# Are algebraic numbers analogous to group elements with finite order?

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.

• 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. – Glen O May 3 '13 at 11:54
• @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! – user50229 May 3 '13 at 11:58
• @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. – Easy May 3 '13 at 11:59
• @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.) – user50229 May 3 '13 at 12:06
• @user50229, it just means it has finite order. – 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.

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).

• 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. – Pete L. Clark May 3 '13 at 16:33
• @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? – Jack Schmidt May 3 '13 at 18:20
• (I've edited my answer optimistically.) – Jack Schmidt May 3 '13 at 18:21
• 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. – Pete L. Clark May 3 '13 at 20:59
• Unfortunately this answer is way above my head! – user50229 May 5 '13 at 18:49