For a prime number \(p\) denote by \(\mathbb{Z}/p\mathbb{Z}\) the set of integers \(\{0,1,2,3...,p-1\}\) with addition and multiplication modulo \(p\). By this we mean that \(p-1+4\) add up to \(3\) modulo \(p\) and the product of two residues \(a\) and \(b\) is viewed as the residue of \(ab\) modulo \(p\). Note that the set of the elements of \(\mathbb{Z}/p\mathbb{Z}\) contains \(0\) and every element has an additive inverse, namely for an element \(a\) there exists an element \(b = p-a\) in \(\mathbb{Z}/p\mathbb{Z}\) such that \(a+b = 0\) modulo \(p\). Such a set with addition, multiplication, \(0\) and additive inverses is called a Ring.
For example For \(p=5\) we have \(\mathbb{Z}/5\mathbb{Z}\) as the set: \(\{0,1,2,3,4\}\), where \(2+3 = 0\) so the additive inverse for \(2\) modulo \(5\) is \(3\), \(3+4= 2\), \(2\times 3 = 1\), \(3\times 4 = 2\).
Prove that for a prime number \(p\) the ring \(\mathbb{Z}/p\mathbb{Z}\) contains a multiplicative inverse for any non zero element. Namely, for any \(a \neq 0\) modulo \(p\) there exists \(b\) modulo \(p\), such that \(ab = 1\) modulo \(p\). A ring with this property of multiplicative inverses is called a Field.