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My question is:

Let $R$ be a ring with the unity $e$ and $a \in R$. If $a^{\circ}\triangleq\{x\in R \;| \;ax=0\}=\{0\}$, as the following conter-example given by Matthias Klupsch, we know that it is not left invertible in general. Then what additional conditions on $R$ or on $a$ will make that $a$ is left invertible?

If we have known $a$ is right invertilbe, then what will happen?

${\bf Notes:}$ Let $A$ and $B$ be two sets, and $f : A \to B$ be a mapping, then it is well-known that:

(1) $f$ is injective iff $f$ is left invertible.

(2) $f$ is surjective iff $f$ is right invertible.

Thanks!

  • 8
    If you take $R = \mathbb{Z}$, you have $2^\circ = \{0\}$, but $2$ is not left-invertible. Generalizing this example, if your statement would be correct, then every integral domain would be a field.2012-10-27
  • 2
    Maybe you meant the converse?2012-10-27
  • 1
    Thanks very much! But if we already have $a^{\circ}\triangleq\{x\in R \;| \;ax=0\}=\{0\}$, then what additional conditions will make that $a$ is left invertible?2012-10-27
  • 1
    For commutative rings, an element is invertible if and only if it is not contained in any maximal ideal.2012-10-27
  • 0
    Why the tag [banach-algebras]?2012-10-27

3 Answers 3