For an odd prime $p$, let $\zeta:=e^{\frac{\pi i}p}$ and choose odd $1<n<p$. Further let $q(x)$ and $r(x)$ be integer polynomials such that $r(x)$ has no common factor with $x^n+1$, and $\xi$ any root of unity (not necessarily related to $\zeta$).

I have observed that the minimal polynomial of $$(1+\zeta^n)\frac{q(\xi)}{r(\xi)}$$ always seems to have all its coefficients divisible by $p$, except either the first or the last one.

Likewise for the minimal polynomial of $(1+\zeta^n)a$, where $a$ is any real algebraic number.

How to prove this?

  • $\begingroup$ Shouldn't $n$ be odd? $\endgroup$ Sep 17, 2015 at 16:49
  • $\begingroup$ @IlyaBogdanov Yes,sorry, I forgot to mention that. $\endgroup$
    – Wolfgang
    Sep 18, 2015 at 7:38

2 Answers 2


So $\zeta=-\xi$, where $\xi$ is a primitive $p$'th root of unity. Let $\pi=1-\xi$. The minimal polynomial of $(1+\zeta^n)a$ is a product of terms of the form $X-(1+\zeta^{nj})\alpha_{ij}$, where $j$ is odd and $i$ runs over some index set (maybe depending on $j$). Note that $j$ is odd because the Galois conjugates of $\zeta$ are the odd powers of $\zeta$. Reducing mod $\pi$, you get a product of terms of the form $$ X-(1+(-\xi)^{nj}) \equiv X-(1+(-1)^{nj})\alpha_{ij} \equiv X-(1+(-1)^{n})\alpha_{ij}\pmod\pi. $$ So if $n$ is odd, the minimal polynomial of $(1+\zeta^n)a$, reduced mod $\pi$, is a power of $X$, and since the polynomial has $p$-integral rational coefficients, they are all divisible by $p$. (Hmmm... I guess I've assumed that $a$ is $p$-integral. If $a$ has negative $p$-adic valuation, a similar argument should work, where now one may find that the constant term of the minimal poly in $\mathbb Z[X]$ is the one not divisible by $p$.)

  • $\begingroup$ I find the conflict in notation unfortunate. If instead you just used $\zeta$, and reminded people that $\zeta = -1 \bmod (1 + \zeta)$, I would find your presentation much easier to read. Gerhard "Also Needs Fewer Greek Letters" Paseman, 2015.09.17 $\endgroup$ Sep 17, 2015 at 20:37
  • $\begingroup$ @GerhardPaseman I guess I'm just too used to taking $\pi$ to be 1 minus a $p$'th root of unity, with the norm of $\pi$ generating the ideal $p\mathbb Z$. But you're probably right that for this problem it would be easier to just use $1+\zeta$ with $\zeta$ a primitive $2p$'th root of unity. $\endgroup$ Sep 17, 2015 at 23:47
  • $\begingroup$ I am still trying to understand the crucial "The minimal polynomial of $(1+\zeta^n)a$ is a product of terms of the form $X-(1+\zeta^{nj})\alpha_{ij}$". I believe you are right, but how do you know that the Galois conjugates of ζ are sufficient to "capture it all"? And what role does the p-adic valuation of a play here? $\endgroup$
    – Wolfgang
    Sep 18, 2015 at 10:58
  • $\begingroup$ @Wolfgang Let $u$ and $v$ be algebraic numbers. Let $u_1,\ldots,u_n$ be the Galois conjugates of $u$, and let $v_1,\ldots,v_m$ be the Galois conjugates of $v$. Then $\{u_iv_j : 1\le i\le n,1\le j\le m\}$ is a complete set of Galois conjugates of $uv$, although there may, of course, be repeated values. This is clear, since $\sigma(uv)=\sigma(u)\sigma(v)$ for any $\sigma\in\text{Gal}(\overline{\mathbb Q}/\mathbb Q)$. So in the OP's case, the conjugates of $(1-\zeta^n)a$ all have the form $(1-\zeta^{nj})b$, where $b$ is some Galois conjugate of $a$. $\endgroup$ Sep 18, 2015 at 11:02

The second statement is false as stated. Let $p=3$, $n=1$, $a=1/\sqrt{3}$. The minimal polynomial of $(1+\zeta)a$ is $x^4-x^2+1$. Joe Silverman's argument is inapplicable since the $p$-adic value of $(1+\zeta)a$ is 0.

  • $\begingroup$ Good point. I did say that my argument was for $a$ being an algebraic integer (or at least, being $p$-adically integral). As you note, if one puts a denominator in $a$ that exactly kills the $p$-adic valuation of $1+\zeta$, the conclusion fails. Nice example! $\endgroup$ Sep 18, 2015 at 13:10

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