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Do the $n$-th roots of unity of an arbitrary field form a cyclic group?

Or stated differently, can we always find a primitive $n$-th root of unity? Because if we have this element we can generate the group $\langle\zeta_n\rangle$, and we are done.

In particularly I'm interested in the case where $n$ is not a prime.

Qiaochu Yuan
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Jens Wagemaker
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    Yes, because $x^n = 1$ can have at most $n$ solutions for any $n$. Say if the roots of unity is isomorphic as a group to $\mathbb Z/n\mathbb Z \times \mathbb Z/nn'\mathbb Z$, then $x^n=1$ has $n^2$ solutions. – Hw Chu Apr 20 '18 at 14:58
  • @HwChu I understand that it can have at most $n$ solutions, because we have a polynomial ring over a field. But I don't get the point that you are making, can you elaborate? – Jens Wagemaker Apr 20 '18 at 15:02
  • I mean, for instance if $F$ is a field whose root of unity is isomorphic to $\mathbb Z/n\mathbb Z \times \mathbb Z / nn' \mathbb Z$, then the equation $x^n - 1=0$ has at least $n^2$ roots (count the $n$-torsions in $\mathbb Z/n\mathbb Z \times \mathbb Z / nn' \mathbb Z$), and this is absurd. – Hw Chu Apr 20 '18 at 15:10
  • I didn't cover torsion groups yet. Is it a theorem that the group of $n$-th roots of unity are always isomorphic to a group $\mathbb{Z} / n \mathbb{Z} \times \mathbb{Z} / n n' \mathbb{Z}$? You say nothing about $n'$, what is this, or does some theorem state the existence? – Jens Wagemaker Apr 20 '18 at 15:19
  • It should be isomorphic to a subgroup of that. $\mathbb Z/ n\mathbb Z$ is the largest you can get. – Hw Chu Apr 20 '18 at 15:21
  • Do you have a reference? – Jens Wagemaker Apr 20 '18 at 15:22
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    Every finite subgroup of a field's multiplicative group is cyclic. – Angina Seng Apr 20 '18 at 15:42
  • Yes, see https://mathoverflow.net/questions/54735/collecting-proofs-that-finite-multiplicative-subgroups-of-fields-are-cyclic. – lhf Apr 20 '18 at 19:01
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    Your two questions are not the same! The group can be cyclic without there existing a primitive $n$th root of unity (it just won't be cyclic of order $n$). – Eric Wofsey Apr 20 '18 at 20:52
  • Already answered in Stackexchange, e.g. https://math.stackexchange.com/questions/50791/if-xm-e-has-at-most-m-solutions-for-any-m-in-mathbbn-then-g-is-cycl/50795#50795 – Andrea Mori Mar 15 '25 at 10:16

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Some comments quote a theorem that any finite subgroup of the multiplicative group of any field is finite cyclic and hence has a generator. This theorem applies to $n$-th roots of unity and so they form a cyclic subgroup and have a generator. But is that generator a primitive $n$-th root of unity?

The problem of existence of $n$-th roots of unity depends on the field. For example, if the field has $4$ elements, then all $3$ non-zero elements are $3$-th roots of unity and, aside from $1$ itself, all roots of unity are $3n$-th roots of unity. So if $k=3n>3$, then there exist $k$-th roots of unity, but none of them are primitive, because if they were, then number of $k$-th roots of unity would be $k$.

Somos
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    I don't see how this answers the question. As the question is merely about showing the existence of the $n$-th primitive roots. But as someone already pointed out it would follow from the general theorem that every finite subgroup of the multiplicative group of a field is finite. – Jens Wagemaker Apr 20 '18 at 19:36
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    Or, even more concretely, there are no primitive $k$th roots of unity in $\Bbb R$ for $k\ge3$. – Greg Martin Aug 05 '23 at 06:43