Every Non-zero Element in Z_n Has Multiplicative Inverse If and Only If N Is a Prime Number

 

Let $n$ be a natural number greater than $1$. Suppose $({\mathbb{Z}}_n,+,\cdot )$ be a ring of integers mod $n$. Then, every element $a\in {\mathbb{Z}}_n$ has multiplicative inverse if and only if $n$ is a prime number. In other words,  $({\mathbb{Z}}_n,+,\cdot )$ is a field if and only if $n$ is a prime number.

Proof:

If $n$ is a composite number, we can write $n=a\cdot b$ with $a,b$ are natural numbers less than $n$. Clearly, $a,b$ are not multiple of $n$. Therefore, the ring $({\mathbb{Z}}_n,+,\cdot )$ has zero divisors $\overline{a},\overline{b}\in {\mathbb{Z}}_n$ such that $\overline{a}\cdot \overline{b}=\overline{n}=\overline{0}$ and $\overline{a}\ne \overline{0}\ne \overline{b}$. Since zero divisors don't have multiplicative inverse (why?), then we can conclude $\overline{a},\overline{b}$ also have no multiplicative inverse. Hence, if $n$ is a composite number then there exist some element of  ${\mathbb{Z}}_n$ that has no multiplicative inverse.

Now, if $n$ is a prime number, take any nonzero element $\overline{a}\in {\mathbb{Z}}_n$. Since $\overline{a}$ is nonzero, $a$ is not a multiple of $n$. Therefore, the greatest common divisor of $a$ and $n$ is $gcd(a,n)=1$. By Bézout's identity, there exists $x,y\in \mathbb{Z}$ such that $ax+ny=1$. Taking mod $n$ of both sides, we obtain $ax\equiv 1$. Thus, there exists $\overline{x}\in {\mathbb{Z}}_n$ such that $\overline{a}\cdot \overline{x}=\overline{1}$. In other words, $\overline{a}$ has multiplicative inverse in ${\mathbb{Z}}_n$. Since $\overline{a}$ is arbitrary, then this is also true for any element of  ${\mathbb{Z}}_n$ if $n$ is a prime number. $◼$