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.
In the case of $\mathbb{Q}[\sqrt{2}]$, we have not only $\sqrt{2}$ and its multiplicative inverse, but everything necessary to retain closure under the operations. There is a notational point to make here:
- $F[a]$ is defined to be the set $\{f(a) \ | \ f(x) \in F[x]\}$.
- $F(a)$ is defined to be the "smallest" extension field of $F$ that contains $a$.
But notice that we are talking about $\mathbb{Q}[\sqrt{2}]$ as a field as in the second point! What gives?
Theorem: When $a$ is algebraic over a field $F$, then $F[a] = F(a)$.
Proof:
Since $F[a]$ is a ring, most field properties already hold. What is left is to demonstrate the existence of multiplicative inverses. To do this, we take advantage of the Euclidean algorithm:
Let $f(x) \in F[x]$ be the minimal polynomial for $a$. Every polynomial without $a$ as a root will correspond to a nonzero element in $F[a]$, and moreover, every such polynomial will be relatively prime to $f(x)$. That is, given such a $g(x)$, there exists polynomials $h(x)$ and $k(x)$ such that:
$$f(x)h(x) + g(x)k(x) = 1$$
Since $a$ is a root of $f(x)$, evaluating the above at $a$ gives:
$$g(a)k(a) = 1$$
So given any nonzero $g(a) \in F[a]$, there exists some $k(a)$ that serves as its multiplicative inverse. This is to say: every nonzero element in $F[a]$ has a multiplicative inverse. We can conclude that, if $a$ is algebraic over $F$, then $F[a]$ is a field and $F[a] = F(a)$.
Final comments:
What makes an algebraic adjunction to a field special? Unlike transcendental adjunctions, algebraic ones are finite. That is, if $a$ is algebraic over $F$, then $F[a]$ can be viewed as a vector space over $F$ spanned by finitely many basis "vectors".
For example, $\mathbb{Q}[\sqrt{2}]$ is a finite extension of degree $2$, meaning any basis contains $2$ basis vectors. One possible basis is $\{1, \sqrt{2}\}$, and so $\mathbb{Q}[\sqrt{2}] = \{a + b\sqrt{2} \ | \ a, b \in \mathbb{Q} \}$.
Best Answer
The reason is much more general: consider multiplication by $a +b\sqrt2$ in the $\mathbf Q$-vector space $\mathbf Q[\sqrt2]$. If is non-zero, it is an injective endomorphism of this space. As we're in finite dimension, an injective endomorphism is surjective. Hence $1$ is attained, i.e. there exists $a'+b'\sqrt2\in\mathbf Q[\sqrt2]$ such that $$(a +b\sqrt2)(a'+b'\sqrt2)=1.$$
The most general result is this: