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So I'm a bit confused with this... I'm learning some field theory and I've just learnt about groups of units. With rings this makes sense. However say $F$ is a field then isn't every element in $F$ besides the additive identity (call it $0$) a unit by definition? I mean in a field every element has a multiplicative inverse... So wouldn't the group of units in $F$ just be everything besides $0$. I know we talk about this with respect to finite field.. but I mean wouldn't the same logic apply?

Sasha
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    Yes, this is of course true. The units of a any field $K$ are given by $K^{\times}$, which is $K\setminus 0$. This follows from the definition of a unit, i.e., $x$ is a unit iff $xy=1$ for some $y$. – Dietrich Burde Apr 16 '19 at 15:25
  • oh ok... just cause there is this theorem that if $F$ is a field then every finite subgroup of $(F^x,\cdot)$ is cyclic... wouldn't that just mean that $F-0$ is cyclic?... I'm just not sure why this theorem is the useful? – Sasha Apr 16 '19 at 15:28
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    Indeed, this is correct; but note that this group has information about the field. For example, the multiplicative group of $\mathbb{C}$ is divisible (every element has an $n$th root for every $n$), while the same is not true for the multiplicative group of $\mathbb{R}$ or $\mathbb{Q}$; and the multiplicative group of a finite field is cyclic, which has important implication to the theory of finite fields and their extensions. – Arturo Magidin Apr 16 '19 at 15:28
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    Sure, $(F^{\times},\cdot)$ is cyclic for a finite field. – Dietrich Burde Apr 16 '19 at 15:28
  • @Sasha: It is very useful for lots of things related to the study of finite fields, how they are contained in each other, the uniqueness of a field of order $p^n$, etc. It’s too early for you to see its importance, but it is important. – Arturo Magidin Apr 16 '19 at 15:29
  • Ah.. ok thank you I was just very confused on why this definition for fields was so important.. – Sasha Apr 16 '19 at 15:30
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    For every definition there may be a "trivial example", which does not mean that the definition "does not make sense", or is useless. – Dietrich Burde Apr 16 '19 at 15:36
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    The inference from "every finite subgroup of $F-{0}$ is cyclic" to "$F-{0}$ is cyclic", though trivial for finite $F$, can fail for infinite fields. Both the field of rational numbers and the field of real numbers give counterexamples. – Andreas Blass Apr 16 '19 at 17:20
  • I'm sure @AndreasBlass knows/suspects that the converse is also true: If $F\setminus{0}$ is cyclic, then $F$ is finite. Linking it just in case. And highly recommend Blue's answer. – Jyrki Lahtonen Apr 17 '19 at 04:20

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