When people say, "K is an extension of k with dimension n", do they mean as an algebra or as a vectorspace?

Question

For instance, consider k(x), (the fraction field of k[x]). k(x) has dimension 2 as an algebra over k, but dimension \omega as a vectorspace over k. Which one are they talking about, and how can I tell? If you want a for-instance, I'm reading Perrin's Algebraic Geometry right now, and on page 15 on the bottom he writes, "Lemma 4.2. Let k be an uncountable algebraically closed field and let K be an extension of k whose dimension is at most countable. Then K = k."

Answer 1

The word "dimension" isn't usually used for the size of a minimal generating set for an algebra relative to some subalgebra (i.e. $\{1,x\}$ is a generating set for $\mathbb{k}(x)$ as an algebra over $\mathbb{k}$, but I wouldn't call it a "basis").

"Dimension" in the context of an extension field means vector space dimension (unless the author is doing something really weird).

The lemma you refer to says that if $\mathbb{K}$ is a field containing an uncountable algebraically closed field $\mathbb{k}$ and if as a vector space over $\mathbb{k}$, $\mathbb{K}$ is at most countable in dimension, then $\mathbb{k}=\mathbb{K}$.

The proof goes pretty quick. If $\mathbb{K}$ is algebraic over $\mathbb{k}$, then they are equal since $\mathbb{k}$ is algebraically closed. So if $\mathbb{K} \not= \mathbb{k}$, then $\mathbb{K}$ must contain a transcendental (over $\mathbb{k}$) element, say $x$. The collection of elements of the form $\dfrac{1}{x-a}$ for $a \in \mathbb{k}$ clearly belong to $\mathbb{K}$ ($x \not\in \mathbb{k}$ so $x-a \not=0$ and so $x-a$ is invertible). Perrin makes a quick argument showing that the elements $1/(x-a)$ are linearly independent over $\mathbb{k}$. But since we get a different element for each of the uncountably many elements of $\mathbb{k}$, it must be that the dimension of $\mathbb{K}$ (as a vector space over $\mathbb{k}$) is at least uncountable.