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 $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$.

Here is another take:
Let $\alpha \in \mathbb{Q}[\sqrt{2}, \sqrt{3}]$, $\alpha\ne0$, and consider $\phi: \mathbb{Q}[\sqrt{2}, \sqrt{3}] \to \mathbb{Q}[\sqrt{2}, \sqrt{3}]$ given by $\phi(x)=\alpha x$. Then $\phi$ is an injective $\mathbb{Q}$-linear transformation. Since $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$ is a finite-dimensional vector space over $\mathbb{Q}$, the injectivity of $\phi$ implies its surjectivity. In particular, $1$ is in the image of $\phi$, which proves that $\alpha$ is invertible in $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$.
This argument works in general to prove that $ \mathbb{Q}[S]$ is a field when $S$ is a finite set of algebraic numbers.
