Let $p$ be an odd prime and consider the $p$-th cyclotomic field $\mathbb{Q}(\zeta_p)$ and its quadratic subfield $\mathbb{Q}(\sqrt{\pm p})=:K$.

I am interested in the minimal polynomial of a root of unity $\zeta_p$ over $K$ - I understand this may only be answered uniquely up to the automorphism $\sqrt{\pm p} \mapsto -\sqrt{\pm p}$.

I have looked at a few examples, but I have not been able to deduce a pattern in the coefficients of the minimal polynomial. So, my question is, is there a general formula for the coefficients of this polynomial in the form $a+b\sqrt{p}$?

Edit: In fact, it would be sufficient for me to know the value of this minimal polynomial at $-1$, and also I could restrict to the case where $p \equiv 1 \pmod{4}$, however this seems equally challenging.


I assume w.l.o.g. $p \equiv 1 \pmod 4$. The minimum polynomial of $\zeta_p$ over $\mathbb{Q}(\sqrt{p})$ is the product of conjugates $\prod_{\tau\in H} (x-\zeta_p^\tau)$ where $H$ is the (unique) subgroup of index $2$ in $G = \operatorname{Gal}(\mathbb{Q}(\zeta_p) : \mathbb{Q})$ corresponding to $\mathbb{Q}(\sqrt{p})$.

Explicitly, if $G = \langle \sigma \rangle$ then $H = \langle \sigma^{2} \rangle$ consists of all the squares. More explicitly let $\sigma: \zeta_p \mapsto \zeta_p^l$ be a generator; then $\sigma^2: \zeta_p \mapsto \zeta_p^{l^2}$. Hence the minimum polynomial is $\prod_{k=1}^{(p-1)/2} (x-\zeta_p^{l^{2k}})$.

Illustration, in Sage:

sage: p = 17
sage: L.<zeta_p> = CyclotomicField(p)
sage: K.<z> = L.relativize(L.subfield(L(p).sqrt(), 'sqrt_p')[1])
sage: R.<x> = PolynomialRing(L)
sage: g = z.minpoly(x)
sage: G = L.galois_group()
sage: sigma = G.gen().as_hom()
sage: f = prod( (x - (sigma^(2*k))(zeta_p)) for k in range(0, (p-1)/2) )
sage: f == g

The coefficients are elementary symmetric polynomials in the roots. The constant coefficient is $$N^{\mathbb{Q}(\zeta_p)}_{\mathbb{Q}(\sqrt{p})}(\zeta_p) = \prod_{k=1}^{(p-1)/2} \zeta_p^{l^{2k}} = \zeta_p^{\sum l^{2k}} = 1,$$ since the sum of quadratic residues modulo $p \equiv 1 \pmod 4$ vanishes: $\sum l^{2k} \equiv 0 \pmod p$.

The coefficient of $x^{(p-3)/2}$ is $$-\operatorname{Tr}^{\mathbb{Q}(\zeta_p)}_{\mathbb{Q}(\sqrt{p})}(\zeta_p) = -\sum_{k=1}^{(p-1)/2} \zeta_p^{l^{2k}} = \frac{1 \pm \sqrt{p}}{2},$$ where the sign depends on the embedding; this is $-\frac{1}{2}$ times the quadratic Gauss sum $g(1;p) - 1$.

The coefficient of $x^{(p-5)/2}$ is $\frac{1}{2}\left[(\sum \zeta)^2 - (\sum \zeta^2)\right]$, where the sums are over the roots of the minimum polynomial. The first sum we have just seen; the second is $\frac{1}{2}$ times the Gauss sum $g(2; p) - 1 = \left(\frac{2}{p}\right) \sqrt{p} - 1$: $$\frac{1}{2}\left[\left(\frac{-1 \mp\sqrt{p}}{2}\right)^2 - \left(\frac{2}{p}\right) \frac{\sqrt{p}}{2} + \frac{1}{2}\right] = \frac{p+3}{8} + \left(\pm 1 + \left(\frac{2}{p}\right)\right)\frac{\sqrt{p}}{4}.$$

For the other coefficients, mimic the above by using that elementary symmetric polynomials can be expressed in terms of power sums and that the power sums are Gauss sums.

Moreover, since $-1$ is a quadratic residue mod $p \equiv 1 \pmod 4$, we have that $\zeta_p \mapsto 1/\zeta_p$ belongs to $H$; the (unique) minimum polynomial satisfies $p(x) = x^{(p-1)/2} p(1/x)$, so the polynomial is palindromic.

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    $\begingroup$ There is nothing non-trivial in this answer, and as I said, I have looked at several examples - of course using a computer algebra system. Then again, maybe the phrasing of my question was too subtle. I am looking for a formula which will give me the coefficient of the minimal polynomial at some $x^i$ in the form $a+b\sqrt{p}$. $\endgroup$ – Oliver Braun Apr 11 '16 at 6:10
  • $\begingroup$ Dear Oliver, it was not clear to me what was trivial to you. The trivial expression for the roots of the minimum polynomial leads to Gauss sums, as you can see in the updated answer. $\endgroup$ – Ricardo Buring Apr 11 '16 at 19:33
  • $\begingroup$ Thanks, the hint about Gauss sums is quite useful. Or, at least, as useful as it gets, I suppose. I mean, the expressions for the coefficients are still quite complicated, but probably there's no better way. Anyways, thank you. I'll accept your answer now. $\endgroup$ – Oliver Braun Apr 12 '16 at 13:54

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