1
$\begingroup$

I'm working through Gallian's Ch. 13 on Integral Domains, the second chapter in the book dealing with topics concerning rings. I feel this was a good book on intro group theory (which is what a lot of universities use it for), but I feel it's a little dicey on intro ring theory content (I doubt anybody uses it for that).

Anyways the ring theory part is a little bare, and so I didn't have much to work with in proving every element of $Z_{n}$ is a unit or a zero divisor. This book proves the more general result about 'every finite commutative ring' afterwards so I can't use that either. I think I have a solid alternative proof (using kind of a pigeon-hole argument) though even without all the fancy techniques the other solutions out there have. Would love for comments and critiques, especially on the second half of it.


Let $n\geq4$ (the $n<4$ cases are trivial; moreover they are prime and handled elsewhere).

First suppose some arbitrary $k\in\mathbb{Z}_{n}$ is both a unit and a zero divisor. Hence there exists some $l\in\mathbb{Z}_{n}\setminus \{0\}$ such that $kl=lk=0$. Now since $k$ is a unit, we have that $k^{-1}\in\mathbb{Z}_{n}$. It follows that \begin{align*}0 &=k^{-1}0 \\ &=k^{-1}kl \\ &=l,\end{align*}a contradiction. Next suppose $k\in\mathbb{Z}_{n}\setminus \{0\}$ is neither a unit nor a zero-divisor. Hence for all $l\in\mathbb{Z}_{n}\setminus\{0,1\}$, we have $$kl\neq 0,\text{ (since $k,l\neq 0$, and since $k$ is not a zero-divisor)}$$ $$kl\neq 1, \text{ ($k$ is not a unit)}$$ $$kl\neq k. \text{ ($l\neq 1$)}$$Hence the $n-2$ remaining choices for $l \in \mathbb{Z}_{n}\setminus \{0,1\}$ can yield at most $n-3$ unique products $kl\in \mathbb{Z}_{n}\setminus \{0,1,k\}$. In any case we must have that there exist integers $l_{1}>l_{2}\in\mathbb{Z}_{n}\setminus \{0,1\}$ such that $kl_{1}=kl_{2}$. It follows that \begin{align*}0 &= kl_{1}-kl_{2} \\ &= k(l_{1}-l_{2}),\end{align*}which is impossible since the nonzero element $k$ is not a zero divisor and $l_{1}-l_{2}>0$ by construction. Thus a contradiction. QED.

$\endgroup$

1 Answer 1

5
$\begingroup$

The argument is fine, but you slightly over-complicate the reasoning for the contradiction when $k$ is neither a unit nor a zero divisor.

It is enough to note that the map $K:\mathbb Z_n\rightarrow \mathbb Z_n$ defined by $K(l)=kl$ is not injective, as its range is a subset of $\mathbb Z_n- \{1\}$ (since $k$ is not a unit).

It immediately follows that for some distinct $l_1,l_2\in \mathbb Z_n$, $kl_1=kl_1\Rightarrow k(l_1-l_2)=0$ and so $k$ is a zero divisor, a contradiction (just like you concluded).

This approach also comes with the (very slight) advantage of not having to deal with $n=2,3$ as special cases.

$\endgroup$
3
  • $\begingroup$ Really appreciate the feedback! $\endgroup$ Aug 14, 2020 at 12:55
  • $\begingroup$ The same proof works in general: math.stackexchange.com/a/60974/589 $\endgroup$
    – lhf
    Aug 16, 2020 at 14:09
  • $\begingroup$ @lhf Yes fair observation. Thanks for linking this. $\endgroup$ Aug 16, 2020 at 17:53

You must log in to answer this question.

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