Exercise 3.1.1

Proof. Let $f,g,h\in F[x],f\ne 0,f=\mathit{gh},I=⟨g⟩$.

(a)

$\text{∙}$ Suppose that $g=\lambda \in F$ is a constant. As $f\ne 0$ and $f=\mathit{gh}$, then $g\ne 0$, so $\lambda \ne 0$.

Let $p\in F[x]$ any polynomial. Then $p=\lambda (\frac{1}{\lambda }p)=(\frac{1}{\lambda }p)g\in ⟨g⟩$, thus $F[x]\subset ⟨g⟩$. Moreover $⟨g⟩\subset F[x]$, so

$$F[x]=⟨g⟩=I.$$

$\text{∙}$ Conversely, if $F[x]=I=⟨g⟩$, then $1\in ⟨g⟩$, so $1=\mathit{gu},u\in F[x]$, hence $0=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)+\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(u)$, therefore $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)=0$, so $g\in F$ is a nonzero constant.

$$g\in F^{\text{∗}}⟺⟨g⟩=F[x].$$

(b)

$\text{∙}$ If $h=\mu \in F$ is a constant, then $\mu \ne 0$ (since $f\ne 0$), and $f=\mathit{\mu g},\mu \in F^{\text{∗}}$.

If $p\in ⟨f⟩$, then $p=\mathit{uf},u\in F[x]$, thus $p=\mathit{\mu ug}\in ⟨g⟩$, so $⟨f⟩\subset ⟨g⟩$.

If $p\in ⟨g⟩$, then $p=\mathit{qg},q\in F[x]$, thus $p={\mu }^{-1}\mathit{qf}\in ⟨f⟩$, so $⟨g⟩\subset ⟨f⟩$.

$$⟨f⟩=⟨g⟩=I.$$

$\text{∙}$ Conversely, if $⟨f⟩=⟨g⟩,$ then $g\in ⟨f⟩$ , $g=\mathit{vf},v\in F[x]$, thus $g=\mathit{vgh}$. As $f=\mathit{gh}\ne 0,g\ne 0$, thus $1=\mathit{vh}$, therefore $h\in F^{\text{∗}}$ is a constant.

$$h\in F^{\text{∗}}⟺I=⟨f⟩.$$