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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.

Ottavio
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Merlin
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    Sketch: Given a polynomial for $a$ you can make one (not necessarily minimal) for $a^2$ easily enough. If $p(x)$ is a polynomial for $a^2$ then $q(x)=p(-x)$ is one for $-a^2$ and that gives you one for $ia$. – lulu Dec 10 '20 at 19:27
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    $p(-ix)\overline{p(-ix)}$ has real coeff and should annulate $ia$, not necessarily minimal though. – zwim Dec 10 '20 at 19:45
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    $a$ is algebraic over $\Bbb Q$, and $i$ as well, so also their product - see this post. – Dietrich Burde Dec 10 '20 at 19:49
  • I was about to post an answer, but I realized it is essentially the idea from @lulu's comment. I used the squaring trick twice to get a polynomial $s(x)\in\Bbb{Q}[x]$ such that $s(a^4)=0$. Then $s(x^4)$ works (but is never irreducible as it is divisible by both $p(x)$ and $p(-x)$). – Jyrki Lahtonen Dec 11 '20 at 22:28

2 Answers2

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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.

Atticus Stonestrom
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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.

achille hui
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