I am trying to prove that if $k\subseteq E\subseteq F$ are field extensions, then $$\text{tr.deg}_k F=\text{tr.deg}_k E+\text{tr.deg}_E F.$$
If $A=\{a_1,\ldots, a_n\}$ is a transcendence basis for $E$ over $k$ and $B=\{b_1,\ldots, b_m\}$ is a transcendence basis for $F$ over $E$, then I can show that $A\cup B$ is algebraically independent over $k$. However, I can't seem to prove that it is also a maximal algebraically independent set. This is equivalent to showing that $F$ is algebraic over $k(A,B)$, and it seems that for any $x\in F$ there should be a straightforward way to construct a polynomial in $k[A,B,t]$ (where $A$,$B$ are finite sets of variables) vanishing at $x$ from a polynomial in $E[B,t]$ vanishing at $x$, but I can't find it.
EDIT: I think I got it. I would very much like somebody to check whethever this solution makes sense!
So, we have a tower of extensions
$$k\subseteq k(a_1,\ldots, a_n,b_1,\ldots, b_m)\subseteq E(b_1,\ldots, b_m)\subseteq F.$$
Here $k(a_1,\ldots, a_n,b_1,\ldots, b_m)\subseteq F$ is algebraic iff $k(a_1,\ldots, a_n,b_1,\ldots, b_m)\subseteq E(b_1,\ldots, b_m)$ is algebraic.
Take $\alpha\in E(b_1,\ldots, b_m)$. We want to show that it is algebraic over $k(a_1,\ldots, a_n,b_1,\ldots, b_m)$. But $\alpha$ is a rational function, so it is sufficient to show that any monomial $\beta=eb_1^{d_1} \cdots b_m^{d_m}$ is algebraic over $k(a_1,\ldots, a_n,b_1,\ldots, b_m)$. But $e\in E$ is algebraic over $k(a_1,\ldots, a_n)$, so there is some polynomial $P\in k[x_1,\ldots,x_n,t]$ such that $P(a_1,\ldots, a_n,e)=0$.
Write this polynomial as $$P=P_0+P_1t+\cdots+ P_r t^r$$ with $P_i\in k[x_1,\ldots,x_n]$. Denote $Y=y_1^{d_1} \cdots y_m^{d_m}$. Then $Q\in k[x_1,\ldots,x_n,y_1,\ldots,y_m,t]$ defined as
$$Q=P_0Y^r+P_1Y^{r-1}t+\cdots+P_rt^r$$
vanishes at $(a_1,\ldots, a_n,b_1,\ldots, b_m,\beta)$: if we plug in this point into $Q$, we get
$$P_0(a_1,\ldots, a_n)(b_1^{d_1} \cdots b_m^{d_m})^r+P_1(a_1,\ldots, a_n)(b_1^{d_1} \cdots b_m^{d_m})^re+\cdots$$
$$+P_r(a_1,\ldots, a_n)(b_1^{d_1} \cdots b_m^{d_m})^re^r=P(a_1,\ldots, a_n,e)(b_1^{d_1} \cdots b_m^{d_m})^r=0,$$
and we are finally done (I hope).
Edit: Actually, by the same logic we just need to check that each $\beta_i$ is algebraic over $k(\alpha_1,\ldots,\alpha_n,\beta_1,\ldots,\beta_m)$, which is painfully obvious.
Best Answer
It is also sufficient to show that $e$ is algebraic over $k(a_1,\ldots, a_n,b_1,\ldots, b_m)$, which is obvious since $e \in E$ thus $e$ is algebraic over $k(a_1,\ldots, a_n)$. Clearly $b_i^{d_i}$ are all algebraic over $k(a_1,\ldots, a_n,b_1,\ldots, b_m)$. Multiply them together and we can have $\beta$ is algebraic over $k(a_1,\ldots, a_n,b_1,\ldots, b_m)$.