4
$\begingroup$

Let us consider the ring $\mathbb{Z}_n$ where $\bar{m}\in\mathbb{Z}_n$

Could anyone help me prove that $\bar{m}$ is a zero divisor in $\mathbb{Z}_n$ if and only if $m,n$ are not coprime

So far I have:

Assume $\exists \bar{a}: \bar{m} \bar{n}=n \mathbb{Z} \Rightarrow$ for some $b\in\mathbb{Z}:am=bn$

I then assumed $n,m$ were coprime and attempted to use \exists a',b'\in \mathbb{Z}:a'm+b'n=1 to come to a contradiction, however haven't found one, the most promising thing I have found so far is that

a=n(a'b+ab') and b=m(a'b+ab') $\Rightarrow n|a$ and $m|b$

4 Answers 4

4

You're almost there. Suppose that $m$ and $n$ are coprime and that $ma$ is a multiple of $n$. From a'm+b'n=1 you get a'ma+b'na=a. Then, $a$ is a multiple of $n$, that is $\bar a=\bar 0$ and $m$ is not a zero divisor.

  • 0
    Ah god, that's obvious isn't it. So $m,n$ being coprime forces $\bar{a}$ to be zero, thanks for making that connection for me, bugging the hell out of me.2011-09-27
3

HINT $\ $ Recall every element of a finite ring is either a unit or zero-divisor, e.g. see here or here. Thus the contrapositive equivalent of your statement is $\rm\:\gcd(m,n) = 1\:$ iff $\rm\:m\:$ is unit. By Bezout

$\rm \gcd(m,n) = 1\iff\ \exists\ \: j,k\in\mathbb Z:\ j\ m + k\ n = 1\iff \ \exists\ \: j\in \mathbb Z:\ j\ m\equiv 1\ \ (mod\ n)$

  • 0
    As always Bill very enlightening :) Thanks for your help!2011-09-27
1

If $m$ is not coprime with $n$, let $b$ be their greatest common divisor. Then $m \frac{n}{b}$ is an integer multiple of $n$, so equals $0 \mod n$. Therefore $\bar m$ is a zero divisor (since $\frac{n}{b}$ is not equal to $0 \mod n$.

Conversely, if $m$ is a zero divisor, we have $\bar m \bar k = 0 \mod n$ for some $k \not= 0 \mod n$, ie. $mk = nb$ for some integer $b$. We may take $b$ and $k$ to be less than or equal to $n$; since $k\not= 0 \mod n$, $k$ is striclty less than $n$. Therefore the least least common multiple of $n$ and $m$ is strictly less than $nm$. Hence $n$ and $m$ are not coprime.

  • 0
    Thanks! that is of great help!2011-09-27
1

Maybe you can divide this out into two cases, when $n$ is prime and when it is not.

Case 1:

If $n$ is a prime number then $\mathbb{Z}_n$ will be a commutative division ring with a unit, so if $ab \equiv 0 $ mod $p$ for $a,b \in \mathbb{Z}_n$, then this means that by euclid's lemma that $p$ divides $a$ or $p$ divides $b$ so that $a$ or $b$ are equivalent to zero mod $p$.

In particular since any element in $\mathbb{Z}_n$ will be coprime to $n$, no element in $\mathbb{Z}_n$ will be a zero divisor by the result above (Recall that an element $a \neq 0$ in a ring is a zero-divisor if $ab = 0$ for some $b \neq 0$ in the ring).

I think it makes no sense to talk of the other direction for the only time when a number $m$ is not coprime to a prime $n$ is when $m$ is a multiple of $n$, but this is ridiculous for such an $m$ is not in $\mathbb{Z}_n$.

Case 2:

Suppose $n$ is not a prime number. Can you try to work this out for yourself?

  • 0
    I don't agree with not allowing you to use corollaries of theorems that you know.... With regards to determinants you can try to do without them.2011-09-27