3
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.