Prove that if a is a root of $f(x)$ then $\overline{a}$ is also a root

Question

Given a polynomial $ f \in \mathbb R[x] $, how do we prove that:

if $a \in \mathbb C$ is a root of $f$ then also $\overline{a}$ is also a root of $f$ ?

Answer 1

$f\left(\overline{x}\right)=\sum\limits_{k=0}^da_k \left(\overline{x}\right) ^ k = \sum\limits_{k=0}^da_k\overline{x^k} = \sum\limits_{k=0}^d \overline{a_k}\overline{x^k} = \sum\limits_{k=0}^d \overline{a_kx^k}= \overline{\sum\limits_{k=0}^da_kx^k} = \overline{f(x)}=\overline{0}=0$

Answer 2

Hint $\, $ Being a ring hom, $\rm\:x\mapsto \bar x\:$ preserves $\rm\:\color{#c00}{sums\,\ \&\,\ products}.\:$ Also it $\rm\:\color{#0a0}{fixes\ coefficients}\in\Bbb R.\:$ Therefore, by induction, it preserves polynomial terms $\rm\ \overline{f(w)} = f(\overline w),\ \ f(x)\in\Bbb R[x],\ $ since such polynomial terms are compositions of said basic operations. $ $ More explicitly $$ \begin{eqnarray} \rm \overline{f(w)}\: &=&\rm\ \ \overline{a_n w^n +\,\cdots + a_1 w + a_0}\\ &=&\rm\,\ \overline{a_n w^n}\, +\,\cdots + \overline{a_1 w} + \overline a_0\quad by\ \ \ \color{#c00}{\overline{x+y}\, =\, \overline x + \overline y}\ \ \ \forall\ x,y \in \Bbb C\\ &=&\rm\,\ \overline a_n\, \overline w^n+\,\cdots + \overline a_1\overline w + \overline a_0\quad by\ \ \ \color{#c00}{\overline{x\, *\, y}\, =\, \overline x\, *\, \overline y}\ \ \ \forall\ x,y \in \Bbb C \\ &=&\rm\,\ a_n\, \overline w^n + \,\cdots + a_1 \overline w + a_0\quad by\ \ \ \color{#0a0}{\overline a = a}\ \ \ \forall\ a\in \Bbb R\\ &=&\rm\ f(\overline w)\\ \rm So\,\ \ 0 = f(w)\! \ \Rightarrow\ 0 = \bar 0 = \overline{f(w)}\:& =&\ \rm f(\overline w)\quad {\bf QED} \end{eqnarray}$$

Remark $\ $ The analogous polynomial preservation property holds true for any algebraic structure, i.e. since homomorphisms preserve the basic operations (including constants = $0$-ary operations), they also preserve the "polynomial" terms composed of these basic operations. Said equivalently, hom's commute with polynomials.

Answer 3

Write the polynomial as $$ p(x)=\sum_{i=0}^n a_nx^n $$ where $a_n\neq 0$ and $a_i$ is real for every $i$. Suppose that $z\in\mathbb{C}$ is a root of $p$. Then $$ p(z)=\sum_{i=0}^n a_nz^n=0 $$ If you consider this expression in $\mathbb{C}$, it is also true. If you take its conjugate, and since $\overline{a_i}=a_i$ (since they're all real): $$ 0=\overline{0}=\overline{p(z)}=\overline{\sum_{i=0}^n a_nz^n}=\sum_{i=0}\overline{a_nz^n}=\sum_{i=0}\overline{a_n}\overline{z^n}= $$ $$ =\sum_{i=0}a_n\overline{z}^n=p(\overline{z}) $$ and therefore $\overline{z}$ is a root for $p$.