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Let $γ$ be a root of $x^5 − x + 1 = 0$ in an algebraic closure of $\mathbb{Q}$. Find a polynomial with rational coefficients of which $γ +\sqrt2$ is a root

Is it possible to directly modify the polynomial itself so that its root is $\gamma + \sqrt2$ to get the polynomial?

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  • $\begingroup$ There are infinitely many rational numbers, so the tag finite-fields was inappropriate. $\endgroup$ Jan 16, 2020 at 6:39

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You’ll use $\gamma$’s polynomial, yes. If $\alpha = \gamma + \sqrt2$, then $\alpha-\sqrt2$ solves the polynomial; substitute. Open the powers with the binomial theorem. You’ll have terms with $\alpha$ and $\sqrt 2$. Group terms with $\sqrt 2$, leave them alone in one side of the equation and square. Finally, this polynomial over $\alpha$ has rational coefficients and we found what we wanted.

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$1$, $\gamma$, $\gamma^2$, $\gamma^3$, $\gamma^4$, $\sqrt{2}$, $\gamma\sqrt{2} $, $\gamma^2\sqrt{2} $, $\gamma^3 \sqrt{2} $, $\gamma^4 \sqrt{2} $ form a basis for the $\mathbb{Q}$ vector space $\mathbb{Q}(\gamma, \sqrt{2})$.

This means that $\{1, (\gamma + \sqrt{2}), (\gamma + \sqrt{2})^2, ... , (\gamma + \sqrt{2})^{10} \} $ is not linearly independent over $\mathbb{Q}$. Write these numbers down in terms of the basis, and use linear algebra to find a linear combination of these things that is $0$, which will be your polynomial.

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    $\begingroup$ Why is $[\mathbb Q(\gamma, \sqrt2) \colon \mathbb Q] = 6$? $\endgroup$ Jan 15, 2020 at 11:20
  • $\begingroup$ I made a mistake. You need $\gamma \sqrt{2}$ and so on as well. $\endgroup$ Jan 15, 2020 at 12:32
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You can compute the resultant of $x^5 − x + 1$ and $(y-x)^2-2$ to eliminate $x$: $$ y^{10} - 10 y^8 + 38 y^6 + 2 y^5 - 100 y^4 + 40 y^3 + 121 y^2 + 38 y - 17 $$ The resultant can be computed by hand but it is a large determinant and is best found with a computer, for instance with Resultant in Mathematica and WolframAlpha.

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