Exercise 3.2.1

Proof.

(a)

Let $f=\underset{i=0}{\overset{n}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}{a}_i{x}^i,g=\underset{j=0}{\overset{m}{\sum }}{b}_j{x}^i\in ℂ[x]$.

By definition of the product of polynomials,

$$\mathit{fg}=\underset{k=0}{\overset{n+m}{\sum }}{c}_k{x}^k,\mathrm{with}{c}_k=\underset{i+j=k}{\sum }{a}_i{b}_j=\underset{i=0}{\overset{k}{\sum }}{a}_i{b}_{k-i}$$

Then, using the fact that conjugation is a field automorphism in $ℂ$,

$$\begin{array}{llll}\hfill \overline{\mathit{fg}}& =\underset{k=0}{\overset{n+m}{\sum }}\overline{c_k}{x}^k& \hfill & \\ \hfill & =\underset{k=0}{\overset{n+m}{\sum }}\overline{\underset{i+j=k}{\text{∑}}{a}_i{b}_j}{x}^k& \hfill & \\ \hfill & =\underset{k=0}{\overset{n+m}{\sum }}\underset{i+j=k}{\sum }\overline{a_i}\overline{b_j}{x}^k& \hfill & \\ \hfill & =\underset{i=0}{\overset{n}{\sum }}\overline{a_i}{x}^i\underset{j=0}{\overset{n}{\sum }}\overline{b_j}{x}^j& \hfill & \\ \hfill & =\bar{f}\bar{g}.& \hfill & \end{array}$$

(b)

If $f\in ℂ[x]$ and $\alpha \in ℂ$, $$\begin{array}{llll}\hfill \bar{f}(\alpha )=0& \Rightarrow \underset{i=0}{\overset{n}{\sum }}\overline{a_i}{\alpha }^i=0& \hfill & \\ \hfill & \Rightarrow \overline{\underset{i=0}{\overset{n}{\sum }}\stackrel{¯}{a_i}{\alpha }^i}=\bar{0}=0& \hfill & \\ \hfill & \Rightarrow \underset{i=0}{\overset{n}{\sum }}{a}_i{\bar{\alpha }}^i=0& \hfill & \\ \hfill & \Rightarrow f(\bar{\alpha })=0.& \hfill & \end{array}$$