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Show that $(9-5\sqrt 3)(2-\sqrt 2) $ is not a square in $\mathbb Q (\sqrt 2, \sqrt 3) $; equivalently, prove that $\mathbb Q (\sqrt 2, \sqrt 3, u)$ is of degree $2$ over $\mathbb Q (\sqrt 2, \sqrt 3) $ where $u $ is a number such that $u^2 = (9-5\sqrt 3)(2-\sqrt 2) $.

This question is a small section of what is required for this question: Showing that $\mathbb{Q}(\sqrt{2},\sqrt{3}, \sqrt{(9 - 5\sqrt{3})(2-\sqrt{2})})$ is normal over $\mathbb{Q}$, and finding its Galois group.

The answer of the above question reads as follows: "Note that $Gal(\mathbb Q(\sqrt 2, \sqrt 3)/ \mathbb Q(\sqrt 3))$ has order $2$, generated by $σ_{1,0} ∈ Gal(\mathbb Q(\sqrt 2, \sqrt 3)/\mathbb Q)$, where $σ_{1,0}$ fixes $\sqrt 3$ and sends $\sqrt 2$ to its negative. Let $N$ denote the norm of the extension $\mathbb Q(\sqrt 2, \sqrt 3)/\mathbb Q(\sqrt 2)$; if $α^2 = x^2$ for some $x ∈ \mathbb Q(\sqrt 2, \sqrt 3)$, then $N(α^2) = N(x^2) = N(x)^2 ∈ \mathbb Q(\sqrt 3)$, i.e. $N(α^2)$ is a square in $\mathbb Q(\sqrt 3)$. We note $N(α^2) = (9 − 5√3)^2(2)$so if $N(α^2)$ is a square in $\mathbb Q(\sqrt 3)$, then so is $2$, a contradiction.

However, as a person totally unfamiliar with field norms, I was wondering if there is any other way to solve this, or at least explaining the similar idea of the proof in simpler notions. Brute-forcingly writing $u^2 = (a+b\sqrt2+c\sqrt3+d\sqrt6)^2$ with all letters as rationals and expanding it seems a lot tedius - may take hours to solve everything, and there's no gaurantee that this would lead to a contradiction. Standard result in Galois theory&Group theory are welcome.

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  • $\begingroup$ Your sentence beginning “Brute-forcingly…”: why do you take a quotient that has form $w^2/z^2=(w/z)^2$, when you know that what’s in my parentheses is itself in form $A+B\sqrt2+C\sqrt3+D\sqrt6$ for rationals $A,B,C,D$? $\endgroup$ – Lubin Jun 3 '18 at 2:39
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    $\begingroup$ You shouldn't be "afraid" of the norm (= product of conjugates), a map which intervenes naturally here because it respects multiplication, and so allows to avoid brute force and tedious calculation, as you say, because you don't mix addition and multiplication. In your hint, you norm down from $\mathbf Q(\sqrt 2, \sqrt 3)$ to $\mathbf Q(\sqrt 3)$ to get a contradiction, but you could equally well norm down to $\mathbf Q(\sqrt 2)$ to conclude in the same way (just try). $\endgroup$ – nguyen quang do Jun 3 '18 at 5:27
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Here is a more "hands-on" (and systematic) approach. Suppose that $$u^2=(2-\sqrt{2})(9-\sqrt{3})\tag{1}$$

for some $u\in{\mathbb Q}(\sqrt{2},\sqrt{3})$. You can write $u=v+w\sqrt{3}$ with $v,w\in{\mathbb Q}(\sqrt{2})$. Expanding in (1) yields

$$ (v^2+3w^2)+(2vw)\sqrt{3}=9(2-\sqrt{2})-(2-\sqrt{2})\sqrt{3} \tag{2} $$

Since $1$ and $\sqrt{3}$ are linearly independent over ${\mathbb Q}(\sqrt{2})$, it follows that

$$ v^2+3w^2=9(2-\sqrt{2}), \ 2vw = -(2-\sqrt{2}) \tag{3} $$

The second equality yields $w=-\frac{2-\sqrt{2}}{2v}$, and reinjecting this into the first equality we obtain

$$ v^2+3\bigg(\frac{2-\sqrt{2}}{2v}\bigg)^2=9(2-\sqrt{2}) \tag{4} $$

which can be rewritten as $$ v^4-9(2-\sqrt{2})v^2+3\bigg(\frac{2-\sqrt{2}}{2}\bigg)^2=0 \tag{5} $$

or $$ v^4-9(2-\sqrt{2})v^2+3(1-\sqrt{2})=0 \tag{6} $$ Complating the square, we find that this is equivalent to

$$ \bigg(v^2-\frac{9(2-\sqrt{2})}{2}\bigg)^2=78(\sqrt{2}-1) \tag{7} $$

Writing $v^2-\frac{9(2-\sqrt{2})}{2}$ as $a+b\sqrt{2}$ with $a,b\in\mathbb{Q}$, as in the deduction of (3) we obtain

$$ a^2+2b^2=-78, 2ab=78 \tag{8} $$

and the first equality is impossible for rational $a,b$.

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