Can you determine the transcendence degree of an algebra by looking at a generating set? Let $K$ be a field and $A$ be a $K$-algebra generated (as $K$-algebra) by a set $S$. The transcendence degree of $A$ is$$
\operatorname{trdeg}(A) = \sup\{|T| : T \subset A,\, T \text{ algebraically independent}\} \quad .
$$
Setting
$$
N = \sup\{|T| : T \subset S,\, T \text{ algebraically independent}\}\quad\phantom{.}
$$
(here $T$ is a subset of $S$ instead of $A$), does the equality
$$
N = \operatorname{trdeg}(A)
$$
always hold?
So far I've got that obviously $N \leq \operatorname{trdeg}(A)$ and if $A$ is contained in a finitely generated $K$-algebra equality must hold.
 A: $\mathbf{Another \; result}$: Let $\mathcal{A}$ be the smallest spanning of $A$(the basis of $A$ as a $K$ vector space). The equality $N=$trdeg$(A)$ is equivalent by saying that there is a transcendence base in a base for the vector space $A$.                                                         
Now if algebraically independent implies linearly independent, it obviously implies that a transcendence base is in a base for the vector space $A$, because all transcendence bases are also algebraically independent(we are going to call this statement (*)).                                              
Let $B:=\{b_{1};...;b_{m} \}$ be a set of algebraically independent elements of $A$. This, by definition happens iff we choose $f \in K[X_{1};...;X_{M}]$ such that $f(b_{1};...;b_{m})=0$, then $f=0$.                                                    
Let $f:=a_{1}T_{1}+...+a_{m}T_{m} \in K[X_{1};...;X_{M}]$ such that \begin{equation}f(b_{1};...;b_{m})=0 \Leftrightarrow a_{1}b_{1}+...+a_{m}b_{m}=0 \Rightarrow a_{1}=...=a_{m}=0\end{equation}.
So we see that $B$ is indeed linearly independent.                                        
By the statement (*), we indeed obtain that every transcendence base is in a vector space base.                                                                    
$\mathbf{Edit}$: Let's take trdeg$(A)=2$. As in the first result, we must prove that there is a transcendence base in $S$. Let $T:=\{t;t'\}$ be a transcendence base. Let's assume that $S:=L \cup \{t\}$, where $L$ is a set of algebraic elements. Then $A=K(L;t)=(K(t))(L) \cong (K(X_{1}))(L)=(K(L))(X_{1})$ but $K(S;t') \cong (K(L))(X_{1};X_{2})$ and $K(S;t')=A$ since $t' \in A$ ,but $(K(L))(X_{1}) \subsetneq (K(L))(X_{1};X_{2}) $, so $A \subsetneq K(S;t')$ which is a contradiction. The case where trdeg$(A)=n$ with $2 \leq n$ is an analogy of trdeg$(A)=2$.                                                     
