Let $a$ be an algebraic number, prove that $ia$ is also algebraic Of course this is a particular case of the closure under addition and multiplication of algebraic numbers (which proof would also be appreciated), but I would prefer a simple transformation on the already known polynomial giving $a$ as a root in order to obtain a polynomial giving $ia$ as a root. Thank you in advance for your help.
 A: Lemma: for any $p\in\mathbb{C}[x]$, we have $p\overline p\in\mathbb{R}[x]$, where $\overline p$ is obtained by complex conjugating every coefficient of $p$.
Proof: By separating out the real and imaginary parts of its coefficients, we can write $p$ as $p_{\text{Re}}+ip_{\text{Im}}$, where $p_{\text{Re}},p_{\text{Im}}\in\mathbb{R}[x]$. Then $\overline p=p_{\text{Re}}-ip_{\text{Im}}$, whence \begin{align}p\overline{p}&=(p_{\text{Re}}+ip_{\text{Im}})(p_{\text{Re}}-ip_{\text{Im}})\\&=p_\text{Re}^2+p_{\text{Im}}^2\in\mathbb{R}[x],\end{align} as desired.

Now, since $a$ is algebraic, let $p\in\mathbb{Q}[x]$ be an annihilating polynomial for $a$. By the lemma above, the polynomial $q(x):=p(-ix)\overline{p(-ix)}\in\mathbb{R}[x]$. Also, we have $p(-ix),\overline{p(-ix)}\in\mathbb{Q}[i][x]$, so in fact $q\in\mathbb{R}[x]\cap\mathbb{Q}[i][x]=\mathbb{Q}[x]$. We further have $$q(ia)=p(-i^2a)\overline{p}(-i^2a)=p(a)\overline{p}(a)=0,$$ so that $q$ is a polynomial with rational coefficients, annihilating $ia$, as desired.
A: Let $b = ia$.
Since $a$ is algebraic, there is $p(x) = \sum\limits_{k=0}^{\deg p} \alpha_k x^k \in \mathbb{Q}[x]$ such that $p(a) = 0$.
For $\ell = 0,\ldots,3$, let $p_\ell(x)$ be the polynomial $\sum\limits_{k=0}^{\left\lfloor\frac{\deg p - \ell}{4}\right\rfloor} \alpha_{4k+\ell} x^{4k+\ell} \in \mathbb{Q}[x]$.
It is easy to see

*

*$p(x) = p_0(x) + p_1(x) + p_2(x) + p_3(x)$

*$p_\ell(b) = p_\ell(ia) = i^\ell p_{\ell}(a)$
This implies
$$\begin{align}
&\;(p_0(b) - p_2(b)) - i (p_1(b) - p_3(b))\\ = &\; (p_0(a) + p_2(a)) + (p_1(a) + p_3(a))\\ = &\; p(a) = 0\end{align}$$
Multiply by $(p_0(b) - p_2(b)) + i (p_1(b) - p_3(b))$, we find $b$ is a root of $q(x) \in \mathbb{Q}[x]$ where
$$q(x) \stackrel{def}{=} (p_0(x) - p_2(x))^2 + (p_1(x) - p_3(x))^2$$
As a result, $b = ia$ is also algebraic.
