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109.20 Transcendence degree

Exercise 109.20.1. Let $k \subset K \subset K'$ be field extensions with $K'$ algebraic over $K$. Prove that $\text{trdeg}_ k(K) = \text{trdeg}_ k(K')$. (Hint: Show that if $x_1, \ldots , x_ d \in K$ are algebraically independent over $k$ and $d < \text{trdeg}_ k(K')$ then $k(x_1, \ldots , x_ d) \subset K$ cannot be algebraic.)

Exercise 109.20.2. Let $k$ be a field. Let $K/k$ be a finitely generated extension of transcendence degree $d$. If $V, W \subset K$ are finite dimensional $k$-subvector spaces denote

\[ VW = \{ f \in K \mid f = \sum \nolimits _{i = 1, \ldots , n} v_ i w_ i \text{ for some }n\text{ and }v_ i \in V, w_ i \in W\} \]

This is a finite dimensional $k$-subvector space. Set $V^2 = VV$, $V^3 = V V^2$, etc.

  1. Show you can find $V \subset K$ and $\epsilon > 0$ such that $\dim V^ n \geq \epsilon n^ d$ for all $n \geq 1$.

  2. Conversely, show that for every finite dimensional $V \subset K$ there exists a $C > 0$ such that $\dim V^ n \leq C n^ d$ for all $n \geq 1$. (One possible way to proceed: First do this for subvector spaces of $k[x_1, \ldots , x_ d]$. Then do this for subvector spaces of $k(x_1, \ldots , x_ d)$. Finally, if $K/k(x_1, \ldots , x_ d)$ is a finite extension choose a basis of $K$ over $k(x_1, \ldots , x_ d)$ and argue using expansion in terms of this basis.)

  3. Conclude that you can redefine the transcendence degree in terms of growth of powers of finite dimensional subvector spaces of $K$.

This is related to Gelfand-Kirillov dimension of (noncommutative) algebras over $k$.


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