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For the sake of a proof using contradiction ( to be used somewhere), Lets assume that an infinite cyclic field $F$ of non zero units exists with characteristic $\neq 2$ . In this infinite cyclic field of non zero units , suppose an element '$u$' exists satisfying $2u \neq 0$ or $-u \neq u$.

Does this always imply that $-u = u^t$ for some finite $t$? If $F$ is infinite, then, is it not possible that $-u$ is obtained at an $\infty$ index of u? How shall we justify that there always exists a finite $t$ such that $-u = u^t$ ?

Thank you for your help..

MathMan
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    In the related question it was assumed that the group of units of the field is generated by an element $u$. Therefore all the units are of the form $u^t, t\in\Bbb{Z}$. $-u$ is one of those units. It is basic field theory that if the characteristic of the field is $\neq2$, then $-u\neq u$ for all $u\neq0$. – Jyrki Lahtonen Jul 05 '14 at 20:31
  • Char $F \neq 2 \implies \exists$ at least one element $u$ such that $2u \neq 0$. How do we know that $u$ must be the generator of this infinite field of non zero units? – MathMan Jul 05 '14 at 20:35
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    No! The assumption that $F^*$ is cyclic implies the existence of an element $u$ such that all the non-zero elements of $F$ are powers of $u$. The assumption that Char $F\neq2$ implies that $x\neq-x$ for ALL non-zero $x\in F$. – Jyrki Lahtonen Jul 05 '14 at 20:39
  • But if this property of characteristic two fields is difficult to digest, please look at the simpler solution by Blue. There we only need $-1\neq1$. Finiteness follows from the assumption that $-1$ must be a power of the generator. – Jyrki Lahtonen Jul 05 '14 at 20:42
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    May be you accepted my hints to your previous question on this theme too soon? It may pay to wait until all is clear? I guess styles vary here. – Jyrki Lahtonen Jul 05 '14 at 20:44
  • Ohh! I get it now. :-) The definition of characteristic conveys a different meaning though. Like - char $R$ is defined as the smallest positive integer $n$ for which $nx=0~\forall~x \in R$. Hence, i guess I mistakenly concluded that $char R \neq 2 \implies~\exists~$ atleast one element $u \in R$ s.t. $2u \neq 0$ – MathMan Jul 05 '14 at 20:45
  • I did have a look at Blue's solution as well. But, honestly, I found your solution more intuitive :-) – MathMan Jul 05 '14 at 20:49
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    Characteristic is defined as the additive order of $1$ (except infinite order is weirdly enough called characteristic zero). So char $\neq2$ by definition implies that $1+1\neq0$, or equivalently, $1\neq-1$. Multiplying this by any non-zero element $x$ gives $x\neq-x$. – Jyrki Lahtonen Jul 05 '14 at 20:49
  • yeah .. fine for me now .. – MathMan Jul 05 '14 at 20:53
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    (@Jyrki - I think of the characteristic of unital $R$ as the unique wlog nonnegative integer principally generating the kernel of the unique map $\Bbb Z\to R$. For nonunital I suppose we can replace "kernel" with "annihilator," as a $\Bbb Z$-module.) – anon Jul 05 '14 at 22:13

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