Exercise 3.2.7

Proof. By definition, a field $F$ is algebraically closed if every nonconstant polynomial is product of linear factors in $F[x]$.

$\text{∙}$ If $F$ is algebraically closed, and if $f\in F[x]$ is not a constant, this product of linear factors is not empty, so $f$ is divisible by a linear factor $\mathit{ax}+b,a,b\in F$. Hence $f$ has a root $\alpha =-b∕a$ in $F$.

$\text{∙}$ Suppose that every nonconstant polynomial has a root in $F$

We show by induction on $d$ that every polynomial $f\in F[x],d=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)>0$, is product of linear factors in $F[x]$

If $d=1$, $f=\mathit{ax}+b,a\ne 0$, is product of one linear factor.

Let $f\in F[x],d=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)>1$. Then $f$ has by hypothesis a root $\alpha \in F$, so $f=(x-a)g$, where $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)<d$. By the induction hypothesis, $g$ is a constant or is product of linear factors, so it is the same for $f$, and the induction is done.

Conclusion: If $F$ is an algebraically closed field, then the two following propositions are equivalent:

(i)

Every nonconstant polynomial in $F[x]$ is product of linear factors.

(ii)

Every nonconstant polynomial in $F[x]$ has a root in $F$.