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I'm fearful about putting this forward, because it seems the answer should be elementary. Certainly, the Weak Approximation Theorem allows every system of simultaneous inequalities among archimedean absolute values to be satisfied. But equality combined with inequality?

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8 Answers 8

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Yes. Take $$ \alpha=\sqrt{2-\sqrt{2}}+i\sqrt{\sqrt{2}-1}. $$ Neither of the conjugates $$ \sqrt{2+\sqrt{2}}\pm \sqrt{\sqrt{2}+1} $$ have absolute value 1.

It is impossible, however, if $\mathbb{Q}(\alpha)/\mathbb{Q}$ is abelian, since then all automorphisms commute with complex conjugation.

This was all stolen from Washington's Cyclotomic Fields book.

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Yes. E.g. consider the two complex roots of $x^2-(1-\sqrt{2})x+1.$ They are algebraic integers of degree 4 on the unit circle, but their algebraic conjugates are the roots of $x^2-(1+\sqrt{2})x+1$, so they are irrational real numbers.

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There was a similar question on math.stackexchange: http://math.stackexchange.com/questions/4323/are-all-algebraic-integers-with-absolute-value-1-roots-of-unity/4332#4332 and my answer below is partly taken from that.

For any non-real algebraic number x inside C, its absolute value |x| is algebraic, so x/|x| is an algebraic number on the unit circle other than $\pm 1$. Most algebraic numbers do not have all their conjugates of equal absolute value, so it's quite easy to generate examples of what you ask for.

There is a constraint on the minimal polynomial over Q of an algebraic number on the unit circle other than $\pm 1$. Let x be an algebraic number with absolute value 1. Then x and its complex conjugate x* = 1/x have the same minimal polynomial. Writing f(T) for the minimal polynomial of x over Q, with degree n, the polynomials T^nf(1/T) and f(T) are irreducible over Q with common root x*, so the polynomials are equal up to a scaling factor: T^nf(1/T) = cf(T). Setting T = 1, f(1) = cf(1). Assuming x is not rational (i.e., x is not 1 or -1), f has degree greater than 1, so f(1) is nonzero and thus c = 1. Therefore T^nf(1/T) = f(T), so f(T) has symmetric coefficients (the coefficients of $T^i$ and $T^{n-i}$ are equal). In particular, its constant term is 1. Moreover, the roots of f(T) come in reciprocal pairs (since 1 and -1 are not roots), so n is even.

Partial conclusion: an algebraic number in C other than 1 or -1 which has absolute value 1 has even degree over Q and its minimal polynomial has constant term 1. In particular, if x is an algebraic integer then it must be a unit in the number field it generates.

There are no examples of algebraic integers with degree 2 and abs. value 1 that are not roots of unity since a real quadratic field has no elements on the unit circle besides 1 and -1 and the units in an imaginary quadratic field are all roots of unity (and actually are only 1 and -1 except for Q(i) and Q(w)). Thus the smallest degree x could have over Q is 4 and there are examples with degree 4: the polynomial T^4 - 2T^3 - 2T + 1 has two roots on the unit circle and two real roots (one between 0 and 1 and the other greater than 1). In Victor's answer, the two quadratic polynomials he writes down have product T^4 - 2T^2 + T^2 - 2T + 1. In Cam's answer, the algebraic integers he writes down have minimal polynomial T^8 - 12T^6 + 6T^4 - 12T^2 + 1. Note these polynomials are all symmetric, as they must be by the argument I gave above.

Fact: polynomials f(T) of even degree n = 2m with symmetric coefficients are the same as polynomials of the form f(T) = T^mg(T + 1/T), where g has degree m. (This is a nice little exercise by induction.) Moreover, roots of f on the unit circle other than 1 or -1 correspond in a 2-to-1 way to real roots of g in the interval (-2,2) by $e^{i\theta} \mapsto e^{i\theta} + e^{-i\theta} = 2\cos \theta$. (Note $e^{-i\theta}$ is also a root of $f$ on the unit circle and it leads to the same real root of $g$ since $\cos(-\theta) = \cos(\theta)$.) This is how you can count the number of different roots of $f(T)$ on the unit circle: convert it into $g(T)$ and count the real roots of $g(T)$ in $(-2,2)$ using sign changes. For example, taking for f(T) Lehmer's polynomial of degree 10 in cfranc's answer, the corresponding polynomial g(T) is T^5 + T^4 - 5T^3 - 5T^2 + 4T + 3. (Although f(T) has symmetric coefficients, there is no reason g(T) needs to have symmetric coefficients.) This polynomial g has all real roots and 4 of them are in (-2,2), so Lehmer's polynomial has 2*4 = 8 roots on the unit circle.

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I'm a little confused: several times you use the "fact" that 1 is not a root of f, but plenty of other algebraic numbers on the unit circle have 1 as a root of their minimal polynomial, like the cube roots of unity. These have odd degree and the coefficients have non-symmetric minimal polynomial. Did you want some other assumptions on the algebraic number under consideration? –  Noah Stein Sep 14 '10 at 18:25
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@Noah: The minimal polynomial of the (nontrivial) cube roots of unity is the quadratic $x^2+x+1$ and does not have 1 as a root. More generally, the only number whose minimal polynomial has 1 as a root is 1. –  Andreas Blass Sep 14 '10 at 19:29
    
Thanks -- I forgot to turn my brain on. Now the argument makes much more sense. –  Noah Stein Sep 14 '10 at 19:40

Another nice example is Lehmer's number $\lambda \simeq 1.17628$, which is the largest real root of the monic irreducible polynomial $x^{10} + x^9 - x^7-x^6-x^5-x^4-x^3+x+1$. This polynomial has a second real root inside the unit circle, and the remaining roots lie on the unit circle.

The Mahler measure of a monic irreducible polynomial is the absolute value of the product of the roots with norm $\geq 1$. The Mahler measure of an algebraic integer is the Mahler measure of its minimal polynomial. It is believed that Lehmer's number has minimal Mahler measure. For more on Lehmer's number check out this "What is" paper.

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Something wrong here. The conjugates of a root of unity are roots of unity of the same order (the roots of one cyclotomic polynomial). Therefore either $\lambda$ is of modulus one (clearly not the case), or the polynomial is not irreducible (over ${\mathbb Q}$). I opt for the third possibility: you did not mean 'root of unity', but only 'complex number on the unit circle' –  Denis Serre Sep 14 '10 at 14:14
    
Yup, sorry. Fixed now. –  cfranc Sep 14 '10 at 14:27

One interesting set of example is Salem numbers. These are real algebraic integer all of whose conjugates has absolute value less than or equal to one and at least one of the conjugates lies on the unit circle.

So you have a scenario of, one which flew over the cuckoo's nest, at least one in the edge and the rest are all in the nest (caged ?) .

http://en.wikipedia.org/wiki/Salem_number

Also if all its conjugates lies on the unit circle then it has to be a root of unity, its known as Kronecker Theorem.

(Good) effective version of Kronecker's theorem?

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If $f(x)$ is a monic polynomial in ${\mathbb Z}[x]$ and all roots have absolute value 1 (i.e. lie on the circle $S^1$ of radius 1), then all roots are roots of unity (satisfy $x^k=1$). Indeed, if $x_1,...,x_n$ are the roots of $f$, then for every integer $k$, $x_1^k,...,x_n^k$ are all roots of some monic polynomial $f_k$ of degree $n$ over $\mathbb Z$. The coefficients of $f_k$ are (by Vieta's formulas) $\pm$ elementary symmetric polynomials of $x_1^k,...,x_n^k$. Since the absolute values of $x_i^k$ are 1, the coefficients of $f_k$ are bounded independently of $k$. Hence there are finitely many different polynomials $f_k$, so $x_i^k=x_i^l$ for some $k\ne l$ and $i=1,2,...,n$, hence $x_i^{l-k}=1$. Now if you take any algebraic number on $S^1$ which is not a root of unity, some of its conjugates will not belong $S^1$.

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One more example. Let $\alpha_1$ be the real, and $\alpha_2$ and $\alpha_3$ the nonreal, roots of $x^3-x-1$. Then the six numbers $\alpha_i/\alpha_j$, $i\ne j$, are a complete set of conjugates, exactly two of which ($\alpha_2/\alpha_3$ and $\alpha_3/\alpha_2$) lie on the unit circle.

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One more comment on this frequently answered question: Let $a_0 x^{2n} + a_1 x^{2n-1} + \cdots + a_{n} x^{n} + \cdots a_1 x + a_0$ be a palindromic polynomial with real coefficients and $2k$ isolated roots on the unit circle. Then any sufficiently small perturbation of this polynomial to another real palindromic polynomial also has $2k$ roots on the unit circle.

Proof: Notice that $e^{\pm i \theta}$ is a root of this polynomial if and only if $a_0 \cos (n \theta) + a_1 \cos ((n-1) \theta) + \cdots + a_{n/2-1} \cos \theta + a_0/2=0$. Write $f(\theta)$ for the right hand side of this equation. Our hypothesis is that $f$ has $k$ isolated roots, $\theta_1$, $\theta_2$, ..., $\theta_k$. Then we can find $\epsilon>0$ and disjoint intervals $(a_i, b_i)$ around each $\theta_i$ such that (1) We have $f((a_i, b_i)) \supseteq (-\epsilon, \epsilon)$. (2) On $(a_i, b_i)$, we have $|f'|<\epsilon$ (3) Off of the $(a_i, b_i)$, we have $|f|>\epsilon/2$.

Then, if our perturbation is small enough that $f$ and $f'$ change by less than $\epsilon/4$ everywhere, then the perturbed $f$ still has one root in each $(a_i, b_i)$, and no roots elsewhere. QED

Why do I point this out? Take any of the above examples and perturb its coefficients slightly, while keeping them rational and palindromic. Then you get another example! This observation destroys most attempts to classify such polynomials by number theoretic criteria.

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Nice. Since I sort my answers on "newest," most of the "above" examples are actually below.... –  Gerry Myerson Sep 15 '10 at 0:18
    
I try to remember not to use the words above and below on MO for this reason, but it's a hard habit to break. –  David Speyer Sep 15 '10 at 0:20
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Writing the palindromic polynomial of degree 2n as x^n*g(x+1/x), those 2k distinct roots in x on the unit circle correspond in a 2-to-1 way to k distinct real roots of g(y) in [-2,2], so your comment amounts to saying that any small perturbation of a polynomial with real coefficients doesn't change the number of its distinct real roots in an (open) interval, and that is more intuitive. However, couldn't there be a problem at endpoints -2 and 2 for g (corr. to roots 1 and -1 for f), as a perturbation may move those roots of g outside [-2,2], so f may lose the roots 1 and -1. What do you think? –  KConrad Sep 15 '10 at 0:43
    
First of all, I acknowledge that I didn't address the endpoints. I claim that a palindromic polynomial of even degree can't have single roots at $1$ or $-1$! If it had a single root at $1$, then the constant term would be negative the leading term, not equal to it. And, once you know that there is no root at $1$, if there were a root at $-1$ then it would have odd degree. –  David Speyer Sep 15 '10 at 1:21
    
Ah, right, I forgot about the distinctness of roots at endpoints. So your argument reduces exactly to the issue of perturbing a real polynomial with distinct roots in (-2,2). –  KConrad Sep 16 '10 at 4:31

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