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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?
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From the context, it means that $(\text{Arg }z) / \pi$ is irrational. –  Erick Wong Jun 18 '12 at 5:30

3 Answers 3

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.

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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
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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.

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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.

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