The quadratic subfield of $\mathbb{Q}(\zeta_p)$ is given by $\mathbb{Q}(\sqrt{p^*})$, where $p^*$ is the choice of $\pm p$ which is $1$ mod $4$. By some elementary Galois theory, the cyclotomic polynomial $\Phi_p = \frac{x^p-1}{x-1}$ factors into two irreducible polynomials of degree $\frac{p-1}{2}$ over this quadratic subfield, which one can describe as $$ P_{QR}(x) = \prod_{k\ \text{QR}} (x-\zeta_p^k) $$ $$ P_{QNR}(x) = \prod_{k\ \text{QNR}} (x-\zeta_p^k) $$ where $k$ ranges over the (nonzero) quadratic residues and quadratic nonresidues modulo $p$ respectively.

Is there an explicit description for the coefficients of these polynomials in terms of $\sqrt{p^*}$? The first coefficient, the one in front of $x^{\frac{p-3}{2}}$, is $\frac{1\mp\sqrt{p^*}}{2}$ by the statement of Gauss sums.


1 Answer 1


Some trivial observations. We have $$P_{QR}(1/x) x^{(p-1)/2} = \prod_{QR} (1 - x \zeta^k),$$ $$P_{QNR}(1/x) x^{(p-1)/2} = \prod_{QNR} (1 - x \zeta^k),$$ which are easier to work with. On the other hand, if $p \ge 5$, the product of $\zeta^k_p$ over quadratic residues is one, and the product over non-residues is also one. Hence we can write $$\begin{aligned} P_{QR}(1/x) x^{(p-1)/2} = & \ \prod_{QR} ( 1 - x \zeta^k) = \prod_{QR} ( \zeta^{-k} - x) \\ = & \ (-1)^{(p-1)/2} \prod_{QR} (x - \zeta^{-k}) \\ = & \ \begin{cases} P_{QR}(x), & p \equiv 1 \mod 4 \\ - P_{QNR}(x), & p \equiv 3 \mod 4 \end{cases} \end{aligned}$$ because $(-1/p) = (-1)^{(p-1)/2}$. The same swapping occurs for $P_{QNR}$. Let's consider the highest powers of $P_{QR}(x)$ and $P_{QNR}(x)$, or equivalently the lowest powers of $P_{QR}(1/x) x^{(p-1)/2}$ and $P_{QNR}(1/x) x^{(p-1)/2}$. We have $$\frac{P_{QR}(1/x) x^{(p-1)/2}}{P_{QNR}(1/x) x^{(p-1)/2}} = \prod ( 1 - x \zeta^k)^{ \left( \frac{k}{p}\right)}.$$

As mentioned there is the Gauss sum: $$\sum \left( \frac{k}{p}\right) \zeta^{k} = \sqrt{p^*},$$ and similarly $$\sum \left( \frac{k}{p}\right) \zeta^{nk} = \left( \frac{n}{p} \right) \sqrt{p^*},$$ applying $[n] \in (\mathbb{Z}/p\mathbb{Z})^{\times} = \mathrm{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$ to both sides; when $p|n$ we interpret the RHS as being zero, and this is still correct, although we won't actually care about terms this deep into the power series above. We deduce that $$- \log \left( \frac{P_{QR}(1/x) x^{(p-1)/2}}{P_{QNR}(1/x) x^{(p-1)/2}} \right) = \sum_{n=1}^{\infty} \frac{x^n}{n} \sum \left( \frac{k}{p}\right) \zeta^{nk} = \sqrt{p^*} \sum_{n=1}^{\infty} \left( \frac{n}{p}\right) \frac{x^n}{n}.$$ Naturally $$- \log \left( P_{QR}(1/x) x^{(p-1)/2} P_{QNR}(1/x) x^{(p-1)/2} \right) = \log \left( \frac{1 - x^p}{1 - x} \right) = \log(1-x) + O(x^p) = - \sum \frac{x^n}{n},$$ and so (for example) $$\log(P_{QR}(1/x) x^{(p-1)/2}) = \frac{1}{2} \sum \frac{x^n}{n} \left(1 - \left( \frac{n}{p}\right) \sqrt{p^*}\right) + O(x^p),$$ You can now formally expand this out to get the first few terms. For example, the first non-zero term is $$\frac{1 - \sqrt{p^*}}{2},$$ and the second is $$ \frac{3 + p^* - 2 \sqrt{p^*}\left(1 + \left( \frac{2}{p}\right) \right) }{8}$$ For example, if $p \equiv 3,5 \mod 8$ so $(2/p) = -1$, this is $$\frac{3 + p^*}{8}.$$ Note the conditions on $p$ ensure that this is an algebraic integer, as it has to be. As you keep going, you get more and more terms involving the quadratic residues $(n/p)$ for small $n$, and it becomes messier and messier, and dependent on $p$ modulo higher integers. The third term, for example, is $$\frac{15 - 9 \sqrt{p^*} + 3 p^* - p^* \sqrt{p^*} + 6(p^* - \sqrt{p^*})(2/p) - 8 \sqrt{p^*} (3/p)}{48}.$$

  • $\begingroup$ Also, one can internet-search on "Aurefeuillian factorization". $\endgroup$ Apr 2, 2020 at 21:59
  • $\begingroup$ Thanks, this is great!! $\endgroup$ Apr 2, 2020 at 22:50

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