Prove that Q($\sqrt{2}$, $\sqrt{3}$) is a field Prove that $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \{a+b\sqrt{2} +c\sqrt{3} +d\sqrt{6}\ |\ a,b,c,d \in \mathbb{Q}\}$ is a field.
I am doing the subfield test, but having trouble in showing how to express the inverse in such a form. Anyone can help?
 A: HINT: $$(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})(a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6})=(a+d\sqrt{6})^2-(b\sqrt{2}+c\sqrt{3})^2:=r+s\sqrt{6}$$ where $r,s\in\mathbb{Q}$.   Then $$(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})(a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6})(r-s\sqrt{6})=r^2-6s^2=\alpha$$ and $\alpha$ is rational. Hence you have got the inverse.
A: According to the usual notation, $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is a field, the minimum subfield of $\mathbb{C}$ containing $\sqrt{2}$ and $\sqrt{3}$ (and $\mathbb{Q}$, of course).
The set $R = \{a+b\sqrt{2} +c\sqrt{3} +d\sqrt{6}\mid a,b,c,d \in \mathbb{Q}\}$ is trivially a subring of $\mathbb{C}$: just compute sums and products.
So what's necessary is to find inverses of nonzero elements. Doing the computation directly is unnecessarily complicated. You can observe that this set is
$$
R=F[\sqrt{3}]
$$
where $F=\mathbb{Q}[\sqrt{2}]=\{a+b\sqrt{2}\mid a,b\in\mathbb{Q}\}$. Indeed,
$$
a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}=(a+b\sqrt{2})+(c+d\sqrt{2})\sqrt{3}.
$$
Suppose $F$ is a subfield of $\mathbb{C}$ and $t\in\mathbb{C}$ is algebraic over $F$. Then the subring $F[t]$ generated by $t$ and $F$ can be written as
$$
\{a_0+a_1t+\dots+a_{n-1}t^{n-1}:a_0,\dots,a_{n-1}\in F\}
$$
so it's a finite dimensional vector space over $F$. Hence each of its element is algebraic over $F$. If $u\in F[t]$, it has a minimum polynomial $X^m+b_{m-1}X^{m-1}+\dots+b_1X+b_0$ with $b_0\ne0$. Thus
$$
u^m+b_{m-1}u^{m-1}+\dots+b_1u+b_0=0
$$
and
$$
u^{m-1}+b_{m-1}u^{m-2}+\dots+b_1+b_0u^{-1}=0
$$
and
$$
u^{-1}=-b_0^{-1}(u^{m-1}+b_{m-1}u^{m-2}+\dots+b_1)\in F[t]
$$
Thus $F[t]$ is a field.
Now apply this twice: $\mathbb{Q}[\sqrt{2}]=F$ is a field; $\sqrt{3}$ is algebraic over $F$, so $F[\sqrt{3}]$ is a field. End.

If you're interested in an explicit expression for the inverse, here it is. First, $1,\sqrt{2},\sqrt{3},\sqrt{6}$ are linearly independent over $\mathbb{Q}$, because $\mathbb{Q}(\sqrt{2},\sqrt{3})$ has dimension $4$ over $\mathbb{Q}$ and those four element generate it as a vector space.
Thus an element of the form $a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$ is zero if and only if $a=b=c=d=0$. Suppose this is not the case. Then
$$
(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})(u+x\sqrt{2}+y\sqrt{3}+z\sqrt{6})=1
$$
means
$$
au+ax\sqrt{2}+ay\sqrt{3}+az\sqrt{6}+
bu\sqrt{2}+2bx+by\sqrt{6}+2bz\sqrt{3}+
cu\sqrt{3}+cx\sqrt{6}+3cy+3cz\sqrt{2}+
du\sqrt{6}+2dx\sqrt{3}+3dy\sqrt{2}+6dz=1
$$
hence
$$
\begin{cases}
au+2bx+3cy+dz=1\\
bu+ax+3dy+3cz=0\\
cu+2dx+ay+2bz=0\\
du+cx+by+az=0
\end{cases}
$$
Solve the linear system and you'll have your answer.
A: Consider doing this step-wise. Since $\mathbb{Q}(\sqrt{2})$ is obviously a ring, to show it is a field, we just need to find an inverse for $a + b \sqrt{2}$. Let $a, b \in \mathbb{Q}$:
$$(a + b \sqrt{2})(a - b \sqrt{2}) = a^2 - 2b^2 \implies (a + b \sqrt{2})(\frac{a}{a^2 - 2b^2} - \frac{b}{a^2 - 2b^2} \sqrt{2}) = 1$$
Now, given an element of $\mathbb{Q}(\sqrt{2}, \sqrt{3})$, say $a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6}$, we rearrange it into elements of $\mathbb{Q}(\sqrt 2)$:
$$
\begin{align*}
a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6} &= (a + b \sqrt{2}) + (c + d \sqrt{2})\sqrt{3} \\
&= \alpha + \beta \sqrt{3}, \qquad \alpha, \beta \in \mathbb{Q}(\sqrt{2})
\end{align*}
$$
To show it has an inverse, apply the same process as before:
$$
\begin{align*}
(\alpha + \beta \sqrt{3})(\alpha - \beta \sqrt{3}) &= \alpha^2 - 3 \beta^2 \\
(\alpha + \beta \sqrt{3})(\frac{\alpha}{\alpha^2 - 3 \beta^2} - \frac{\beta}{\alpha^2 - 3 \beta^2} \sqrt{3}) &= 1
\end{align*}
$$
Since $\mathbb{Q}(\sqrt 2)$ is a field, $\frac{\alpha}{\alpha^2 - 3 \beta^2}$ and $\frac{\beta}{\alpha^2 - 3 \beta^2}$ are elements of $\mathbb{Q}(\sqrt 2)$, and so $\frac{\alpha}{\alpha^2 - 3 \beta^2} - \frac{\beta}{\alpha^2 - 3 \beta^2} \sqrt{3}$ is in $\mathbb{Q}(\sqrt 2, \sqrt 3)$.
All you need to do now is verify that you aren't dividing by $0$ anywhere.
