Recall that a complex linear algebraic group $G$ is an affine algebraic group (i.e. an affine group variety over $\mathbb C$). It's a well-known fact that there is always a closed embedding $$G \hookrightarrow \operatorname{GL}(n, \mathbb C).$$ My concern is that this seems to imply that $G$ is finite-dimensional as a variety, just because $\operatorname{GL}(n, \mathbb C)$ is for any particular integer $n$ (it has codimension 1 in $\mathbb A^{n^2}$, hence dimension $n^2-1$). This doesn't mesh well with my intuition, because I vaguely know that if $R$ is a non-Noetherian $\mathbb C$-algebra, then $\operatorname{Spec} R$ should be infinite-dimensional as a variety. Unfortunately I don't know any of these examples well-enough to try to construct a group operation on one of them to figure out what's happening.
I'd appreciate it if someone could explain why it should be true that any $G$ should be finite-dimensional; for example if having the group law places some restriction on the size of $G$ that I'm not seeing. It's also possible that my claim is wrong or doesn't make sense in some way; if so I'd also be happy if someone could point out the mistake. Thanks.
EDIT: To clarify on Qiaochu's comment: a linear algebraic group $G$ for me is an $k$-variety $G$ equipped with the structure of a group such that both the group multiplication $G \times G \to G$ and the inversion $G \to G$ are morphisms of varieties. But I think the issue was actually that I don't know the definition of variety: these need to have finite type. Thanks all.