transcendence basis of field extensions of $\mathbb{Q}$ In some exercice, I see the following: Let $K = \mathbb{Q}(X_1 ,\dots , X_n )$ and $k = \mathbb{Q}(e_1 , \dots, e_n )$, where $(e_i)$ are the elementary symmetric polynomials. It states: "Since K is a finite extension, it is algebraic over k, therefore K and k have the same transcendence degree over $\mathbb{Q}$." I see no such result in my course, and the closest thing I found is this result: since K is finite, therefore algebraic over k, the family $S=(e_1 , \dots , e_n )$ contains a transcendence basis of K over $\mathbb{Q}$. Could you please help me find the right direction? Thank you.

A set $\\{t_1,\dots,t_n\\}$ of elements of an extension field $k/F$ is a transcendency basis if

1. there is no nonzero polynomial $f(X_1,\dots,X_n)\in F[X_1,\dots,X_n]$ such that $f(t_1,\dots,t_n)=0$;
2. $k$ is algebraic over $F(t_1,\dots,t_n)$.



Note that condition 1 implies that every $t_i$ is transcendental over $F$ (but it is stronger than this).

It follows that if $K$ is an algebraic extension of $k$, then $\\{t_1,\dots,t_n\\}$ is also a transcendency basis of $K$ over $F$, because the elements are still algebraically independent and $K$ is algebraic over $F(t_1,\dots,t_n)$ (algebraic over algebraic).

Condition 1 is usually expressed by saying that $t_1,\dots,t_n$ are algebraically independent. For infinite sets, substitute condition 1 with “every finite subset is algebraically independent”.