Showing that $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$ contains multiplicative inverses 
Why must $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$ -- the set of all polynomials in $\sqrt{2}$ and $\sqrt{3}$ with rational coefficients  -- contain multiplicative inverses?

I have gathered that every element of $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$ takes form $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{2}\sqrt{3}$ for some $a,b,c,d \in \mathbb{Q}$, and I have shown that $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt{3}]$ are both fields, but it's not clear to me why this allows for general multiplicative inverses to exist in $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$.

It comes from general facts. One way to do this is the following.
First note that $\sqrt{2}$ is algebraic over $\mathbb{Q}$, with minimal polynomial $m = x^{2} - 2$, which is irreducible in $\mathbb{Q}[x]$ for general reasons.
Then $K = \mathbb{Q}[\sqrt{2}]$ is a field. You can explicitly compute inverses in $K$. If $0 \ne f(\sqrt{2}) \in F$, for some $f \in \mathbb{Q}[x]$, then $(f, m) = 1$, so Euclid's algorithm yields $u, v \in \mathbb{Q}[x]$ such that
$$
u f + v m = 1,
$$
and now evaluate for $x = \sqrt{2}$ to get
$$
u(\sqrt{2}) \cdot f(\sqrt{2}) = 1,
$$
so $u(\sqrt{2})$ is the required inverse.
Now repeat with $\mathbb{Q}[\sqrt{2}, \sqrt{3}] = K[\sqrt{3}]$ to get that this is a field too.
