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The theorem statement is: a finite field of characteristic $p$ has $p^n$ elements. I found this very simple proof in the book "Ling, San; Xing, Chaoping; Coding Theory - A First Course". But I don't understand the highlighted step. Doesn't that require $(b_2-a_2)|(a_1-b_1)$?

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Jorge Lopez
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  • If one thinks of $b_2-a_2$ not as an integer, but as an element of the field $\Bbb F_p=\Bbb Z/p\Bbb Z$ then that's OK. – Angina Seng Jun 05 '20 at 02:01
  • It's a field, every element has an inverse. Integers don't have inverses in the ring of integers, but they do in the ring of integers mod $p$. For instance, $2^{-1}\equiv 3$ in ${\bf Z}_5$ even though of course $\frac{1}{2}$ is not an integer. – anon Jun 05 '20 at 02:09
  • Yes you are right. I missed that. I was assuming those were integers. – Jorge Lopez Jun 05 '20 at 02:13
  • @runway44 of course you meant every nonzero element has an inverse. – Gerry Myerson Jun 05 '20 at 02:28
  • This is unnecessarily technical and blurs the view on what is really going on. Maybe you should switch to a better coding theory book (there are very nice choices). The usual proof is to see that the integer multiples of $1$ form a subfield $P$ of order $p$ (the prime field of $P$), and that $F$ is a vector space over $P$. Both are direct consequences of the field axioms. By linear algebra (two vector spaces over the same field are isomorphic if and only if they have the same dimension), $F$ is isomorphic to $P^n$ for some positive integer $n$. Hence $#F = #(P^n) = p^n$. – azimut Dec 20 '23 at 19:12

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Hint: The field is a vector space of finite dimension over its prime sub-field.