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.)

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