I need to find minimal polynomial of $\alpha = \sqrt 2 + \sqrt [3] 3 $ over $\mathbb Q$ and prove that my result is minimal polynomial. How do I do that?
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see here http://math.stackexchange.com/questions/422233/how-to-find-a-minimal-polynomial-field-theory – Dr. Sonnhard Graubner Oct 27 '14 at 16:06
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Constructing a degree 4 rational polynomial satisfying $f(\sqrt{2}+\sqrt{3}) = 0$ asks the same question about a very similar number and my answer there describes a straightforward general algorithm. – MJD Oct 27 '14 at 16:17
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The way to find the minimal polynomial is to start with the supposition that $$ x - \left( \sqrt{2} + \sqrt[3]{3} \right) = 0 $$ then move the $\sqrt[3]{3}$ to the right side of the equality and cube, getting $$ x^2 + 6x - \sqrt{2} \left( 3x^2+2 \right) = 3 $$ Now isolate the terms involving $\sqrt{2}$ on the right and square, getting (after some grouping of terms) $$ x^6 - 6x^4 - 6x^3 + 12x^2 - 36x + 1 = 0 $$ from which you read off the minimal polynomial.
It is easy to verify that this is a polynomial of $\alpha$ over $\Bbb{Q}$. To prove minimality you need to invoke the fact that $\gcd(2,3) = 1$, which implies that a minimal polynomial of a square root and a cube root of two numbers sharing no common factors must be of degree at least $6$.
hardmath
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Mark Fischler
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To prove minimality you need to invoke the fact that gcd(2,3)=1, which implies that a minimal polynomial of a sqare root and a cube root off two numbers sharing no common factors must be of degree at least 6.
– Dark Archon Oct 27 '14 at 20:40 -
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It can be done in two steps:
$\left(\alpha - \sqrt{2}\right)^3 = \left(\sqrt[3]{3}\right)^3 = 3$, then you get:
$p(\alpha)\sqrt{2} = q(\alpha)$, and square both sides again:
$2p(\alpha)^2 - q(\alpha)^2=0$, where $p(\alpha), q(\alpha)$ are integer-coefficient polynomials in $\alpha$, and this is the minimal polynomial you seek.
DeepSea
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