Finite and Infinite Field Extensions of Complex I have a couple questions regarding field extensions $L/\mathbb{C}$, first I believe that if $L/\mathbb{C}$ is a finite extension $([L:\mathbb{C}]) = n$ then this will imply that $L = \mathbb{C}$, i.e., there are no finite degree extensions of the complex numbers which 'enlarge' the complex numbers.  Here is my attempt at this statement: If $L$ is an $n$-dimensional vector space over $\mathbb{C}$ then for a basis $\{l_1, l_2, ..., l_n\} \subset L$, an arbitrary element $x$ in $L$ must be of the form
$$
x = c_1l_1 + c_2l_2 + ... + c_nl_n
$$
for a suitable choice of complex scalars $c_1, c_2, ..., c_n$.  I believe the remaining portion of this proof follows somehow from the fact that $\mathbb{C}$ is an algebraically closed field, but I'm unsure of how to proceed.  $\mathbb{C}$ being algebraically closed implies that for any nonzero polynomial $f(x) \in \mathbb{C}[x]$, all of the roots of $f$ are contained within $\mathbb{C}$.  Unfortunately, I can't use this fact from the above representation of an arbitrary element in $L$ since I would need to use some polynomial in $n$ indeterminates ($f(x) \in \mathbb{C}[x_1, x_2, ..., x_n]$), and I'm not sure if algebraic closure implies anything about polynomial roots in $n$ indeterminates.  Also, I'm unsure of how I would even apply this if it were true - to get a root, I would need to set $x = 0$ but then that wouldn't really be proving anything.
Secondly, I believe that there are extensions of $\mathbb{C}$ which are uncountable.  Namely the field of rational functions in $n$ indeterminates with coefficients in $\mathbb{C}$.  Would such an uncountable degree extensions be referred to as a transcendental extension?  And if the degree of the extension were instead countable, would this say anything about the field extension $L$?
 A: Q1: Yes, this is true. More generally we have the following standard fact.

Proposition: Any finite extension $L/K$ of fields is algebraic.

Proof. Let $\alpha \in L$ be any element. If $\dim_K L = [L : K] = n$ then $\{ 1, \alpha , \alpha^2, \dots \alpha^n \}$ consists of $n+1$ vectors and so must be linearly dependent. A linear dependence between these gives a polynomial $f$ of degree at most $n$ such that $f(\alpha) = 0$, so $\alpha$ is algebraic. $\Box$

Corollary: If $K$ is algebraically closed and $L/K$ is a finite extension, then $L = K$.

Q2: Uncountability is not the definition of a transcendental extension; an extension is transcendental if it is not algebraic, or equivalently if it has at least one transcendental element. Over $\mathbb{C}$ these two conditions are equivalent, but over a countable or finite field they are not; for example $\mathbb{Q}(x)$ is a countable transcendental extension of $\mathbb{Q}$.
Q3: $\mathbb{C}$ has no countable extensions. This is because a transcendental extension must contain $\mathbb{C}(x)$ as a subfield, and $\mathbb{C}(x)$ has uncountable dimension over $\mathbb{C}$ because, for example, the elements $\frac{1}{x - a}, a \in \mathbb{C}$ are linearly independent. (This is a nice exercise.)
This is specifically because $\mathbb{C}$ is itself uncountable. More generally, for $K$ a field, $K(x)$ is countable-dimensional over $K$ if $K$ is finite or countable, and otherwise its dimension is the cardinality of $K$. (This is also a nice exercise. You can do it by proving partial fraction decomposition.)
