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Let $F$ be a field of nonzero characteristic. Artin's book [1] proves the fact that $F$ has prime characteristic along the following lines:

Suppose $F$ has characteristic $m$, where $m$ is not prime. Denote the sum of $n$ copies of $1$ by $\overline n$. Then, $1$ generates a cyclic subgroup of $(F,+)$ with distinct elements $\overline 1 (= 1),\overline 2,......,\overline{m-1}$, and $\overline m =0$. Now, as $m$ is not prime, let $m=rs$, where $1<r,s<m$. Now, in the multiplicative group $(F',*)$, $\overline r$ and $\overline s$ are both members (as neither is $0$), but $\bar r \bar s=0$, which contradicts the fact that $(F',*)$ is a group.

Seems easy to enough to understand, but I am not sure why all the stuff about a cyclic subgroup is there in the middle (in bold).

[1] M. Artin, Algebra (Second Edition), Pearson.

Student
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  • Distinctness of those elements is used later, when concluding that $\overline{r}$ and $\overline{s}$ are non-zero. I guess you could do without mentioning cyclicity explicitly. Probably it is to invoke an earlier discussion about listing elements of a cyclic group and a way of deciding when two elements of a cyclic group are equal. – Jyrki Lahtonen Sep 19 '20 at 12:27
  • More or less this and others linked to that one. I guess there is some variation in the emphasis/details. – Jyrki Lahtonen Sep 19 '20 at 12:28
  • Hmmmmmm, I think i get where you are coming from, but isn't the fact that characteristic is defined as $m$ enough to conclude that $\overline r$ and $\overline s$ are nonzero? Because $r,s < m$. – Student Sep 19 '20 at 12:29
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    That's more or less all there is to it. I wouldn't lose sleep over this. You seem to understand the proof and the result. The author chose to mention cyclic groups here for some reason. Shrug :-) – Jyrki Lahtonen Sep 19 '20 at 12:33
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    Probably the point is that one needs to understand what is meant by $\overline{n}$, and the fact that $\overline{m}=0$, which means the relevant elements are $0$, $1$, $\overline{2}$, $\ldots$, $\overline{m-1}$. The fact that these elements form a cyclic group would be enough of an afterthought to Artin that it is probably just mentioned in passing. – halrankard2 Sep 19 '20 at 13:11

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