This is my attempt at proving the following statement: "Prove that every nonzero $f(x) \in F[x]$ has a unique monic associate in $F[x]$." Please note that $F$ denotes a field. Firstly I have attempted to prove that every nonzero $f(x)$ has a monic associate, and then the uniqueness of said monic associates.
Let $f(x) \in F[x]$. If $f(x)$ is monic we have that $f(x) = 1_{F}f(x)$ where $1_{F}$ denotes the identity element of $F$. Assume that $f(x)$ is not monic and denote the leading coefficient of $f(x)$ by $a$. Then since $F$ is a field, $a^{-1}$ exists and we can write $f(x)=aa^{-1}f(x)=a(a^{-1}f(x))$ by associativity. This proves the existence of monic associates.
For uniqueness, assume that $f(x)=bg(x)$ and $f(x)=cp(x)$ where $g(x),p(x) \in F[x]$ are monic. This implies that $bg(x)=cp(x)$. Comparing coefficients, in particular we must have that $b1_{F}=c1_{F}$ which implies $b=c$. This implies that $bg(x)=bp(x)$. Since every field is an integral domain, this implies $g(x)=p(x)$ which proves uniqueness.
I am not sure if my attempt is correct or not. Any feedback on my attempted proof would be greatly appreciated. Thanks in advance.