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?

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.
