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Let $F$ be a field and $F^\times$ be its group of units. If $F^\times$ is cyclic show that $F$ is finite.

I'm a bit stuck. I know that I can represent $F^\times = \langle u \rangle$ for some $u \in F^\times$ and that we must have that $|F^\times| = o(u)$, I tried assuming $o(u) = \infty$, but I'm not sure exactly where to go from there.

I was wondering if I could get a hint.

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    $\begingroup$ Does unit means the usual "$u$ is a unit if it has an inverse"? If so, $F=\{0\}\cup F^*$. This means that you only need to prove that the cyclic group $F^*$ must be finite. $\endgroup$ – user141763 Apr 14 '14 at 15:27
  • $\begingroup$ Yeah, that's it. $\endgroup$ – Gruzzulinger Apr 14 '14 at 15:29
  • $\begingroup$ But I'm not sure how exactly using $\mathbb{ Q }$ would help. I realize that it can't be cyclic, but how could you set up an isomorphism? $\endgroup$ – Gruzzulinger Apr 14 '14 at 15:34
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As $F=F^\times\cup\{0\}$, the field $F$ is finite if and only if $F^\times$ is finite. The real question here is not "can the field be finite if its group of units is cyclic," but "can the field be infinite?"

Suppose $F^\times=\langle g\rangle$ is infinite generated by $g$. The characteristic must be $2$, since otherwise the element $-1\ne1$ has order $2$, but there is no element of order $2$ in $\Bbb Z$. Clearly $F=\Bbb F_2(g)$, by noting the containment goes both ways. If $g$ is transcendental then $g$ and $g+1$ are multiplicatively independent, so $F^\times$ must be generated by more than one element, a contradiction. If $g$ is algebraic over $\Bbb F_2$, then $\Bbb F_2(g)$ is finite-dimensional over $\Bbb F_2$, hence $F$ is finite, another contradiction.

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    $\begingroup$ +1 I like this better than anything proffered for the "duplicate". Splitting it to cases according to whether $g$ is algebraic/transcendental is "obvious" only in retrospect :-) $\endgroup$ – Jyrki Lahtonen Jun 18 '14 at 10:23
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    $\begingroup$ @DonAntonio: If $-1\neq1$, then $-1=g^k$ for some $k\in\Bbb{Z}\setminus\{0\}$. As $g^0=1=(-1)^2=g^{2k}$ then $k$ must have order two in the (additive) group $\Bbb{Z}$. $\endgroup$ – Jyrki Lahtonen Jun 18 '14 at 11:14
  • $\begingroup$ I think I got it now, @JyrkiLahtonen: "since otherwise" means "if $\;char\neq 2\;$ then $\;-1\neq 1\;$ and ...etc. ". Thanks. $\endgroup$ – DonAntonio Jun 18 '14 at 13:05
  • $\begingroup$ I have one query here please. I haven't come across the notation $\mathbb F_2(g)$ as of yet. Can somebody tell me what it means? I know that $F(a)$ means the smallest field containing $F$ and $a$ $\endgroup$ – MathMan Aug 16 '14 at 19:17
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    $\begingroup$ @VHP You just said what it is: the smallest field containing $\Bbb F_2$ and $g$. $\endgroup$ – blue Aug 16 '14 at 23:02
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Hints: Let $u$ be a generator for the multiplicative group $F^*$

  • If $\mathrm{char}F\neq2$, then $-u\neq u$ is a unit, and hence $-u=u^t$ for some integer $t$. Show that this implies that the order of $u$ is finite (Seth has an even better idea here).
  • If $\mathrm{char}F=2$, then you can show that the even powers of $u$ form a subfield $K$ (think: Frobenius automorphism). Unless $u$ is of a finite odd order this subfield is a proper one. If $K$ is infinite, then it easily follows ($F$ is a vector space of dimension $2$ over $K$ and in $K^2$ there are infinitely many distinct lines through the origin) that $2=[F^*:K^*]=\infty$ which is absurd.
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  • $\begingroup$ In the case $Char~F \neq 2$, then $\exists u \in F$ such that $-u \neq u$ and hence, $-u=u^t$. But, units of $F$ can form an infinite field, then how do we know that there exists a finite $t$ such that $-u = u^t.~-u$ can be very well be obtained at an infinite index. How do we justify this? $\endgroup$ – MathMan Jul 5 '14 at 19:38
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    $\begingroup$ @VHP: The cyclic group generated by $u$ consists of the elements $u^n,n\in\Bbb{Z}$. I don't know of any algebraic structure, where $u^\infty$ would be defined. $\endgroup$ – Jyrki Lahtonen Jul 5 '14 at 20:26
  • $\begingroup$ uhm okay, i just seem to be a little confused with the infinite order of the group. So, $u$ need not be the generator of $F$ right? then, is it fair to say that $-u$ is some power of $u$? $\endgroup$ – MathMan Jul 5 '14 at 20:32
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    $\begingroup$ @VHP: Well, I said that Blue's solution is simpler :-). It uses $-1$ as opposed to $-u$. Anyway. We are assuming that $F^*$ is cyclic, and that implies that there exists an element $u$ such that ALL the non-zero elements of the field are of the form $u^n$ for SOME integer $n$. The goal is to show that this implies that some non-zero power of $u$ is equal to $1$. $\endgroup$ – Jyrki Lahtonen Jul 5 '14 at 20:37
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Here's a suggestion : considering $u-1$, show that $u$ is algebraic over the caracteristic subfield (that is, the subfield generated by $1$). You then only have to rule out the case that the caracteristic subfield is $\Bbb Q$.

Unless $F=\Bbb Z/2\Bbb Z$ (and is thus finite), we must have $u-1\in F^*$ so that there is an integer $n\in \Bbb Z$ with $u-1=u^n$ (necessarily$n\neq 0,1$). From this it follows that $u$ annihiliates a polynomial with coefficients in the caracteristic subfield $\Bbbk$ (if $n<0$, just multiply the equation by a large power of $u$), that is, $u$ is algebraic. Thus $F=\Bbbk[u]$ is finite dimensional over $\Bbbk$. If $\Bbbk$ is finite we are done. But if it weren't, we'd have $\Bbbk=\Bbb Q$ and $\Bbb Q^*$ would be cyclic as a subgroup of the cyclic group $F^*$, which it isn't. Therefore there exists some prime number $p$ with $\Bbbk=\Bbb Z/p\Bbb Z$ and $F$ is finite.

One can directly show that the caracteristic $p$ of $F$ is nonzero. The case $p>2$ is easily dealt with, and only the case $p=2$ remains

Indeed, $\Bbb Q^*$ isn't cyclic, which prohibits it form being a subgroup of $F^*$, thus $p\neq 0$. If $p\geq 3$, then there exists $n$ with $u^n=2$. Since $2\in\Bbbk^*$, we have $2^{p-1}=1$, i.e. $u^{(p-1)n}=1$ and $F^*$ is finite. $F$ is finite because $F=F^*\cup\lbrace 0\rbrace$. So the argument above (considering $u-1$) is only necessary for the case $p=2$.

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Such a field cannot have characteristic $0$, since then $\mathbb{Q}^{\times}$ would be a subgroup of the multiplicative group.

If the field is of characteristic $p$, then the embedding of $\mathbb{F}_p$ makes the field an $\mathbb{F}_p$-algebra, and so it must be a quotient of $\mathbb{F}_p[x]$ (the free $\mathbb{F}_p$-algebra on one generator). But every quotient of $\mathbb{F}_p[x]$ by a nonzero ideal is finite. Therefore, such a field does not exist.

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