Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

While answering this question on mathoverflow, I stumbled across a question that I expect may be easily answered by someone knowing a bit more algebra than me.

Let's make it really specific.

Consider the polynomial equation $X^4-X^3-X^2-X-1=0$. It has two real roots and a pair of complex roots.

How can one show that the complex roots are not roots of a real number?
share|cite|improve this question
From the context, it means that $(\text{Arg }z) / \pi$ is irrational. – Erick Wong Jun 18 '12 at 5:30
up vote 4 down vote accepted

Let $p(x)$ be a fixed monic polynomial with integer coefficients and degree $d$. If $r$ is one of its roots and $r^n = b$ is real and non-negative (and $n$ is minimal with this property), then $b$ is a product of algebraic integers, hence an algebraic integer. We therefore have $$r = \zeta_n \sqrt[n]{b}$$

for some primitive $n^{th}$ root of unity $\zeta_n$. Since $\sqrt[n]{b}$ is real, $\bar{r} = \zeta_n^{-1} \sqrt[n]{b}$ is also a root of $p$, and so it follows that $$\frac{r}{\bar{r}} = \zeta_n^2$$

lies in the splitting field $F$ of $p$. But $\zeta_n^2$ generates a cyclotomic field of degree $\varphi \left( \frac{n}{\gcd(2, n)} \right)$, and so this must divide the degree of $F$, which divides $d!$. It is known that there are only finitely many numbers with a given totient, so there are only finitely many possibilities for $n$, and so for fixed $p$ this indeed reduces to a finite problem as Gerry says.

When $p(x) = x^4 - x^3 - x^2 - x - 1$, we compute that $\bmod 2$ we have $p(x) \equiv \frac{x^5 - 1}{x - 1}$, which is irreducible (the smallest finite field over $\mathbb{F}_2$ which has elements of order $5$ is $\mathbb{F}_{2^4}$), so $p$ is irreducible and its splitting field has degree dividing $4! = 24$. So $\zeta_n^2$ lies in its splitting field only if $$\varphi \left( \frac{n}{\gcd(2, n)} \right) | 24.$$

If $q$ is a prime dividing $n$, then $q - 1 | 24$, so we can only have $q = 2, 3, 5, 7, 13$. Of these, the only odd prime which also divides $24$ is $3$ and it only does so once, so $3$ can occur with multiplicity at most $2$ and the other odd primes occur with multiplicity at most $1$. Since $2^3 | 24$, the prime $2$ occurs with multiplicity at most $5$.

Summarizing, to prove that $p(x)$ does not have a root which is the root of a real number, it suffices to prove that $r^{2^5 \cdot 3^2 \cdot 5 \cdot 7 \cdot 13}$ is not real for any of the roots $r$, and this can be done by a finite calculation. Of course this is a large exponent, and the actual size of the possible values of $n$ is smaller, but the possible values of $n$ are somewhat tedious to list out.

Here's another idea for ruling out values of $n$. Recall that if $K \subset L$ is an inclusion of number fields, then the discriminant $\Delta_K$ of $K$ divides $\Delta_L$. The discriminants of the cyclotomic fields $\mathbb{Q}(\zeta_n)$ are known (see Wikipedia, although the general formula is somewhat complicated), and in particular every odd prime divisor of $n$ divides them, so by computing the discriminant of $p$ we can rule out some prime factors.

WolframAlpha tells me that the discriminant of $x^4 - x^3 - x^2 - x - 1$ is $-563$. This is prime so it must be the discriminant of the splitting field, and this already rules out all of the possible values of $n$ above.

share|cite|improve this answer
There's something wrong with the first display equation. $r$ is a primitive root of unity times a real root of $b$. – anthonyquas Jun 18 '12 at 6:12
Whoops. Thanks for catching that. – Qiaochu Yuan Jun 18 '12 at 6:14
Also, I don't think it's true that a transposition and a 4-cycle necessarily generate $S_4$. I know it's true for $S_p$ for $p$ prime, but $(1\ 3)$ and $(1\ 2\ 3\ 4)$ generate a dihedral group. – anthonyquas Jun 18 '12 at 6:18
Whoops again! Ignore that part then. I mostly need that $p$ is irreducible. – Qiaochu Yuan Jun 18 '12 at 6:21
ok... thanks a lot Qiaochu. This is quite helpful... – anthonyquas Jun 18 '12 at 6:23

I'd like to see an elegant way to do this, but here's a reasonably concrete way:

If $z$ is either complex root, then since $z$ is algebraic, so is $|z| = \sqrt{z\bar{z}}$, and so is $z/|z|$. We can thus find the minimal polynomial of $z/|z|$ and then the question reduces to whether it is an $n$th root of unity (which amounts to comparing it against a finite list of cyclotomic polynomials).

For this particular case it looks like the minimal polynomial is $X^{10} - X^9 + 2X^7 + 7X^6 + 8X^5 + 7X^4 + 2X^3 - X + 1$, which does not match $\Phi_{11}(X)$ or $\Phi_{22}(X)$, the only cyclotomics of degree 10.

share|cite|improve this answer
Thanks Erick. It's nice to see a solution, but as you say one feels there should be an elegant solution. – anthonyquas Jun 18 '12 at 6:05

If $f(x)$ has a root that is $\root n\of b$ for some real $b$, then it will have an irreducible factor in common with the minimal polynomial for $\root n\of b$. This places very strong restrictions on $n$ and should reduce it to a finite problem in all cases.

I am assuming that, as in the example, the polynomial is monic with integer coefficients, making its roots algebraic integers, which restricts the possibilities for $b$, as it, too, must be an algebraic integer.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.