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So I want to prove that if I have $L/K/F$ then given that the transcendence degrees are finite, then $$tr_F(L)=tr_K(L)+tr_F(K)$$

$\bf{Definition:}$ We say that a set $X=\{x_i\}_{i\in I}$ is algebraically independent over $F$ if $f\in F[\{t_i\}_{i\in I}]$ such that $f((x_i)_{i\in I})=0$ implies that $f=0$.

Now I basically started like this: Let $B_1=\{x_1,...,x_m\}$ be a transcendence basis for $L$ over $K$, and $B_2=\{y_1,...,y_n\}$ a transcendence basis for $K$ over $F$. Let $B_3=\{x_1,...,y_n\}$.

We shall prove that $B_3$ is a transcendental basis for $L$ over $F$. Say $f(t_1,...,t_{m+n})$ is a polynomial with coefficients in $F$, such that $f(x_1,...,y_n)=0$. Then define:$$h(t_1,...,t_m)=f(t_1,...,t_m,y_1,...,y_n)$$Then $h$ is a polynomial with coefficients in $K$ since $B_2\subset K$, and $h(x_1,...,x_m)=0$. This means that $h=0$ by the algebraic independence of $B_1$. Then, we can let $$g(t_{m+1},...,t_{m+n})=f(x_1,....,x_m,t_{m+1},...,t_{m+n})$$However, I cannot use the fact that $g$ will be zero because $g$ might have coefficients in $F$. Any hints?


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I think you mean "because $g$ might have coefficients in $K$". :) – m_l Apr 20 '12 at 9:22
Don't we also need to prove that we can not have a transcendence basis of $L/F$ larger than $B_3$? – MathManiac Jun 13 at 6:32
up vote 3 down vote accepted

The usual trick is as follows: You write, with some abuse of notation, $ f(s_1,\ldots,s_m,t_1,\ldots,t_n) = \sum_{i \in \mathbb{N_0}^m } f_i(t_1,\ldots,t_n) s^i \in F[t_1,\ldots,t_n][s_1,\ldots,s_m]$.

Similar to your argument regarding $h$, you can conclude that $f_i(y_1,\ldots,y_n) = 0$ for all $i \in \mathbb{N_0}^m$. But the $f_i$ are polynomials with coefficients in $F$.

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