# Major misunderstanding about field extensions and transcendence degree

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?

• Let $I = \{ \frac{1}{f(t)} | f(t) \in k[t]$ irreducible $\}$. Then $k[t]$ is a finitely generated $k$-algebra but its fraction field $k(t) = k[t, I]$ is not finitely generated because $I$ is always infinite. $k[t]$ is a finitely generated $k$-algebra of transcendance degree $1$, its fraction field is the purely transcendental extension of degree $1$. Both are infinite dimensional $k$-vector spaces. Once allowing things like choice and transfinite induction then every field extension is a tower of algebraic and purely transcendental extensions. – reuns Jan 7 '19 at 6:38
• "transcendence degree" and "dimension" are two different things. – Slade Jan 7 '19 at 6:45
• @Slade They are when the notion of dimension is as a vector space over the ground field, but there are geometrical definitions of inherent dimension which do relate closely to transcendence degree. Think of the dimension of a surface embedded in some space - sometimes we want this to be two whatever the dimension of the space. – Mark Bennet Jan 7 '19 at 7:39
• Where have you encountered the term "finitely generated $k$-algebra of transcendence degree $1$"? – Pierre-Yves Gaillard Jan 9 '19 at 12:05

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.
• Thank you so much, this clears up a lot of confusion. I had always taken "finitely generated" to mean "as a $k$-algebra". So just to confirm I have things right now, when we say "a finitely generated field extension of $k$ of transcendence degree $1$", we mean an extension $L/k$ such that we can choose $a_{1}, a_{2}, \ldots , a_{n} \in L$ so that only one of them is transcendental and $L = k(a_{1}, \ldots , a_{n})$? – Joe Jan 8 '19 at 2:16
• And in asking this I am assuming that if an element $a \in L$ is algebraic over $k$, then we say it is algebraically dependent on itself, so the set $\{ a\}$ is not algebraically independent? – Joe Jan 8 '19 at 2:23
• It is not too hard to show that being f.g. and of transcendence degree $1$ is in fact equivalent to being a finite extension of a purely transcendental subextension of transcendence degree $1$. That is, there is a transcendental element $b$ and elements $a_1,\dots, a_n$ in $L$ such that $L=K(b,a_1,\dots,a_n)$ and $L=K(b)(a_1,\dots,a_n)$ is finite over $K(b)$. – asdq Jan 9 '19 at 11:45