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111.20 Transcendence degree

Exercise 111.20.1. Let K'/K/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 111.20.2. 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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