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Ok, I actually worked this out as I was typing it up. But my solution seems kind of inelegant and involves a lot of tedious algebra that I've omitted here. Can anyone think of an easier method?

(1) Reducibility over $\mathbb{Q}(\sqrt{6})$

Quite straightforward, as the substitution $y = x^2$ and the quadratic formula on $y^2 - 10y + 1$ give:

$x^4-10x^2+1 = (x^2-5-2\sqrt{6})(x^2-5+2\sqrt{6})$.

(2) Reducibility over $\mathbb{Q}(\sqrt{2})$

Further reduction gives: $(x-\sqrt{5-2\sqrt{6}})(x+\sqrt{5-2\sqrt{6}})(x-\sqrt{5+2\sqrt{6}})(x+\sqrt{5+2\sqrt{6}})$.

It's also not that hard to show $\sqrt{5-2\sqrt{6}} = \sqrt{2}-\sqrt{3}$ and $\sqrt{5+2\sqrt{6}} = \sqrt{2}+\sqrt{3}$ so the reduction becomes:

$(x-\sqrt{2}-\sqrt{3})(x+\sqrt{2}+\sqrt{3})(x-\sqrt{2}+\sqrt{3})(x+\sqrt{2}-\sqrt{3})$.

If we label these a, b, c, d left to right then:

$a.c = x^2-2\sqrt{2}x -1$ and $b.d = x^2+2\sqrt{2}x -1$ so it's reducible over $\mathbb{Q}(\sqrt{2})$.

(2) Reducibility over $\mathbb{Q}(\sqrt{3})$

$a.d = x^2-2\sqrt{3}x+1$ and $b.c = x^2+2\sqrt{3} +1$ so it's reducible over $\mathbb{Q}(\sqrt{3})$.

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1 Answer 1

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Well, a possibility is to write it as a difference of two squares and thus factors into a sum and difference. For e.g.: $$x^4-10x^2+1 = (x^2-5)^2 - 24$$ and noting that $24=4\cdot6$ is a square in $\mathbb{Q}(\sqrt{6})$, so this is the difference of two squares.

Similarly, $x^4-10x^2+1 = (x^2-1)^2 - 8x^2$ and note $8$ is a square in $\mathbb{Q}(\sqrt{2})$.

Again, $x^4 - 10x^2 + 1 = (x^2+1)^2 - 12x^2$ and $12$ is a square in $\mathbb{Q}(\sqrt{3})$.

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  • $\begingroup$ Thank you, that's just the kind of solution I was looking for! $\endgroup$
    – Lewy
    Apr 5, 2013 at 16:02
  • $\begingroup$ @Lewy Glad to have been of help... $\endgroup$
    – Macavity
    Apr 5, 2013 at 16:35

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