What are the units in $\mathbb{Q}(\sqrt{2}, \sqrt{3})$? 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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*Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$?
 A: 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 $\frac{\sqrt{2}+\sqrt{6}}{2}=\frac{1+\sqrt{3}}{\sqrt{2}}$ 
A: The fundamental unit of $\mathbb{Q}(\sqrt{6})$ is $$5+2\sqrt{6}$$ Does this help ?
