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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

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    $\begingroup$ HINT: What is the multiplicative order of $-1$? $\endgroup$ – Brian M. Scott Sep 23 '15 at 15:45
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    $\begingroup$ Does an infinite cyclic group have any element of order $2$? $\endgroup$ – Brian M. Scott Sep 23 '15 at 15:50
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    $\begingroup$ @BrianM.Scott : You are assuming that the characteristic is not 2. $\endgroup$ – Nate Sep 23 '15 at 16:01
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    $\begingroup$ @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. $\endgroup$ – Brian M. Scott Sep 23 '15 at 16:03
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    $\begingroup$ -1 =1 in characteristic 2 $\endgroup$ – Nate Sep 23 '15 at 16:03
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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 algebaic 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.

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    $\begingroup$ 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 $k- F_p$) $\endgroup$ – quid Sep 23 '15 at 16:26
  • $\begingroup$ @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 . $\endgroup$ – user118494 Sep 23 '15 at 19:13
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    $\begingroup$ @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.) $\endgroup$ – quid Sep 23 '15 at 20:16
  • $\begingroup$ @Nate : Ok . Got it. $\endgroup$ – user118494 Sep 24 '15 at 7:49

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