Exercise 4.1.2

Proof. Suppose that $f,g\in F[x]$ are monic, and $f\mid g,g\mid f$.

$f=\mathit{gh},h\in F[x]$ and $g=\mathit{fl},l\in F[x]$, so $f=\mathit{fhl}$, where $f\ne 0$ since $f$ is monic, thus $\mathit{hl}=1$, and so $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(h)+\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(l)=0$, $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(h)=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(l)=0$.

Therefore $h=\lambda \in F^{\text{∗}}$, $f=\mathit{\lambda g}$. In particular, $f,g$ have the same degree $d$.

Write $f=\underset{k=0}{\overset{d}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}{a}_k{x}^k,g=\underset{k=0}{\overset{d}{\sum }}{b}_k{x}^k$.

As $f,g$ are monic, $a_d=b_d=1$, and $a_d=\lambda b_d$, so $\lambda =1$, and $f=g$.

Conclusion: If $f$ and $g$ are monic polynomials in $F[x]$ each of which divides the other, then $f=g$ □