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?

  • 2
    $\begingroup$ 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. $\endgroup$
    – reuns
    Jan 7 '19 at 6:38
  • 1
    $\begingroup$ "transcendence degree" and "dimension" are two different things. $\endgroup$
    – Slade
    Jan 7 '19 at 6:45
  • $\begingroup$ @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. $\endgroup$ Jan 7 '19 at 7:39
  • $\begingroup$ Where have you encountered the term "finitely generated $k$-algebra of transcendence degree $1$"? $\endgroup$ 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.

  • $\begingroup$ 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 $L = k(a_{1}, \ldots , a_{n})$? $\endgroup$
    – Joe
    Jan 8 '19 at 2:16
  • $\begingroup$ 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? $\endgroup$
    – Joe
    Jan 8 '19 at 2:23
  • $\begingroup$ 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 $K(b)$. $\endgroup$
    – asdq
    Jan 9 '19 at 11:45

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