0

Let $R$ be a finite ring with unity. Let $x \in R$. Prove that $x$ is a Left Zero Divisor $\iff$ x is a Right Zero Divisor.

My attempt Suppose $x$ is a LZD. Then, $\exists y \in R$ such that $xy = 0$. Since, $R$ has unity, let $y = 1 \Rightarrow xy = x \cdot 1 = x = 1 \cdot x = yx = 0$. Thus, $x$ is a RZD. And, the converse is similar. Is this correct?

user26857
  • 53,190
Jellyfish
  • 1,723
  • 3
    Can I let whacka's rep = 1,000,000? Since the number one million exists. It's important you actually understand the meaning of what you're saying. Just because there exists a $y$ for which $xy=0$ doesn't mean you get to choose what $y$ is! Indeed, if $xy=0$ were true for $y=1$, that would mean $x=0$! But the term "zero divisor" is restricted to nonzero elements, so this doesn't make any sense. – anon Apr 17 '15 at 21:50
  • @Ilham, I thought since $R$ has unity and $y \in R$, maybe $y$ could be equal to $1$. – Jellyfish Apr 17 '15 at 21:53
  • @abcd1234 I haven't tried this, but a suggestion.. Try using http://math.stackexchange.com/questions/60969/every-nonzero-element-in-a-finite-ring-is-either-a-unit-or-a-zero-divisor

    and this argument in the question of the next link, which is flawed when finiteness is not assumed:

    http://math.stackexchange.com/questions/1228884/x-is-a-left-zero-divisor-iff-x-is-a-right-zero-divisor

    Show that every non-zero element in the finite commutative ring that is not a zero divisor is a unit, and hence has an inverse.

     – Ilham Apr 17 '15 at 21:55
  • You should require that $y \ne 0$. And why are you taking $y=1$? This does not imply that $xy=0$. And you did not use that the ring is finite. You can take the Ring of endomorphisms over an infinite dimensional vectorspaces as counterexample, you have to look at monos and epis. – user60589 Apr 17 '15 at 21:55
  • The set of human beings has Emma Watson and my girlfriend is in the set of human beings, so maybe Emma Watson is my girlfriend? Thus, when I am in the process of logically proving to my friend that I am awesome, I can cite the fact that Emma Watson is my girlfriend! This is not how logic works. – anon Apr 17 '15 at 22:14
  • If my mockery of your logic is being seen as a mockery of you as a person, then indeed perhaps it is time I stopped talking. I'll shut up. – anon Apr 17 '15 at 22:18
  • $x$ is a left zero divisor if $\exists y \neq 0 \in R$ s.t $xy = 0$. Therefore, I thought since $y \neq 0$, $y$ could be $1$. – Jellyfish Apr 17 '15 at 22:22
  • @whacka Please don't spread a non-sense like "the term "zero divisor" is restricted to nonzero elements". See http://en.wikipedia.org/wiki/Zero_divisor#Zero_as_a_zero_divisor or learn some basic commutative algebra. – user26857 Apr 18 '15 at 07:34

2 Answers2

2

Let $x$ be a left zero divisior but not a right zero divisior.

Then set $f:R\to R$ by $f(r)=rx$. Notice that $f$ is one to one as, $r_1x=r_2x\implies (r_1-r_2)x=0\implies r_1-r_2=0$ as $x$ is not a right zero divisior.

Then $f$ must be onto as $R$ is finite which means there exist $r$ s.t. $1=rx$.

Since $x$ is left zero divisior, $xy=0\implies rxy=0 \implies 1y=y=0$ Contradiction. I left other direction to you.

mesel
  • 15,125
1

Your method doesn't really work, since you cannot assume that $y =1 $. In fact, $1$ cannot be a zero divisor at all, since $1 \cdot x = 0$ is equivalent to $x = 0$.

You should examine the set $\{r \cdot x: r\in R\}$, for $x$ a left zero-divisor. Is it possible that $1 = rx$ if $xy = 0$ for some nonzero $y$? What can you infer from $r_1x = r_2x$ for $r_1\ne r_2$?

Rolf Hoyer
  • 12,619