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I was taught that the non-zero elements of a finite field forms a cyclic multiplicative group.

But it is not true that a finite field always forms a cyclic group under addition.

Can anyone explain why? (Ideally with counter example and if possible some insight of what is missing in the additive structure that led to this, so that I can understand how to search for a relevant counter example?)

To be honest the only examples of finite field I know of are the $\mathbb{Z}_p$ with $p$ prime.

Thank you

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  • $\begingroup$ You already understand the problem: Try to construct a finite field with four elements, say $\mathbb{F}_4 = \mathbb{F}_2/ \langle x^2 + x + 1\rangle$. $\endgroup$ – JavaMan Aug 13 '17 at 3:55
  • $\begingroup$ Thanks. This is very helpful as it reminds me of the fact that finite fields can be of order power of primes also. $\endgroup$ – kfp22 Aug 13 '17 at 4:00
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No: suppose that $F$ is a finite field of order $q=p^d$ with $p$ prime. Since $F$ has characteristic $p$, it follows that $px=0$ for all $x\in F$, so every element of the additive group has order at most $p$. In particular the additive group can only be cyclic for $d=1$.

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  • $\begingroup$ That is very clear indeed... thanks a lot $\endgroup$ – kfp22 Aug 13 '17 at 3:57

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