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 $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$.
Best Answer
Let $K$ be an extension field of the field $F$ and let $0\ne a\in K$ be algebraic over $F$; then $F[a]$ contains the inverse of $a$.
Indeed, if $f(X)=c_0+c_1X+\dots+c_{n-1}X^{n-1}+X^n$ is the minimal polynomial of $a$ over $F$, then it's irreducible, so $c_0\ne 0$ and $$ c_0+c_1a+\dots+c_{n-1}a^{n-1}+a^n=0. $$ Multiply by $c_0^{-1}a^{-1}$ to get $$ a^{-1}=-c_0^{-1}(c_1+\dots+c_{n-1}a^{n-2}+a^{n-1}) $$ so $a^{-1}\in F[a]$.
If now $0\ne b\in F[a]$, you also have $F[b]\subseteq F[a]$. Since $F[b]$ is an $F$-subspace of $F[a]$, it's finite dimensional over $F$; therefore $b$ is algebraic over $F$ and, by what we showed above, $b^{-1}\in F[b]\subseteq F[a]$.
Now you can apply this to $\mathbb{Q}[\sqrt{2},\sqrt{3}]$, which is equal to $F[\sqrt{3}]$ where $F=\mathbb{Q}[\sqrt{2}]$, which is a field. Then also $F[\sqrt{3}]$ is a field by the same reason. You can take $K=\mathbb{C}$, of course.