Adjoining a number to a field 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.
 A: 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 $\mathbb{Q}[\sqrt{2}] = \{a + b\sqrt{2} \ | \ a, b \in \mathbb{Q} \}$.
A: The notation "$A[x]$" refers to polynomials in $x$ with coefficients in $A$ and so could in theory not contain inverses. In practice, you can show that an algebra over a field $k$ that is a finite dimensional $k$-vector space and also an integral domain must itself be a field and so we do contain inverses. You are then correct to point out that in cases where we are not dealing with a finite dimensional vector space (such as with a transcendental extension) it may indeed occur that the resulting object is not a field.
