# Proof verification: transcendence degree additive in towers

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.

so it is sufficient to show that any monomial $$\beta=eb_1^{d_1} \cdots b_m^{d_m}$$ is algebraic over $$k(a_1,\ldots, a_n,b_1,\ldots, b_m)$$
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 $$k(a_1,\ldots, a_n,b_1,\ldots, b_m)$$.
Let $$F/E/k$$ be a tower of extensions and assume $$E/k$$ is algebraic. If $$B\subseteq F$$ is any subset, then $$E(B)/k(B)$$ is algebraic.
Certainly, any element $$a\in E$$ is algebraic over $$k$$, and hence over $$k(B)$$. Also any element $$b\in B$$ is obviously algebraic over $$k(B)$$. But algebraic elements of $$F$$ over $$k$$ form a field, so every element of $$E(B)$$ is algebraic over $$k$$.
Back to your question, you know $$E/k(A)$$ is algebraic, so $$E(B)/k(A\cup B)$$ is algebraic by what I said. You also know $$F/E(B)$$ is algebraic, so $$F/k(A\cup B)$$ is algebraic and you are done. This does not construct an specific polynomial though.