0

Theorem: Every finite field has order $p^n$.

Proof:

Assume by contradiction a field $F$ of order $k$ such that $k$ has two distinct primes (of orders $m$ and $n$ dividing it), namely $p$ and $q$. Since $<F, +>$ is an (abelian) group of order k, we have a Sylow $p$-subgroup of order $p^m$. By Cauchy's theorem (existence of an element of prime order if it divides the group order) we have an element of order $p$ in the subgroup, hence in the group. i.e., $a +_p a = 0$ (where $+_p$ is addition to itself p times). The same logic with another distinct prime gives $b +_q b = 0$. WLOG take $p < q$

Consider $c \in F$ such that $c*a=b$ (field multiplication). This exists because $F/\{0\}$ possesses group structure. Then do $c(a +_p a) = c*0=0$ since the zero element of a field "absorbs multiplicatively". But then the LHS becomes $(b +_p b) = 0$ by distributivity and the construction of the element $c$. This implies that $b$ is of order $p$, which it cannot be because it was constructed to be within a Sylow $q$-subgroup and $p$, being different from $q$, can divide no power of $q$, so $b$ being of order $p$ contradicts Cauchy's theorem.

Therein lies the problem (as I see it) with a field with order not a prime power.

José Carlos Santos
  • 427,504
BRSTCohomology
  • 1,156
  • "Addition to itself $p$ times" is more commonly referred to as "multiply by $p$". And it doesn't matter whether you mean $p\in \Bbb Z$, or the representation of $p$ in your ring (in other words, the integer $p$ multiplied with the unit in the ring). – Arthur Jan 22 '18 at 08:59
  • 1
    The proof may be valid, but it is unnecessarily complicated, because we can do without Cauchy and without Sylow, see this duplicate. But indeed, your proof is just along the same lines as the one by "caffeinemachine" at the duplicate. – Dietrich Burde Jan 22 '18 at 08:59
  • IMO, the proof of the Sylow theorem is stronger than the result you are trying to prove and the argument is circular. –  Jan 22 '18 at 09:09
  • You're right, in fact there was absolutely no need to invoke Sylow's theorem and I have no idea why I did it. If using Cauchy is considered a good approach, I'm satisfied as it looks like my proof is solid if I just ctrl+F+deleted all references to Sylow. – BRSTCohomology Jan 22 '18 at 09:12

1 Answers1

1

The proof looks right, but I don't understand why you felt the need of using Sylow's theorems. Why did you not apply Cauchy's theorem directly to $(F,+)$?

There are some problems concerning your notation. You should have written $F\setminus\{0\}$ instead of $F/\{0\}$. And I don't think that there is a need of introducing the $+_p$ notation. You could have just written $pa$ instead $a+_pa$.

José Carlos Santos
  • 427,504
  • Sorry about the set difference mistake (its kinda late). You're right, there was no need to appeal to Sylow's theorem, and I have no excuse for that. I used my weird notation because I sometimes confuse $pa$ with left-multiplication (I am not a fully fledged algebraist). With this, is using Cauchy's theorem a satisfactory and complete approach? (i.e. I did not leave a gap such that my result is either flawed or weaker) – BRSTCohomology Jan 22 '18 at 09:10
  • I saw no gap in the proof. You could avoid the use of Cauchy's theorem. Since the field has order $p^mq^n$, you know that $p^mq^n1=0$ and therefore $p.(p^{n-1}q^m.1)=0$. – José Carlos Santos Jan 22 '18 at 09:13
  • 2
    @MarcusAurelius Even avoiding $pa$, the standard notation for what you wanted would have been $\sum_{i=1}^pa$. The notation $a+_p a$ is not only non-standard, it's also just not very good. It suggests that something like $a+_p b$ is possible, and just becomes unwieldy if $a$ was a more involved expression (which would then need to be duplicated). – Derek Elkins left SE Jan 22 '18 at 10:19