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It is known that the splitting field of $x^{p^n}-x$ over $\mathbb{F}_p$ is $\mathbf{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)\cong\mathbb{Z}/n\mathbb{Z}$ and the splitting field of $\Phi_n(x)$ over $\mathbb{Q}$ is $\mathbf{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})\cong(\mathbb{Z}/n\mathbb{Z})^{\times}$.

Then for a fixed positive integer $n$, is there an explicit separable polynomial such that its Galois group is cyclic of order $n$? What I can know is that finite abelian extension of $\mathbb{Q}$ is contained in a cyclotomic extension of $\mathbb{Q}$. Can somebody give me some reference about this problem? Thanks!

Quay Chern
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    Already for n=3 this is a nontrivial problem: the discriminant of the cubic equation must be a square, and only then is the Galois group cyclic of order 3. – TCiur Sep 02 '24 at 12:58
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    I think https://math.stackexchange.com/questions/2204162/real-cyclic-extensions-of-mathbb-q-of-given-degree provides some hints at constructing such a polynomial, and yes, you start with a cyclotomic extension and then find a subfield. – TCiur Sep 02 '24 at 13:03
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    FWIW, the case $n=99$ was looked at somewhat recently here. The same tools can be used in the general case. Extracting a specific polynomial is a taller order :-) – Jyrki Lahtonen Sep 03 '24 at 03:49

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I will give you an example for $n=3$. Other values of $n$ can be replicated. Consider $K = \mathbb{Q}(\zeta_7)$ which has Galois group $C_6$. There is a subfield $L \subset K$ fixed by the automorphism sending $\zeta_7 \mapsto \zeta_7^6$. It is cyclic of order $3$ and generated by $\zeta_7 + \zeta_7^6$, so it suffices to find the minimal polynomial of this guy. It is given by $$x^3 + x^2 - 2x - 1$$

I don't know of a uniform way to do this for all $n$, but if you want specific values of $n$, this is possible as above.

If $2n + 1$ is a prime, then you might be able to get a uniform solution for these cases by repeating the above. The polynomial will always be the minimal polynomial of $\zeta_{2n+1} + \zeta_{2n+1}^{2n}$.

Oscar Lanzi
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TCiur
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Some cyclotomic polynomials such as $x^4+x^3+x^2+x+1$ for degree $4$ work, but they cover only degrees equal to the Euler totient of an odd-prime power, or degree $1$ or $2$; thus for example no odd degrees greater than $1$.

Oscar Lanzi
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    The Galois group of $\Phi_n(x)$ is isomorphic to $\Bbb{Z}_n^*$, which is cyclic only when $n$ is the power of an odd prime, twice such a number, or a factor of $4$. The idea is ok, as cyclotomic fields are the key (as in your comment under the other answer)! – Jyrki Lahtonen Sep 03 '24 at 18:06
  • Yes, that is, for $n>1$, $(\mathbb{Z}/n\mathbb{Z})^{\times}$ is cyclic if and only if $n=2,4,p^k,2p^k$ where $p$ is an odd prime and $k\geqslant1$ is an integer. – Quay Chern Sep 07 '24 at 16:12