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I'd like to find the four independent units in (the ring of integers of ) $\mathbb{Q}(\sqrt{2}, \sqrt{3}) \subseteq \mathbb{R}$ We also have that $\mathbb{Q}(\sqrt{2}, \sqrt{3}) \simeq \mathbb{Q}[x,y]/(x^2 - 2, y^2 - 3)$, as a field extension.

I just want to find the Norm, $\mathfrak{N}(x)$ for $x = a + b \sqrt{2} + c \sqrt{3} + d\sqrt{6} \in \mathbb{Q}(\sqrt{2}, \sqrt{3})$. The conjugates are like this:

$$ \big(a + b \sqrt{2} + c \sqrt{3} + d\sqrt{6}\big) \big(a - b \sqrt{2} + c \sqrt{3} - d\sqrt{6}\big) \big(a + b \sqrt{2} - c \sqrt{3} - d\sqrt{6}\big) \big(a - b \sqrt{2} - c \sqrt{3} + d\sqrt{6}\big)$$

If we multiply all four of these things together, we obtain a mess. I used sympy: $$ a^4 - 4\,a^2b^2 - 6\,a^2c^2 - 12\,a^2d^2 + 48\,abcd + 4\,b^4 - 12\,b^2c^2 - 24\,b^2d^2 + 9\,c^4 - 36\,c^2d^2 + 36\,d^4 $$ Instead we can rearrange the terms it looks almost manageable: $$ (a^4 + 4\,b^4 + 9\,c^4 + 36\,d^4)- (4\,a^2 b^2 + 6 \, a^2 c^2 + 12\,a^2 d^2 + 12\,b^2c^2 + 24\,b^2 d^2 + 36\, c^2 d^2 ) + (48\, abcd)$$ and Dirichlet's Unit theorem says we can find integers $a,b,c,d \in \mathbb{Z}$ such that this thing $=1$.

Fortunately, I can find two subfields off the bat: $[\mathbb{Q}(\sqrt{2}, \sqrt{3}): \mathbb{Q}(\sqrt{2})] = 2$ and $[\mathbb{Q}(\sqrt{2}, \sqrt{3}): \mathbb{Q}(\sqrt{3})] = 2$ and we get that :

$$ 1, 3 + 2\sqrt{2}, 2 + \sqrt{3} \in \mathbb{Q}(\sqrt{2}, \sqrt{3}) $$ are still units in this quartic field (by Pell Eq). There's one left. Which is it?

Related:

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    $\begingroup$ The notation $\mathbb{Q}(\sqrt{2},\sqrt{3})$ denotes the field over $\mathbb{Q}$ generated by the set $\{\sqrt{2},\sqrt{3}\}$. But you want the ring, not the field, so the correct notation is $\mathbb{Q}[\sqrt{2},\sqrt{3}]$. $\endgroup$
    – quasi
    Mar 29, 2018 at 16:22
  • $\begingroup$ $(1+\sqrt{2})$ is a fundamental unit of $\mathbb{Q}(\sqrt{2})$ (of norm $-1$, that's OK). There are three fundamental units. $\endgroup$
    – orangeskid
    Mar 29, 2018 at 16:31
  • $\begingroup$ @quasi, but the two notations denote in this case the same thing. $\endgroup$
    – Mariano Suárez-Álvarez
    Mar 29, 2018 at 16:41
  • $\begingroup$ You don't really want the units in $Q(\sqrt2,\sqrt3)$ but in its ring of integers, probably, and a good start would be finding that ring! $\endgroup$
    – Mariano Suárez-Álvarez
    Mar 29, 2018 at 16:42
  • $\begingroup$ @Mariano Suárez-Álvarez:: Right, so the OP presumably wants $\mathbb{Z}[\sqrt{2},\sqrt{3}]$ (and still square brackets). $\endgroup$
    – quasi
    Mar 29, 2018 at 16:48

2 Answers 2

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Just a check that Pell unit provided by @Rene Schipperus is indeed independent of the other Pell units. For assume that we have $$(3+2\sqrt{2})^m(2+\sqrt{3})^n(5+2\sqrt{6})^p=1$$ Applying the Galois maps $\sqrt{2}\mapsto - \sqrt{2}$ ,$\sqrt{3}\mapsto - \sqrt{3}$ we get $$(3-2\sqrt{2})^m(2-\sqrt{3})^n(5+2\sqrt{6})^p=1$$ so $$(3+2\sqrt{2})^m(2+\sqrt{3})^n=(3-2\sqrt{2})^m(2-\sqrt{3})^n$$ and therefore $$\left(\frac{3+2\sqrt{2}}{3-2\sqrt{2}}\right) ^m=\left(\frac{2-\sqrt{3}}{2+\sqrt{3}}\right)^n$$ so both must be in $\mathbb{Q}$, since $\mathbb{Q}(\sqrt{2})\cap \mathbb{Q}(\sqrt{3})=\mathbb{Q}$. Now, from $\left(\frac{3+2\sqrt{2}}{3-2\sqrt{2}}\right) ^m \in \mathbb{Q}$, applying the Galois map $\sqrt{2}\mapsto -\sqrt{2}$ we conclude that $$\left(\frac{3+2\sqrt{2}}{3-2\sqrt{2}}\right) ^m=\left(\frac{3-2\sqrt{2}}{3+2\sqrt{2}}\right) ^m$$ so the common value must be $\pm 1$. We conclude that $m=0$. Similarly we get $n=0$.

ADDED: Using WolframAlpha I got this result. The surprise was $$2+\sqrt{3}=\left(\frac{\sqrt{2}+\sqrt{6}}{2}\right )^2$$ Moreover, the units $$1+\sqrt{2}, \frac{\sqrt{2}+\sqrt{6}}{2}, \sqrt{2}+\sqrt{3}$$ form a fundamental system.

Note that $\frac{\sqrt{2}+\sqrt{6}}{2}=\frac{1+\sqrt{3}}{\sqrt{2}}$

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  • $\begingroup$ After thinking about it for a while, I realize we're still not done since there could another generator $x \in \mathbb{Q}(\sqrt{2}, \sqrt{3})$ such that $\langle 5 + 2 \sqrt{6} \rangle \subseteq \langle x \rangle$. The notes have $x = \frac{\sqrt{2} + \sqrt{6}}{2}$ but a derivation could be nice. $\endgroup$
    – cactus314
    Mar 29, 2018 at 18:13
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    $\begingroup$ @cactus314: Indeed, $5+2\sqrt{6} = (\sqrt{2}+\sqrt{3})^2$. So the three units $1+\sqrt{2}$, $2+\sqrt{3}$, $\sqrt{2}+\sqrt{3}$ generate a subgroup of finite index inside the group of units. The question is whether this index is $1$ or not. $\endgroup$
    – orangeskid
    Mar 29, 2018 at 18:19
  • $\begingroup$ @cactus314: please check the added stuff in the answer. $\endgroup$
    – orangeskid
    Mar 29, 2018 at 18:40
  • $\begingroup$ @Jyrki Lahtonen: Yes, thank you! I am still impressed with WolframAlpha. $\endgroup$
    – orangeskid
    Mar 29, 2018 at 18:49
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The fundamental unit of $\mathbb{Q}(\sqrt{6})$ is $$5+2\sqrt{6}$$ Does this help ?

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  • $\begingroup$ I think so... I can't rule out that $5 + 2 \sqrt{6} \in \langle 3 + 2 \sqrt{2}, 2 + \sqrt{3}\rangle$. $\endgroup$
    – cactus314
    Mar 29, 2018 at 16:36

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