TITLE: Showing that $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$ contains multiplicative inverses
QUESTION [6 upvotes]: 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 $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$.

REPLY [2 votes]: The easy way to see this is to express an element of $\mathbb Q[\sqrt2,\sqrt3]$ as $A+B\sqrt3$, where A and B are members of $\mathbb Q[\sqrt2]$. Then $\dfrac{1}{A+B\sqrt3}=\dfrac{A-B\sqrt3}{A^2-3B^2}$. The denominator, $A^2-3B^2$, is a member of $\mathbb Q[\sqrt2]$, so you already know how to invert that.