There's quite a lot of literature about how to find a generator of the multiplicative subgroup of a finite field $\mathbb{F}$.
A much simpler question: can we find an additive generator $a$? So that $\{ a, a + a, a + a + a, ...\} = \mathbb{F}$?
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1don't forget that $p\times a=0$ if the characteristic of $\mathbb F$ is $p$. – lulu Jul 19 '23 at 02:41
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There only exists additive generations for prime fields, and then it is any non-zero element. For fields with the size of other prime powers, there cannot be an additive generator. – Thomas Andrews Jul 19 '23 at 02:43
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Could you tell us what your thoughts and solution attempts are behind it, so that we can not only solve the task, but also really help you. – Kevin Dietrich Jul 19 '23 at 02:46
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The question is equivalent to asking whether a finite field is a cyclic additive group.
Finite prime fields are cyclic additive groups. Extension fields are not.
Matthias
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2The term "prime field" usually means a field with no proper subfields, so $\mathbf{Q}$ is also a prime field, and its additive group is not cyclic. – Keenan Kidwell Jul 19 '23 at 04:39
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