Cardinality of transcendental basis over field generated by algebraic independent set

Question

Recall that $T$ is a transcendental basis of $\Bbb R$ over $\Bbb Q$ if it is a maximal algebraically independent set. Also, For $B\subset\Bbb R$, the transcendental degree of $\Bbb R$ over $\Bbb Q(B)$ is the cardinality of any transcendental basis of $\Bbb R$ over $\Bbb Q(B)$. Here is my question.

Assume $B\subset T$, then the transcendental degree of $\Bbb R$ over $\Bbb Q(B)$ equals to the cardinality of $T\setminus B$. Is that right? Is $T\setminus B$ a transcendental basis of $\Bbb R$ over $\Bbb Q(B)$?

Answer 1

Suppose $F$ is a subfield of $K$. A subset $T$ of $K$ is a transcendency basis of $K$ over $F$ if and only if

1. $T$ is algebraically independent over $F$;
2. $K$ is algebraic over $F(T)$.

You ask whether $T\setminus B$ is a transcendency basis of $\mathbb{R}$ over $\mathbb{Q}(B)$. It is.

The set is obviously algebraically independent. If $r\in\mathbb{R}$, we know it is algebraic over $\mathbb{Q}(T)=F(T\setminus B)$, where $F=\mathbb{Q}(B)$. Done.