I have some confusion over a proof I have been given to show that $\mathbb{F}_n$, ie the class of integers modulo n admits multiplicative inverses when n is prime.
This is the proof:
We need to show that for all $a \in \mathbb{F}_n$ we can find $b$ such that $[a] \cdot [b] = 1$.
We know by our definition of multiplication on the integers modulo n that $[a] \cdot [b] = [ab].$
So, we are looking for $b$ such that $[ab] = 1$.
Recognize that the equality holds only when $n | (ab-1)$ which suggests that $kn = ab - 1$ and therefore $1 = ab - kn$ for some $k$.
Since $n$ is prime, we can always find such $b$.
We complete the proof.
The part I am confused about is the second to last line of the proof. Why does $n$ being prime suggest that we can always find a $b$ that makes the equality work? Is this because every integer is a composite of primes?
It is a general fact that for integers $a,b$ we can find integers $k,j$ with $$ \gcd(a,b)=ak+bj $$ by the Euclidean algorithm.
Take $a$ to be any integer and $b$ to be prime. Then of course $\gcd(a,p)=1$ and reducing modulo $p$, yields $$ 1\cong [a][k] $$ and $[k]$ is the inverse of $[a]$ modulo $p$.