So presumably this question is very basic, but I'm having some trouble with apparent contradictions in my reasoning.
Let $k$ be a field and $k \subseteq K$ a field extension. We say that $K$ is a finitely generated field extension if it is finitely generated over $k$ as a $k$-algebra. We say that $K$ is a finite field extension if it is finite dimensional as a $k$-vector space. By Zariski's lemma, these are equivalent concepts: A finitely generated field extension is finite.
We say that an element $t \in K$ is transcendental over $k$ if there is no monic polynomial with coefficients in $k$ for which $t$ is a root.
So far, is this correct? I think so. Which brings me to my confusion. I have encountered the term "finitely generated $k$-algebra of transcendence degree $1$". I don't understand how such an extension can exist. If $k \subseteq K$ is a field, and $t \in K$ is a transcendental element over $k$, then the elements $1, t, t^2, t^3, t^4, \ldots $ would be algebraically independent. Indeed if there was a dependency, then $t$ would fail to be transcendental. But then this is an infinite set of generators.
Where is the flaw in my reasoning? How can a transcendence degree $1$ field extension be finitely generated?
Usually a field extension $L/K$ is said to be finitely generated if there are elements $a_1,\dots,a_n\in L$ such that $L=K(a_1,\dots,a_n)$, which means that $L$ is the smallest subfield of $L/K$ containing $a_1 \dots, a_n$. This is not to be confused with the notion of being finitely generated as a $K$-algebra, which requires $L$ to be the smallest sub-$K$-algebra containing these elements. This object is denoted by $K[a_1,\dots,a_n]$ and is not a field in general, see for example $K[X]\subset K(X)$ where $X$ is an indeterminate.