Multiplicative group of an infinite field is not cyclic

Question

Question.

Prove that the multiplicative group of any infinite field can never be cyclic .

$\mathbb R$, $\mathbb Q$, $\mathbb C$ are some infinite fields whose multiplicative groups are not cyclic, I know.

I need some lead as to how to begin the proof.

Sorry for the lack of work on my part (I'm clueless) and any help is appreciated.

Does an infinite cyclic group have any element of order $2$? — Brian M. Scott, Sep 23 '15 at 15:50
@BrianM.Scott : The book this problem is from says to consider fields of char 0 and char $p$ separately . But this logic does not use characteristic of fields anywhere . So is a different approach needed now $?$ — , Sep 23 '15 at 16:01
@BrianM.Scott : You are assuming that the characteristic is not 2. — Nate, Sep 23 '15 at 16:01
@user80631: Yes; on account of your examples I was thinking of fields of characteristic $0$. The same argument works provided that the characteristic is not $2$, as Nate mentions. That leaves only characteristic $2$ to be dealt with separately. — Brian M. Scott, Sep 23 '15 at 16:03
@BrianM.Scott : Pls , help me with that $p\neq 2$ case then . — , Sep 23 '15 at 16:05
@Nate : Ok . Then do you think for char 0 and char $p\neq 2$ the same logic that Brian Scott gave will work $?$ — , Sep 23 '15 at 16:06
I think BrianM.Scott gave more or less the whole idea of the proof in those cases. An infinite cyclic group has no nontrivial elements of finite order, but -1 always has order 2 so the multiplicative group cannot be cyclic. — Nate, Sep 23 '15 at 16:09

score 11 · Accepted Answer · edited Feb 22 '21 at 23:20

11

Ok here is the characteristic 2 case:

Assume $k$ is an infinite field of characteristic $2$ with a cyclic multiplicative group. Note that any element of an algebraic extension of $\mathbb{F}_2$ has finite multiplicative order, so this implies that every element of $k-\{0,1\}$ must be transcendental.

Next let $x$ be a generator for the multiplicative group, which exists as we are assuming it is cyclic. Consider the element $1+x$ of our field. It is nonzero and therefore equal to some power of $x$ since $x$ is a generator. But then $1+x=x^n$ for some $n$, so $x$ is algebraic over $\mathbb{F}_2$, contradicting the above claim.

edited Feb 22 '21 at 23:20

user26857

52,094

answered Sep 23 '15 at 16:18

Nate

11,206

1

Given the hint in the book it might be worth adding that this proof works for any positive characteristic $p$ (replacing $k - {0,1}$ by $k- F_p$) – quid Sep 23 '15 at 16:26
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@Nate : One step is not clear to me . "Any element of algebraic extension of of $\mathbb F_2$ has finite multiplicative order" . Could you please give a little more explanation of that $?$ . Thank you . – user118494 Sep 23 '15 at 19:13
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@user118494 the field generated by an algebraic element will be an extension of $F_2$ of finite degree. Thus, a finite field, in which the order thus ought to be finite. (That there is a larger surrounding field is not relevant.) – quid Sep 23 '15 at 20:16

score 8 · Answer 2 · edited Feb 22 '21 at 23:16

I was studying this for a Galois Theory course and also bumped into this question, and actually for fields of characteristic different from 2 there is a really surprisingly simple proof (spoiler, this will happen because in characteristic $2$ we get $-1=1$).
So suppose $k$ is an infinite field such that char $k \neq 2$ and there is $x$ with $\langle x \rangle = k^*$. This implies that there is an $n \in \mathbb{N}$ (different than $0$ because $-1 \neq 1$) such that $x^n = -1$, which implies that $x^{2n} = 1$ and so $k^*$ is finite, contradiction.

Multiplicative group of an infinite field is not cyclic

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