1
$\begingroup$

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!

$\endgroup$
2

1 Answer 1

3
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ Why does $x\mapsto ax$ is surjective imply that its image contains 1? $\endgroup$
    – Tony Tarng
    Commented Aug 19, 2016 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 $1 \in R$. $\endgroup$ Commented Aug 19, 2016 at 5:11
  • $\begingroup$ @AlexWertheim I got it. Thanks a lot! $\endgroup$
    – Tony Tarng
    Commented Aug 19, 2016 at 5:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .