matrix. Suppose I have Ring Z modulo 6 which is commutative and for this case left ideal is equal to right ideal. My doubt is, if there is any non commutative ring with different right and left ideals? $\endgroup$
    – Amanda
    Commented Jul 23, 2015 at 10:08
  • $\begingroup$ "help me for regular elements not for matrix" What does that mean? The word "regular" is used for lots of things in algebra, but I suspect you don't mean any of them, but rather you think that matrices can't be considered as elements for some reason. Matrices over rings are certainly elements of rings. $\endgroup$
    – rschwieb
    Commented Jul 23, 2015 at 17:42
  • $\begingroup$ For matrix we can show that there is a right and left ideal which is not equal. But my question is for any set whose elements are regular which means its inverse exist, and those elements should not be commutative. $\endgroup$
    – Amanda
    Commented Jul 26, 2015 at 9:29

1 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.


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