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Let $f(x)$ be a polynomial in $\mathbb{Q}[x]$. Prove that if $a + \sqrt2 b$ is a zero of $f(x)$ then so is $a - \sqrt2b$.

Can I treat $a + \sqrt2 b$ like a complex number? Then I can do something like

let $f(x) = \sum a_ix^i$, and $z = a+\sqrt2b$

Then $f(\bar z) = \sum a_i \bar z^i = ... = \bar{f(z)}$.

If not.. what else can I try?

jun
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    Note that $a,b$ are rational. And yes, it is like the case of complex numbers and conjugated, but you need to prove it, – Thomas Andrews Oct 30 '20 at 05:17
  • @ThomasAndrews Thanks for the comment. So you saying I need to prove that $a+\sqrt2b$ acts like a complex number? – jun Oct 30 '20 at 05:19
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    You'd need to first show that the conjugate map is well-defined, which you can do by showing that the $a + \sqrt{2}b$ representation is unique (i.e. use the irrationality of $\sqrt{2}$ to show that $a + \sqrt{2}b = c + \sqrt{2}d$ implies $a = c$ and $b = d$). Then show conjugates are preserved under $+, -, \times$, and conclude the same about polynomials. – user837206 Oct 30 '20 at 05:24

2 Answers2

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Hint: employ conjugation as here or, if that's unknown, give a direct proof, e.g. as below.

$a\pm\sqrt2\,b\,$ is a root of $f(x)\!\iff\! \pm\sqrt 2\,$ is a root of $\,f(a+bx)=: \bar f(x)\in\Bbb Q[x]$

Thus, equivalently, we seek $\,\bar f(\sqrt 2) = 0\,\Rightarrow\, \bar f(-\sqrt 2) = 0$.

Now split $\bar f(x) = g(x^2) + h(x^2)\,x\,\ $ into even + odd parts $ $ (cf. here and here)

so $\ 0 = \bar f(\sqrt 2) = g(2)\ +\ h(2) \sqrt 2\,\Rightarrow\, \color{#c00}{h(2)\!=\!0}\Rightarrow \color{#c00}{g(2)\!=\!0},\,$ by $\,g(2),h(2)\in\Bbb Q\not\ni \sqrt 2$

thus $\,\ \bar f(-\sqrt 2) = \color{#c00}{g(2)\ -\ h(2)} \sqrt 2 = 0. \ \ \ \rm\small QED$

Bill Dubuque
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  • If $,\color{#c00}{h(2)\neq 0},$ then solving $, 0 = g(2)+h(2),\sqrt 2,$ for $,\sqrt 2,$ yields $,\sqrt 2\in \Bbb Q\ \ $ – Bill Dubuque Oct 30 '20 at 09:20
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We can divide $f$ by $\left(x-(a+\sqrt{2}b)\right)\left(x-(a-\sqrt{2}b)\right) = x^2-2ax+a^2-2b^2 \in \Bbb{Q}[x]$ to obtain $$f(x) = \left(x-(a+\sqrt{2}b)\right)\left(x-(a-\sqrt{2}b)\right)q(x) + cx+d$$ for some $q \in \Bbb{Q}[x]$ and $c,d \in \Bbb{Q}$. Plugging in $x = a+\sqrt{2}b$ gives $$0 = f(a+\sqrt{2}b) = c(a+\sqrt{2}b) + d = (ca+d) + \sqrt{2}cb$$ which by irrationality of $\sqrt{2}$ implies $ca+d = cb = 0$. Assuming $b \ne 0$, this implies $c = 0$ and then $d = 0$. Therefore $$f(x) = \left(x-(a+\sqrt{2}b)\right)\left(x-(a-\sqrt{2}b)\right)q(x)$$ so in particular $f(a-\sqrt{2}b) = 0$.

mechanodroid
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