MathOverflow is a question and answer site for professional mathematicians. 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

I was asked whether it was possible to produce a monic polynomial with integer coefficients, constant coefficient equal to $1$, having a real root $r > 1$ and a pair of complex roots with absolute value $r$, which are not $r$ times a root of unity. Bonus if the polynomial did not have roots of absolute value one. An answer (without the bonus) is:

$x^{12} - 4x^{11} + 76x^{10} + 156x^9 - 429x^8 - 2344x^7 + 856x^6 - 2344x^5 - 429x^4 + 156x^3 + 76x^2 - 4x + 1$.

I'd like an answer to the bonus question in the following strengthened form: Is there a unit $r$ in a number field such that $r$ has the same absolute value (bigger than one) at a real and a complex place (of $\mathbb{Q}(r)$ to avoid trivial answers) but no archimedian place where $r$ has absolute value $1$?

share|cite|improve this question
Remark: if you can find a unit with the same absolute value at a real and complex place, then you're surely done, because you can adjoin sqrt(2) to the number field and then multiply r by (1+sqrt(2))^N for some large N to mess up all the arch abs values until none of them are 1. I think this answers the bonus question (unless I slipped up). I was going to use this observation to answer the strong bonus question until I realised the deg 12 poly you posted above was reducible. In the strong form of the bonus q did you still mean to demand that the ratio of the embeddings wasn't a root of unity? – Kevin Buzzard Mar 10 '10 at 15:22
I think you are right about the 1+sqrt(2). My polynomial is reducible and, from that, one can figure out how I found it. Yes, I do want the conjugates of r not to be r times a root of unity for the strong bonus question. – Felipe Voloch Mar 10 '10 at 15:56
I would have had 30 minutes in which to do other things had you just explained yourself how you constructed it ;-) – Kevin Buzzard Mar 10 '10 at 17:54
PS it probably took longer to reverse-engineer it than it did for you to construct it! – Kevin Buzzard Mar 10 '10 at 18:15
Sorry for your lost time! If it serves as consolation, it took me more time than it should have to get the polynomial because I kept doing it wrong. – Felipe Voloch Mar 10 '10 at 18:24

I got an answer from Jeff Vaaler, whose office is next door to mine. Sometimes the internet is not the best source. Although, since he didn't remember the author, I wouldn't have located the paper without mathscinet.

It's impossible to get the "strenghtened form". It follows from the results of:

Ferguson, Ronald, Irreducible polynomials with many roots of equal modulus. Acta Arith. 78 (1997), no. 3, 221--225.

share|cite|improve this answer

The polynomial $f(x)=x^3+3x^2+2x-1=(x+1)^3-(x+1)-1$ has a real root $q$, $0 < q < 1$, and complex conjugate roots $s$ and $t$, with $st=1/q=r^2$, with $r=|s|=|t|>1$. The polynomial $(x^2-1/q)(x^2-1/s)(x^2-1/t)=x^6-2x^4-3x^2-1$ has $r$ as one of its roots. So I think the product $(x^3+3x^2+2x-1)(x^6-2x^4-3x^2-1)$ should work.

What have I missed?

share|cite|improve this answer
That's a nice example and is better (lower degree) than what I could cook up following Kevin's suggestion, namely $f((1+\sqrt(2))x)f((1-\sqrt(2))x)$, where $f$ is the degree 12 polynomial in my original question. That solves the "bonus" but not the "strengthening" (which is just as well as per Ferguson) since your polynomial is reducible. Is that what you were referring to when you asked what were you missing? – Felipe Voloch Mar 12 '10 at 2:21
Since no one had posted an explicit solution to the bonus question I figured there must not be one and so there must be something wrong with my example - I must have missed some condition. I came up with several other "answers" before the one I posted only to realize that I had missed something in each of them. So I was hedging my bets. I'm glad my example passes muster. – Gerry Myerson Mar 12 '10 at 3:33
I made an observation which I think solved the bonus question (take Filipe's solution, which is a product of two irreducible polys, with roots alpha,beta say, and then figure out the min polys of alpha(1+sqrt(2))^N and beta(1+sqrt(2))^N and multiply them together; for N suff large this will work), but was too lazy to do the algebra. – Kevin Buzzard Mar 12 '10 at 10:06

The big problem is that Abs(a) = b implies that a = b x 1c. Whether or not c is a rational number is another question, however. If I gave you $x^3 - 11x^2 + 55x -125$, its roots are x = 5, x = 3+4*i*, x = 3-4*i* In this case, |3+4*i*| = 5, and 5*(-1)ArcTan(3/4)/Pi = 3+4*i*, and ArcTan(3/4)/Pi evaluates to ~ .295... Granted, this doesn't satisfy the requirement that the constant is equal to 1, but the point still stands that any two numbers which share an absolute value only have a factor difference of some power of unity.

share|cite|improve this answer
Gabriel: "root of unity" means "z in the complexes such that z^n=1 for some integer n", so e.g. (3+4i)/5 is not a root of unity. – Kevin Buzzard Mar 11 '10 at 20:47

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.