If a field has a cyclic multiplicative group, is it necessarily finite?
$\begingroup$
$\endgroup$
4
-
$\begingroup$ Is the product group the same as the multiplicative group (en.wikipedia.org/wiki/Multiplicative_group)? $\endgroup$– user5810Aug 29, 2011 at 4:31
-
$\begingroup$ Yes, it has been edited. $\endgroup$– AdterramAug 29, 2011 at 4:38
-
$\begingroup$ Do you mean: If a field is cyclic as a multiplicative group..., or If a field has a cyclic multiplicative subgroup? $\endgroup$– Daniel MansfieldAug 29, 2011 at 5:27
-
2$\begingroup$ It wouldn't be good to use the first expression, since the multiplicative group excludes an element of the underlying set. $ \; $ If the field has a cyclic multiplicative group, then the field's multiplicative group is cyclic, since the field has a unique multiplicative group (namely, it's multiplicative group). $\endgroup$– user5810Aug 29, 2011 at 6:07
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
Assume that the elements of field $F$ are 0, ${a^n}$.
Case 1: If the characteristic is not 2. Then$1 \ne - 1$, therefore$ - 1 = {a^n}(n \ne 0)$ and ${a^{2n}} = 1$. Hence we conclude that it is finite.
Case 2: If the characteristic is 2. First we have $1 + a = {a^s}$and we can assume that $s \ne 1$ , otherwise the field is trivial. Then we conclude that $a$ is algebraic over ${F_2}$and ${\rm{ F= }}{F_2}(a)$. We can set the degree of the minimal polynomial of is n , then all the elements of $F$ can be written as a linear combination of $1,a,{a^2} \cdots {a^{n - 1}}$over${F_2}$ , but the cardinal number of all such element is finite, which concludes the proof.
$\begingroup$
$\endgroup$
1
Yes; it suffices to rule out that the multiplicative group is $\mathbb{Z}$. The field can't have characteristic zero, since $\mathbb{Q}^\times$ is not finitely generated (so not contained in $\mathbb{Z}$). If it has characteristic $p \neq 2$, then it contains $\mathbb{F}_p$, where $\mathbb{F}_p^\times$ has torsion of order greater than 2 and so is also not contained in $\mathbb{Z}$. If $p = 2$, then the field can't contain $\mathbb{F}_4$ for the same reason, so if it's not $\mathbb{F}_2$, it must be transcendental over it. But $\mathbb{F}_2(x)^\times$ is not finitely generated again, so not contained in $\mathbb{Z}$ either.
answered Aug 29, 2011 at 4:42
-
4$\begingroup$ The same argument proves the stronger claim that, if the multiplicative group is finitely generated then the field is finite. $\endgroup$ Aug 29, 2011 at 5:55
$\begingroup$
$\endgroup$
The following proof is a short proof of the stronger claim stated by @Gjergji-Zaimi :
The main ingredient is the fact that if finitely generated $\mathbb{Z}$-algebra is a field, then it is a finite field. (this is essentially equivalent to Nullstellensatz).
If the multiplicative group of a field $F$ is finitely generated then $F$ would be a finitely generetad $\mathbb{Z}$-algebra, hence $F$ is a finite field.
