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Let $R$ be a finite non-zero ring without zero divisors. Denote $R^{*}=R \setminus \{0\}$. Consider the action of the multiplicative semigroup $R^{*}$ on itseslf by left translation - i.e., for $x \in R^{*}$, we have $x.y = xy$ by definition.

I already proved that $R^{*}$ acts on $R^{*}$ by bijections (i.e. for each $x$, the map $f_{x}:R^{*} \to R^{*}$ defined by $f(y) = xy$ is a bijection.

I now have to prove that $R$ has both the left identity and the right identity. I actually had this question posted earlier, and then deleted it because I thought I had figured it out on my own, but I was mistaken. Essentially, what I wound up showing was that for any element $x$ of $R$, $x$ had both a left- and right-inverse.

Now, I am trying to salvage some of what I did and use it to help me prove what it is that I was actually supposed to prove here. To that effect, I was given the hint that to show that $R$ has a left identity, use the fact that $x$ is the left identity if and only if $f_{x} = id$ - i.e., the identity map, $f(y) = x.y = xy = y$.

Here's what I am trying to attempt: Consider any nonzero $y \in R$ and fix $x \neq 0$, $x \in R$. Then, $f(y) = x.y = xy$. Since $R$ is finite, $\exists$ only finitely many distinct powers of $x$. Therefore, $\exists$ some $m$ s.t. $x^{m} = x$. WLOG, let $m>1$. Then, $xy = x^{m}y$, which implies that $x^{m}y - xy = 0\, \implies \, (x^{m}-x)y=0 \, \implies \, x(x^{m-1}-1)y = 0$, and since $x \neq 0$, $y \neq 0$, since $R$ has no zero divisors we must have that $x^{m-1}-1 = 0 \, \implies \, x^{m-1} = 1$. So, all I did was show that there is some element in $R$ that is equal to $1$, but I'm pretty sure this isn't right because 1) I didn't use the hint, and 2) since I tried to use the hint, I should have wound up showing that $x = 1$. Could somebody please help me figure out how to show this correctly?

For proving that $R$ has a right identity, I've got a whole other problem: no hint. See, $f_{x}$ is defined so that $x$ acts on $y$ on the left, so I honestly have no idea how to approach this part.

Please help me figure this out. It's killing me.

Thank you in advance, and please be willing to answer follow-up questions.

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When you write $x^{m-1}-1$, that's the point where you are making a mistake. What is $1$ here? You didn't prove yet that $1$ exists.

But otherwise you are very close to solution. Indeed, let's take non-zero $x$ and find power $m > 1$ such that $x^m = x$. Let $e = x^{m-1}$. I claim that $e$ is both left and right identity.

Indeed, consider some $y \in R$. Let $a = ey$. Now,

$$ey = a \Rightarrow $$ $$xey = xa \Rightarrow $$ $$x(x^{m-1})y = xa \Rightarrow$$ $$x^my = xa \Rightarrow $$ $$xy = xa \Rightarrow $$ $$x(y - a) = 0 \Rightarrow $$ $$ y = a \Rightarrow$$ $$ y = ey$$

That means that $e$ is a left identity. For the right identity everything goes the same way, just consider $yex$ instead of $xey$.

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  • $\begingroup$ thank you so much for that! You really saved the day! $\endgroup$ – ALannister Feb 27 '17 at 17:55
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Actually what you've written can be greatly simplified.

Let $x$ be nonzero. Then there exists a nonzero $e$ such that $ex=x$.

Then $e^2x=ex=x$ as well, and by cancellation $e^2=e$.

Thereafter, $e(y-ey)=(y-ye)e=0$ for any $y\in R$. So, $e$ is a left and right identity.

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