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If $R$ is a finite commutative ring with identity and $a \in R$, prove that $a$ is either a zero divisor or a unit.

If $a$ is a zero divisor we are done. If $a$ is not a zero divisor and $a \neq 0_{R}$, then $ab = 0_{R}$ implies that $b = 0_{R}$ otherwise $a = 0_{R}$. Hence, $R$ is an integral domain and since every finite integral domain is a field (Theorem $3.9$ in my book), $R$ is a field and so $a$ is a unit.

Now I don't know what is wrong with it but it has to be since there is a hard question a little later that one needs to show a nonzero finite commutative ring with no zero divisors is a field and this seems to follow easily from that.

stupidproofs123
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2 Answers2

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Let $a \in R$. We must prove that $a$ is a zero divisor or a unit. If $a$ is $0$ or a zero divisor then we are done. So assume that $a$ is not a zero divisor.

Now consider the set $\{a,a^2,a^3, \ldots\}$. By closure under ring operations this must be a subset of $R$ and hence finite so this means $a^k=a^j$ for some $k< j$. Consequently $$a^{k}(1-a^{j-k})=0.$$

Since $a$ is not a zero divisor so $a^k \neq 0$ and is also not a zero divisor, this means $$a^{j-k}=1.$$ Thus $a$ is invertible.

darij grinberg
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Anurag A
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  • If $a^{k} = a$ for some $k$ is a zero divisor we are done. If $a^{k}$ is not a zero divisor then $1-a^{j-k} = 0$ which implies that $a^ka^{-j} = 1$ and $a^k = a$ is a unit? – stupidproofs123 Jul 18 '19 at 05:46
  • @zongxiangyi we are not claiming that all elements of $R$ are in that set. The question says if $a \in R$ then ..... so we start with $a \neq 0$ and consider this set $\langle a \rangle \subset R$. – Anurag A Jul 18 '19 at 05:49
  • I guess it is not necessarily true that $a^{k} = a$ for some $k$ so I am not sure how to prove this. – stupidproofs123 Jul 18 '19 at 05:52
  • @stupidproofs123 I have added more to my answer to complete the proof. – Anurag A Jul 18 '19 at 05:56
  • I wanted to take that approach but didn't know if $a$ is not a zero divisor implied $a^{k}$ wasn't since $a$ doesn't have to equal some $a^{k}$ with $k > 1$. – stupidproofs123 Jul 18 '19 at 06:02
  • @stupidproofs123 imagine if $a^k \neq 0$ is a zero divisor, then $\exists , b \neq 0$ such that $a^kb =0$. But this also implies $a(a^{k-1}b)=0$ which makes $a$ a zero divisor. – Anurag A Jul 18 '19 at 06:04
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Let $a \in R$. We must prove that $a$ is a zero divisor or a unit.

Consider the map $\varphi_a(r)=ar$ from $R$ to $R$.

One Proof

If $a$ is a unit, then $\varphi_a$ is a bijection. Assume that $a$ is not a unit, $\varphi_a$ cannot be a bijection, otherwise $a\varphi_a^{-1}(1)=1$ and thus $a$ is a unit. So $\varphi_a$ cannot be a injection. There exist two different elements $b,c\in R$ such that $$\varphi_a(b)=ab=ac=\varphi_a(c).$$ Hence $$a(b-c)=0.$$ So $a$ is a zero divisor since $b-c\ne 0$.

Another Proof

If $0\ne a$ is not a zero divisor, then $\varphi_a$ must be a injection, otherwise there exist two difference elements $b,c\in R$ such that $$\varphi_a(b)=ab=ac=\varphi_a(c).$$ Hence $$a(b-c)=0.$$ So it is a contradiction and $\varphi_a$ is a injection. Furthermore, $\varphi_a$ is also a surjection. Therefore $a\varphi_a^{-1}(1)=1$. Thus $a$ is a unit.

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Remark:Please recall the fact that, if $A$ is a finite set and $\varphi$ is a map from $A$ to $A$, then $$\varphi \text{ is a bijection.} \Leftrightarrow \varphi \text{ is a injection.}\Leftrightarrow \varphi \text{ is a surjection.}$$

darij grinberg
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Zongxiang Yi
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