Y by skyguy88
The answer is yes. I split the proof in a proposition, a lemma and a theorem.
Proposition: Let be a positive algebraic number such that is atomic. Suppose that is not finitely generated. Then has infinitely many Betti elements.
Proof: We have that . Let be the minimal pair of . Consider the set We know that , then is a nonempty set of positive integers. Let be its minimal element and let such that and . Consider such that . We have that is not an isolated vertex in if and only in there exists , , such that . It is not hard to see that for some . Then and . This contradictis the minimality of in . Hence is an isolated vertex in . Then is a Betti element of .
Now, for each , consider . It is clear that and , for all . Following exactly the same reasoning as before, we obtain that there exists an isolated vertex in . Therefore is a Betti element of for each . This concludes our proof.
Lemma: Let . We say that, for and , if for all and for some . Then, the set:
has only finitely many elements.
Proof: Let us proceed by induction on . If , we have that , and by the well-ordering theorem we have that has a minimal element . It follows that . Now, assume the Lemma is valid for . Let . Consider the set:
Fix , and such that . Then for every such that , we have that and . Now, let such that . Then we have that for some and some , , where
Let , such that . It is clear that for , . Applying the lemma in , we obtain that the set
has only finitely many elements.
Since we obtain that also has only finitely many elements. Our proof concludes
Theorem: Let be a positive algebraic number such that is atomic and finitely generated. Then has only finitely many Betti elements.
Proof: Let be the minimal polynomial of , . Suppose that is a Betti element of . Then, there exist such that and are not in the same connected component in . We can say that , , where . It is known that , for some . In this case we will say that produces a Betti element.
Consider the set . For and we say that if and only if for each , or .
It is not hard to see that is an equivalence relation that has finitely many equivalence classes. Let be an equivalence class of . Because of the definition of it is not hard to see that for some and , there exists a map such that for , where is the set of positive coeficients of . Now, note that if we have that for , then
It follows that there exists such that , , . Note that connects and in . Then does not produce a Betti element. Using Lemma 3 on it follows that only finitely many polynomials in produce a Betti element. Since there are only finitely many equivalence classes in , we conclude that must have only finitely many Betti elements.
Proposition: Let be a positive algebraic number such that is atomic. Suppose that is not finitely generated. Then has infinitely many Betti elements.
Proof: We have that . Let be the minimal pair of . Consider the set We know that , then is a nonempty set of positive integers. Let be its minimal element and let such that and . Consider such that . We have that is not an isolated vertex in if and only in there exists , , such that . It is not hard to see that for some . Then and . This contradictis the minimality of in . Hence is an isolated vertex in . Then is a Betti element of .
Now, for each , consider . It is clear that and , for all . Following exactly the same reasoning as before, we obtain that there exists an isolated vertex in . Therefore is a Betti element of for each . This concludes our proof.
Lemma: Let . We say that, for and , if for all and for some . Then, the set:
has only finitely many elements.
Proof: Let us proceed by induction on . If , we have that , and by the well-ordering theorem we have that has a minimal element . It follows that . Now, assume the Lemma is valid for . Let . Consider the set:
Fix , and such that . Then for every such that , we have that and . Now, let such that . Then we have that for some and some , , where
Let , such that . It is clear that for , . Applying the lemma in , we obtain that the set
has only finitely many elements.
Since we obtain that also has only finitely many elements. Our proof concludes
Theorem: Let be a positive algebraic number such that is atomic and finitely generated. Then has only finitely many Betti elements.
Proof: Let be the minimal polynomial of , . Suppose that is a Betti element of . Then, there exist such that and are not in the same connected component in . We can say that , , where . It is known that , for some . In this case we will say that produces a Betti element.
Consider the set . For and we say that if and only if for each , or .
It is not hard to see that is an equivalence relation that has finitely many equivalence classes. Let be an equivalence class of . Because of the definition of it is not hard to see that for some and , there exists a map such that for , where is the set of positive coeficients of . Now, note that if we have that for , then
It follows that there exists such that , , . Note that connects and in . Then does not produce a Betti element. Using Lemma 3 on it follows that only finitely many polynomials in produce a Betti element. Since there are only finitely many equivalence classes in , we conclude that must have only finitely many Betti elements.