# $R$ finite commutative ring with identity. Prove that $a \in R$ is either a zero divisor or a unit. [duplicate]

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.

• Just because an element $a$ is not a zero divisor doesn't mean that the whole ring will not have any zero divisors. – Anurag A Jul 18 '19 at 5:32
• I don't understand how it isn't proved. This fits the definition of an integral domain in my book. That if $ab = 0$ then either $a = 0$ or $b = 0$. – stupidproofs123 Jul 18 '19 at 5:34
• In $\Bbb{Z}_6$, the element $5$ is not a zero divisor but that doesn't mean $2$ and $3$ are not zero divisors. – Anurag A Jul 18 '19 at 5:35
• It has to be true for every element $a$, not just those that are zero divisors as I have implied here..oops. – stupidproofs123 Jul 18 '19 at 5:39
• You got it. That's the catch. – Anurag A Jul 18 '19 at 5:40

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.

• 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 5:46
• @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 5:49
• 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 5:52
• @stupidproofs123 I have added more to my answer to complete the proof. – Anurag A Jul 18 '19 at 5:56
• 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 6:02

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 $$\varphi \text{ is a bijection.} \Leftrightarrow \varphi \text{ is a injection.}\Leftrightarrow \varphi \text{ is a surjection.}$$