# 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.

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

• I'm not sure about your second sentence. The set $S$ doesn't have to generate $A$ as a $K$-vector space. – Carlos Esparza Oct 6 '18 at 17:29
• By $S$ generating $A$, do you mean it generates as a vector space or an algerbra? – Mario 04 Oct 6 '18 at 17:31
• I mean as $K$-algebra – Carlos Esparza Oct 6 '18 at 17:32
• 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. – Mario 04 Oct 6 '18 at 17:34