Alternative proof of $p(\alpha)=0$, then is also $p(\alpha^*)=0$ If for monic polynomial $p \in \mathbb{Z}[x]$ and $\alpha \in \mathbb{Q}(\sqrt{d})$ such that $p(\alpha)=0$, then $p(\alpha^*)=0$.
The operation $*$ is defined by:
$$(a+b\sqrt{d})^* = a-b\sqrt{d}$$
$$\alpha^* = \alpha, \ \alpha\in\mathbb{Q}$$
Begining of proof: $$p(x)=x^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0.$$
$$p(\alpha^*) = (\alpha^*)^n + a_{n-1} (\alpha^*)^{n-1} + \dots + a_1\alpha^* + a_0$$
$$=(\alpha^n)^* + (a_{n-1}\alpha^{n-1})^* + \dots +(a_1\alpha)^* + a_0^*\\$$
$$=(\alpha^n + a_{n-1}\alpha^{n-1} + \dots + a_1\alpha + a_0)^* = (p(\alpha))^*=0.$$
My question is: is this proof ok, because if we have polynomials with some degree we usually tend to proof this by using mathematical induction on degree of polynomials. Is this possible in this case?
 A: Let $p(x)=x^n+a^{n−1}x^{n−1}+\dots+a_1x+a_0 = 0$ with root $a+b\sqrt d \in Q(\sqrt d)$.
Take the conjugate on both sides.
$$ \overline { p(x) } = \overline { x^n+a_{n−1}x^{n−1}+\dots+a_1x+a_0 } = \overline 0$$
$$ \overline { p(x) } =  { \overline { x^n } + \overline { a_{n−1}x^{n−1} } + \dots + \overline { a_1x } + \overline a_0 } = 0$$
$$ \overline { p(x) } =  { \overline x^n } + { { a_{n−1} }{  { { \overline x}^{n−1} } } + \dots + { a_1 \overline x } + \overline a_0 } = 0$$
$$\therefore p(a+\sqrt b) = 0 \implies p(a-\sqrt b) = 0$$
A: Let $\alpha_1,\alpha_2\in\mathbb Q\left(\sqrt d\right)$ where $d$ is square-free. Then $\alpha=a_1+b_1\sqrt d,\alpha_2=a_2+b_2\sqrt d$ for some $a_1,b_1,a_2,b_2\in\mathbb Q$. Now,
$$
\left(\alpha_1+\alpha_2\right)^*=\left(a_1+b_1\sqrt d+a_2+b_2\sqrt d\right)^*=\left(\left(a_1+a_2\right)+\left(b_1+b_2\right)\sqrt d\right)^*\\
=a_1+a_2-\left(b_1+b_2\right)\sqrt d=a_1-b_1\sqrt d+a_2-b_2\sqrt d=\alpha_1^*+\alpha_2^*
$$
Thus, we have $\left(\alpha_1+\alpha_2\right)^*=\alpha_1^*+\alpha_2^*$. Similarly,
$$
\left(\alpha_1\alpha_2\right)^*=\left(\left(a_1+b_1\sqrt d\right)\left(a_2+b_2\sqrt d\right)\right)^*=\left(\left(a_1a_2+b_2b_2d\right)+\left(a_1b_2+a_2b_1\right)\sqrt d\right)^*\\
=\left(a_1a_2+b_1b_2d\right)-\left(a_1b_2+a_2b_1\right)\sqrt d\\
\alpha_1^*\alpha_2^*=\left(a_1-b_1\sqrt d\right)\left(a_2-b_2\sqrt d\right)=\left(a_1a_2+b_2b_2d\right)+\left(-a_1b_2-a_2b_1\right)\sqrt d\\
=\left(a_1a_2+b_1b_2d\right)-\left(a_1b_2+a_2b_1\right)\sqrt d
$$
Thus, we have $\left(\alpha_1\alpha_2\right)^*=\alpha_1^*\alpha_2^*$. By induction on these two statements, we obtain
$$
\left(\sum_{i=1}^n\alpha_i\right)^*=\sum_{i=1}^n\alpha_i^*\\
\left(\prod_{i=1}^n\alpha_i\right)^*=\prod_{i=1}^n\alpha_i^*
$$
Thus we also obtain that $\left(n\alpha\right)^*=n\alpha^*$ and $\left(\alpha^n\right)^*=\left(\alpha^*\right)^n$ by setting all the $\alpha_i=\alpha$ in the above two formulae.
Also, for any $q\in\mathbb Q$, we have $q=q+0\sqrt d$, and thus $q^*=q-0\sqrt d=q$.
Let $p\left(x\right)\in\mathbb Q\left[x\right]$ such that $p\left(\alpha\right)=0$ for $\alpha\in\mathbb Q\left(\sqrt d\right)$. Let $p\left(x\right)=\sum_{i=0}^nc_ix^i$
$$
0=0^*=\left(\sum_{i=0}^nc_i\alpha^i\right)^*=\sum_{i=0}^n\left(c_i\alpha^i\right)^*=\sum_{i=0}^nc_i^*\left(\alpha^i\right)^*=\sum_{i=0}^nc_i^*\left(\alpha^*\right)^i
$$
Since $c_i\in\mathbb Q$, we have $c_i^*=c_i$. Then,
$$
0=\sum_{i=0}^nc_i\left(\alpha^*\right)^i=p\left(\alpha^*\right)
$$
Thus for any $p\left(x\right)\in\mathbb Q\left[x\right]$, if $p\left(\alpha\right)=0$ then  $p\left(\alpha^*\right)=0$. Since it is true for any $p\left(x\right)\in\mathbb Q\left[x\right]$, it is also true for monic polynomials with integer coefficients.
