Would you say that the "elements with finite order" in group theory are analogous to "algebraic numbers" in field theory?

I thought this is the case since requiring an algebraic number $\alpha$ to be the root of a polynomial (i.e. requiring a finite combination of terms in $\alpha$ using $+$ and $\times$ to give the identity zero) is like the two-operation equivalent of an element $g$ in group theory having finite order (where there is only one operation $\times$ and we require a term in $g$ which is required to give the identity 1).

However I don't think the analogy is quite complete because in the case of the polynomial we are allowed to also multiply powers of $\alpha$ by other elements of the field to achieve the identity.

  • $\begingroup$ How do you define "analogous" in this case? One could argue that the analogue for "elements with finite order" in group theory is "elements with finite order" in field theory. $\endgroup$ – Glen O May 3 '13 at 11:54
  • $\begingroup$ @Glen O I'm not quite sure, maybe there is a more appropriate word, I just thought they are sort of similar ideas. There might not be any link of course, I am relatively new to algebra! $\endgroup$ – user50229 May 3 '13 at 11:58
  • $\begingroup$ @user50229, are you trying to say that in regards to algebraic numbers being a group, it will be torsion? But it is easy to see there are torsion-free algebraic numbers. $\endgroup$ – Easy May 3 '13 at 11:59
  • $\begingroup$ @Easy Hi Easy, I had not come across the notion of 'torsion' yet, but I can read up about it now that you've mentioned it! (Having looked on Wikipedia it seems this is more advanced than I have covered so far.) $\endgroup$ – user50229 May 3 '13 at 12:06
  • $\begingroup$ @user50229, it just means it has finite order. $\endgroup$ – Easy May 3 '13 at 12:09

There is a general model-theoretic notion of algebraic elements, see here.

If $L/K$ is a field extension, then $a \in L$ is algebraic over $K$ in the usual sense iff it is algebraic in the sense of model theory (applied to the structure $(L,+,*,0,1)$ and the subset of $K$).

If $G$ is a group, then $a \in G$ is algebraic over $\emptyset$ in the sense of model theory iff there is some $n \in \mathbb{Z}$ with $g^n=1$ and $\{h \in G : h^n=1\}$ is finite. Thus, if $G$ is finite, then $n=0$ works and every element is algebraic. This concept is more interesting when $G$ is infinite. Then every algebraic element is torsion. The converse probably does not hold.


Here is an answer to help you get started:

There are two kinds of numbers: algebraic and transcendental. An important theorem shows that a field can be split into those two types as well, so that everything boils down to (repeated) algebraic extensions like K[x]/(f) and transcendental extensions like K(x). For general groups, this is no longer the case. The largest class of groups like that are called polycyclic groups. They also have a normalization theorem where the elements of finite order can be (mostly) gathered together, and then one has the repeated adjoining of "transcendental" extensions, in this case $\mathbb{Z}$.

The transcendence degree of a field corresponds to the Hirsch length of a polycyclic group.

A “purely transcendental” field extensions is just a $K(X)$. However, a “torsion-free” polycyclic group can have many different structures. In other words, for fields $K(x_1)(x_2)\ldots(x_n) = K(X)$ so that repeated transcendental extensions simplify the same way, but for polycyclic groups extended by $\mathbb{Z}$ then $\mathbb{Z}$ then $\dots$ then $\mathbb{Z}$ there are many possibilities, not just $\mathbb{Z}^n$.

If one steps away from polycyclic groups then things go badly. For instance polycyclic groups are finitely generated, so we are ignoring groups that could correspond to field extensions like $\mathbb{Q} \leq \mathbb{C}$. Let's increase the hypotheses on the group structure, but allow infinite generation: abelian groups. Now there is still a normalization lemma, the elements of finite order form a subgroup, but there is no longer any great definition of transcendence basis. As an abelian group, we have rank 1 groups like $\mathbb{Z}$ with a basis, but $\mathbb{Z}[\tfrac12]$, $\mathbb{Z}_{2}=\{ \tfrac{a}{2b+1} : a,b \in \mathbb{Z} \}$, $SF = \{ \tfrac{a}{b} : a,b \in \mathbb{Z}, b \neq 0 \text{ is square-free} \}$, and $\mathbb{Q}$ do not. So the normalization is much less useful. Even worse, putting together rank 1 groups is not the only way to get rank 2 groups, so even the uncountably many types of rank 1 groups (versus 1 type of trdeg 1 field) are not enough to describe the rank 2 groups (versus 1 type of trdeg 2 field).

  • $\begingroup$ Your (otherwise very nice) answer seems to confuse finitely generated field extensions with finitely generated $K$-algebras: in the transcendental case they are quite different. In particular: $K[x]$ is not a field at all, and the theory of transcendence bases for field extensions does not follow from Noether Normalization: rather it is the other way around. $\endgroup$ – Pete L. Clark May 3 '13 at 16:33
  • $\begingroup$ @PeteL.Clark: thanks. The notation for fields was just a typo. The normalization result might be a real confusion on my part. Is my claim just "the existence of a transcendence basis"? Does that result have a name? $\endgroup$ – Jack Schmidt May 3 '13 at 18:20
  • $\begingroup$ (I've edited my answer optimistically.) $\endgroup$ – Jack Schmidt May 3 '13 at 18:21
  • $\begingroup$ The fact that for every field extension $K/F$ there is a subextension $L$ such that $L/F$ is purely transcendental and $K/L$ is algebraic is precisely the existence of transcendence bases, yes. This is not a very deep result: if you look e.g. in $\S$ 12 of math.uga.edu/~pete/FieldTheory.pdf, you can see that it follows within a page of giving the basic definitions involved. $\endgroup$ – Pete L. Clark May 3 '13 at 20:59
  • $\begingroup$ Unfortunately this answer is way above my head! $\endgroup$ – user50229 May 5 '13 at 18:49

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.