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 \( \left ( \mathbb{Z}_n, +,  \cdot \right ) \) 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,  \( \left ( \mathbb{Z}_n, +,  \cdot \right ) \) 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 \( \left ( \mathbb{Z}_n, +,  \cdot \right ) \) has zero divisors \(\bar{a},\bar{b} \in \mathbb{Z}_n \) such that \(\bar{a} \cdot \bar{b} = \bar{n} = \bar{0} \) and \(\bar{a} \neq \bar{0} \neq \bar{b} \). Since zero divisors don't have multiplicative inverse (why?), then we can conclude \(\bar{a},\bar{b}\) 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 \(\bar{a} \in \mathbb{Z}_n \). Since \(\bar{a}\) 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 \(\bar{x} \in \mathbb{Z}_n\) such that \(\bar{a} \cdot \bar{x} = \bar{1} \). In other words, \(\bar{a}\) has multiplicative inverse in \(\mathbb{Z}_n\). Since \(\bar{a}\) is arbitrary, then this is also true for any element of  \(\mathbb{Z}_n\) if \(n\) is a prime number. \( \blacksquare \)