Proof Verification in Ring Theory, Divisors of $0$ in $Z_n$

Question

I am trying to prove the fact that the $0$ divisors of $(Z_n,+,\times)$ are the non-zero elements $m\not=0$ that are NOT coprime with $n$. I see this in the following way:

-If $m$ is a $0$ divisor, then $m$ not coprime with $n$

or equivalently:

-If $m$ coprime with $n$, then $m$ is not a $0$ divisor

I chose the contrapositive.

By definition, $m,n$ coprime if $gcd(m,n)=1$.

By definition $m$ is a $0$ divisor if for $m\not=0$ and $m$ in $(Z_n,+,\times)$ there exists $q\not=0$ and $q$ in $(Z_n,+,\times)$ such that $mq=0$.

Suppose $m,n$ coprime ($gcd(m,n)=1$), that means that $|m|=n$ (the order of $m$ is $n$), which we know from Group Theory (As $(Z_n,+)$ is cyclic, and using Lagrange). So only $m*(k*n)=0$ for $k$ in $(Z_n,+,\times)$ in $(Z_n,+,\times)$, but $k*n=0$ by definition in $(Z_n,+,\times)$, so $m$ is not a divisor of $0$ as there doesn't exist a $q$ satisfying the conditions above as $q=k*n=0$. Fraleigh gives a different proof, but I like to practice proving stuff by myself, and I have serious doubts about the way I proceeded with this. Any help is immensely appreciated!

Answer 1

So, let me see if I get your proof right:

Assume $m$ and $n$ are coprime. This means that the order of $m$ in the group $(\Bbb Z_n, +)$ is equal to $n$. Therefore, if we have $m\cdot l = 0$ for some integer $l$, we must have $l = kn$ for some integer $k$. But $kn = 0$ in $\Bbb Z_n$, so we must have $l = 0$, which means that $m$ is not a zero divisor.

The idea here is good, and it does work. I do see one technical flaw, though: You ought to be very specific about what letters signify integers, and what letters are elements of $\Bbb Z_n$. Along with this you have to know that multiplying an element in $\Bbb Z_n$ with an integer is not the same as multiplying two elements of $\Bbb Z_n$.

It can be fixed, and quite easily (you just have to make sure to separate the integer $k$ from the congruency class $\bar k\in \Bbb Z_n$, and so on). You can even be quite lax about this once you are able to keep it straight. But at the moment, it looks like you aren't entirely conscious about it, and if you let that continue it may lead to problems down the line.

Edit: One way of fixing it is to let every variable be an integer, and then translate statements about integes into statements about the corresponding equivalence classes in $\Bbb Z_n$. Let $m\in \Bbb Z$ be coprime with $n\in \Bbb Z$. Then we have, for any $k\in \Bbb Z$, that $n\mid mk\implies n\mid k$ (for instance, by the fundamental theorem of arithmetic). Now, translating the last sentence directly to $\Bbb Z_n$ (meaning that we change $n\mid x$ into $[x] = 0$), we get that $[mk] = [0]$ implies $[k] = 0$. But $[mk] = [m][k]$, so we have $[m][k]= 0\implies [k] = 0$, and therefore $[m]$ is not a zero divisor.

Answer 2

The sentence "the zero divisors ($\ldots$) are the elements ($\ldots$)" it means $m$ is a zero divisor if and only if it is not coprime with $n$.

Your approach is perfect but it does only one direction.

Also I would maybe rephrase

So only $m*(k*n)=0$ for $k$ in $(Z_n,+,\times)$

as

So $m*q=0$ for $q$ in $\mathbb{Z}$ only when there is $k\in \mathbb{Z}$ such that $q=k*n$