# Find a polynomial with rational coefficients

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?

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

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.
$$1$$, $$\gamma$$, $$\gamma^2$$, $$\gamma^3$$, $$\gamma^4$$, $$\sqrt{2}$$, $$\gamma\sqrt{2}$$, $$\gamma^2\sqrt{2}$$, $$\gamma^3 \sqrt{2}$$, $$\gamma^4 \sqrt{2}$$ 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.
• Why is $[\mathbb Q(\gamma, \sqrt2) \colon \mathbb Q] = 6$? Jan 15, 2020 at 11:20
• I made a mistake. You need $\gamma \sqrt{2}$ and so on as well. Jan 15, 2020 at 12:32
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.