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I was trying to find the minimal polynomial of $\sqrt{i + \sqrt{2}}$ over $\mathbb{Q}$. I went as follows:

$a = \sqrt{i + \sqrt{2}} $

$a^{2} = i + \sqrt{2} $

$a^{4} = 1 + 2\sqrt{2}i $

$(a^{4} - 1)^{2} = -8 $

$a^{8} - 2a^{4} + 9 = 0 $

I'm guessing that $p(x) = x^{8} - 2x^{4} + 9$ is the minimal polynomial. I was then thinking about how to show this (or discover that it is reducible). I could try the rational roots theorem to show that there are no rational roots of $p(x)$. Then any possible polynomial factors would be of degree 2 and 6, 3 and 5, or 4 and 4 with coefficients in $\mathbb{Z}$. I could set up a homomorphism with $\mathbb{Z_{n}}$ and use Gauss's lemma. Either way it seems I'd have to check multiple systems of equations to investigate whether it is irreducible or not. Are there any ways of doing this quicker? It seems like a bit of a slog to have to work through all the equations I'd have to check - e.g. to check for degree 3 and 5 I'd have to check if there exists $a(x) = x^{5} + ax^{4} + bx^{3} + cx^{2} + dx + e$

and

$b(x) = x^{3} + fx^{2} + gx + h$

such that $p(x) = a(x)b(x)$.

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    $\begingroup$ Degree $8$ is the intuitive answer, since $i$ has degree $2$, $\sqrt2$ has degree $2$, and they are linearly independent over $\Bbb Q$, so their sum should have degree $4$. Finally, the square root of that sum should have degree $8$. This is in no way a rigorous proof, though. $\endgroup$ – Arthur Nov 16 '16 at 16:34
  • $\begingroup$ Linear independence here is equivalent to saying $i$ is not in the field extension $Q(\sqrt{2})$ is that right? $\endgroup$ – Franz Nov 16 '16 at 16:39
  • $\begingroup$ Yes, that is the same thing. And that's quite clearly true, since $\Bbb Q(\sqrt2) \subseteq \Bbb R$ and $i \notin \Bbb R$. $\endgroup$ – Arthur Nov 16 '16 at 16:42

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