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.


$\mathbf{Another \; result}$: 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$(A)=2$.

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    $\begingroup$ I'm not sure about your second sentence. The set $S$ doesn't have to generate $A$ as a $K$-vector space. $\endgroup$ – Carlos Esparza Oct 6 '18 at 17:29
  • $\begingroup$ By $S$ generating $A$, do you mean it generates as a vector space or an algerbra? $\endgroup$ – Mario 04 Oct 6 '18 at 17:31
  • $\begingroup$ I mean as $K$-algebra $\endgroup$ – Carlos Esparza Oct 6 '18 at 17:32
  • $\begingroup$ Sorry, I thought as a vector space, but take for example $\mathbb{Q}(\pi; \sqrt{2})$. The transcendence base is still in it. Leave me some time to figure out if it happens in general. $\endgroup$ – Mario 04 Oct 6 '18 at 17:34

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