When I studied algebra, we talked about fields such as $\mathbb{Q}[\sqrt{2}]$, the rational numbers with the square root of two adjoined to the field. Structures like these are called field extensions and are themselves fields.
My question is when we talk about these structures, are we saying we add the square root of two and its (multiplicative) inverse, or just the square root of two and any numbers necessary to keep the closure property of our field satisfied? If the latter is the case, is this part of what makes transcendental numbers so significant? If we only add the transcendental number to the field, and all numbers necessary to keep the closure, there is no way to obtain an inverse for that number. This is unlike the square root of 2, whose inverse can be written as $\frac{\sqrt{2}}{2}$, which would be in the extension by closure.
We didn't go into a lot of details in the algebra class I took so these are just things I've been thinking about on my own.
Best Answer
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:
But notice that we are talking about $\mathbb{Q}[\sqrt{2}]$ as a field as in the second point! What gives?
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 $\mathbb{Q}[\sqrt{2}] = \{a + b\sqrt{2} \ | \ a, b \in \mathbb{Q} \}$.