Exercise 3.2.6

Proof.

(a)

Let $f\in ℝ[x]$, and suppose that $f(\alpha )=0$. Then $\bar{f}=f$, and $\bar{f}(\alpha )=0$. By Ex. 3.3.1(b), this implies $f(\bar{\alpha })=0$.

Conclusion : if $f\in ℝ[x]$,

$$f(\alpha )=0\Rightarrow f(\bar{\alpha })=0.$$

(b)

$\text{∙}$ Suppose that every nonconstant polynomial in $ℂ[x]$ has a root in $ℂ$.

Name $x_1,\cdots ,x_r$ the real roots of $f$ : $f=a{(x-x_1)}^{k_1}\cdots {(x-x_r)}^{k_r}g$, where $a\in ℝ$, and $g\in ℝ[x]$ is monic and has no real root. We show by induction on $d$ that every polynomial $g\in ℝ[x]$ without real root, monic, of degree $d$, is product of monic quadratic real polynomials.

If $d=0$, $g=1$ is the empty product.

We suppose $d>0$, and put the induction hypothesis that every polynomial in $ℝ[x]$ without real root, monic, of degree less than $d$, is product of monic quadratic real polynomials.

Let $g\in ℝ[x]$ a polynomial of degree $d$ without real root. $g$ has by hypothesis a complex root $\alpha$. Then $g=(x-\alpha )g_1,g_1\in ℂ[X]$.

By (a), $\bar{\alpha }$ is a root of $g$. $0=g(\bar{\alpha })=(\bar{\alpha }-\alpha )g_1(\bar{\alpha })$, and $\bar{\alpha }\ne \alpha$, thus $g_1(\bar{\alpha })=0$, $g_1=(x-\bar{\alpha })h,h\in ℂ[x]$, therefore

$$g=(x-\alpha )(x-\bar{\alpha })h,h\in ℂ[x].$$

$u=(x-\alpha )(x-\bar{\alpha })=x^2+\mathit{sx}+t$, where $s=\alpha +\bar{\alpha }\in ℝ,t=\alpha \bar{\alpha }\in ℝ$, thus $u\in ℝ[x]$, and also $h\in ℝ[x]$, since $h$ is the quotient of the Euclidean division of $g$ by $u$.

$g=(x^2-\mathit{sx}+t)h$, where $h\in ℝ[x]$ is monic, of degree less than $d$, without real root. By the induction hypothesis, $h$ is product of monic real quadratic polynomials, thus it is the same for $g$, and the induction is done.

Consequently, $f$ is product of linear or quadratic factors with real coefficients.

$\text{∙}$ Conversely, suppose that every polynomial in $ℝ[x]$ is product of linear or quadratic factors with real coefficients.

Let $f\in ℂ[x]$, with $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)\ge 1$. We will show that $f$ has a complex root.

By hypothesis $f$ has a linear or a quadratic factor.

If $f$ has a linear factor $\mathit{ax}+b$, then $-b∕a$ is a (real) root of $f$, and if $f$ has a factor $a{x}^2+\mathit{bx}+c,a\ne 0$, then Lemma 3.2.3 and Exercise 3.2.2 show that $f$ has a complex root. In both cases, $f$ has a complex root, so every non constant polynomial in $ℂ[x]$ has a complex root.