I've just started learning ring theory, and the book I'm using uses "Every nonzero element in a finite ring is either a unit or a zero divisor" implicitly without explanation. So I came across this proof here on this site:

In a finite commutative ring with unity, every element is either a unit or a zero-divisor. Indeed, let $a\in R$ and consider the map on $R$ given by $x \mapsto ax$. If this map is injective then it has to be surjective, because $R$ is finite. Hence, $1=ax$ for some $x\in R$ and $a$ is a unit. If the map is not injective then there are $u,v\in R$, with $u\ne v$, such that $au=av$. But then $a(u-v)=0$ and $u-v\ne0$ and so $a$ is a zero divisor.

I understand all of the proof, except the part: Hence, $1=ax$ for some $x\in R$ and $a$ is a unit. How is this conclusion reached? Thanks for your time!


If $x\mapsto ax$ is surjective, in particular its image contains $1$. This means exactly that there is some $x$ such that $ax=1$.

  • $\begingroup$ Why does $x\mapsto ax$ is surjective imply that its image contains 1? $\endgroup$ – Tony Tarng Aug 19 '16 at 5:03
  • 2
    $\begingroup$ @TonyTarng: by hypothesis, your ring $R$ has unity. Hence, if the map $x \mapsto ax$ from $R$ to $R$ is surjective, then every element of $R$ has a preimage under this map; this includes $1 \in R$. $\endgroup$ – Alex Wertheim Aug 19 '16 at 5:11
  • $\begingroup$ @AlexWertheim I got it. Thanks a lot! $\endgroup$ – Tony Tarng Aug 19 '16 at 5:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.