Let $R$ be a ring with unity. Show that $x$ is a left zero-divisor if and only if $x$ is a right zero-divisor.

Suppose, $x$ is a left zero divisor. Then, $\exists y \in R$ such that $xy = 0 \Rightarrow (xy)\cdot 1_R = 0 \Rightarrow (xy)(x^{-1}x)= 0 \Rightarrow (xyx^{-1})x = 0$. Hence, $x$ is a right zero divisor.

Now, suppose, $x$ is a right zero divisor. Then,$\exists y \in R$ such that $yx = 0 \Rightarrow 1_R\cdot(yx) = 0 \Rightarrow (xx^{-1})(yx)= 0 \Rightarrow x(x^{-1}yx) = 0$. Thus, $x$ is a left zero divisor.

But, this isn't correct. So, please tell me what's wrong with my proof.

  • 2
    $\begingroup$ If $\;x\;$ is a zero divisor it doesn't exist $\;x^{-1}\;$ $\endgroup$
    – Timbuc
    Apr 10, 2015 at 17:25
  • $\begingroup$ One problem is that for example in your first proof, $xyx^{-1}$ is already zero since $xy=0$. Also, there is no reason that $x$ should be invertible. $\endgroup$
    – rogerl
    Apr 10, 2015 at 17:25
  • $\begingroup$ You have received (good) answers to the question as it was originally asked. Do not edit the question to something different. If you missed an essential condition, ask a new question with that condition, don't edit the question so that the answers become invalid. $\endgroup$ Apr 17, 2015 at 21:31
  • $\begingroup$ Ok then, I'll ask a new question. $\endgroup$
    – Jellyfish
    Apr 17, 2015 at 21:33

4 Answers 4


Your proof is wrong because you use $x^{-1}$ that cannot exist if $x$ is either a left or a right zero divisor.

On the other hand, it's impossible to prove the assert, because it's false.

Consider the ring of endomorphisms $R$ of an infinite dimensional vector space, say $V$ with a countable basis $\{x_n: n\in\mathbb{N}\}$.

Consider the endomorphism $f$ defined by $x_n\mapsto x_{n+1}$. Then if you consider $g$ defined by $g(x_0)=x_0$ and $g(x_n)=0$ for $n>1$, we have $$ g\circ f(x_n)=0 $$ for all $n\in\mathbb{N}$, so $g\circ f=0$ and $f$ is a right zero divisor. However, if $h\ne0$, then $h(x_n)\ne0$ for some $n$ and so $$ f\circ h(x_n)\ne0 $$ as it is readily proved, because $f$ is an injective map. So $f$ is a right zero divisor that is not a left zero divisor.

If you want an example the other way around, just consider $f$ as an element of the opposite ring $R^{\mathrm{op}}$.

  • $\begingroup$ @abcd1234 You can see the example suggested by avid19 $\endgroup$
    – egreg
    Apr 10, 2015 at 20:11
  • $\begingroup$ what about this C* algebraic problem ?math.stackexchange.com/questions/1687407/… $\endgroup$ Mar 10, 2016 at 7:15
  • 1
    $\begingroup$ @AliTaghavi Sorry: I upvoted your question, but I'm no expert in C*-algebras. $\endgroup$
    – egreg
    Mar 10, 2016 at 7:26

Although the claim is incorrect, the following is nonetheless true:

Proposition. Let $R$ denote a ring (with unity). Then if every non-zero element of $R$ has a left multiplicative inverse, then every non-zero element of $R$ has a two-sided multiplicative inverse.

Proof. Suppose $x_0$ is non-zero. Then $(*) \;\; x_1x_0=1$ for some non-zero $x_1$. Hence $(**)\;\; x_2x_1 = 1$ for some non-zero $x_2$. Now consider the product $x_2 x_1 x_0$. We have:

$$x_2 x_1 x_0 = (x_2 x_1) x_0 = 1x_0 = x_0$$

$$x_2 x_1 x_0 = x_2 (x_1 x_0) = x_2$$

So $x_0=x_2$. Hence from $(**)$ we have $x_0x_1=1$. But we already know that $x_1 x_0 = 1$ from $(*)$.


You're assuming that a left/right zero divisor has a left/right inverse. This is false.

To see why, suppose $x$ is a left zero divisor. Then there is $y \neq 0$ such that $xy = 0$. If $x$ had a left inverse $x^{-1}$, then $y = x^{-1}xy = x^{-1}0 = 0$, which is a contradiction. Likewise if $x$ is a right zero divisor with right inverse.


How do you know $x^{-1}$ exists? Also if it did, $xyx^{-1}=(xy)x^{-1}=0x^{-1}=0$. The statement is false, by the way:



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