Q:

Prove that if $f$ and $m$ are polynomials over $K$, and $m$ is irreducible, then hcf$(f,m)$ is either $1$ or $m$.

Proof that if $f$ and $m$ are polynomials over the field $K$, and $m$ is irreducible, then hcf$(f,m)$ is either $1$ or $m$.
I would really appreciate any help.

A:

Hint: 
If  hcf$(f,m)=1$ then there is nothing to prove. Now suppose  hcf$(f,m)\neq 1$ and then show $m|f $.
Let $m=\alpha_n x^n+\alpha_{n-1}x^{n-1}+ \cdot \cdot +\alpha_1x+C $. Since $m$ is irreducible(it has monic divisors $1$ and ${\alpha_n}^{-1}m$) and since hcf$(f,m)\neq 1$ this implies  hcf$(f,m)= {\alpha_n}^{-1} m $. Now since hcf$(f,m)= {\alpha_n}^{-1} m$ we have $({\alpha_n}^{-1} m) |f$ and thus $m|\alpha_nf$. Now you should be able to conclude. 

