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

Let $z$ be a non-zero element of $\mathbb{C}$. Does $z^k + z^{-k}$ belong to $\mathbb{Z}[z + z^{-1}]$ for every positive integer $k$?

Motivation: I came up with this problem from the following question.

Maximal real subfield of $\mathbb{Q}(\zeta )$

share|cite|improve this question
$z^k+z^{-k}$ is a polynomial (with integer coeffiecients) of $z+z^{-1}$. In fact it is a Chebyshev polynomial of the second kind. – PAD Jul 25 '12 at 5:43
What's the reason for the downvotes? – Makoto Kato Jul 27 '12 at 6:05
up vote 10 down vote accepted

Let's go by induction as in @Arturo Magidin's answer. The result holds for $k=0,1$. Assume $z^k+z^{-k} \in \mathbb{Z}[z+z^{-1}]$ for $0\le k\le n$. But $$z^{n+1}+z^{-(n+1)} = (z^n+z^{-n})(z+z^{-1}) - (z^{n-1} + z^{-(n-1)}),$$ and so $z^{n+1}+z^{-(n+1)} \in \mathbb{Z}[z+z^{-1}]$.

share|cite|improve this answer
The result holds for $k=0,1$. – Alexander Thumm Jul 25 '12 at 5:26
@AlexanderThumm: I use $k=1,2$ as the base case. Arturo showed the result held for $k=2$ and I didn't repeat the argument. – user26872 Jul 25 '12 at 5:33
Yes I know. I only wanted to mention, that there would be no argument required if you would start with $k=0,1$. – Alexander Thumm Jul 25 '12 at 5:39
@AlexanderThumm: I admit, that is cleaner ... – user26872 Jul 25 '12 at 5:40
@AlexanderThumm: Edited. Thanks for the input. – user26872 Jul 25 '12 at 5:42

Yes. By the fundamental theorem of symmetric polynomials, $x^k+y^k\in\Bbb Z[x,y]^{S_2}$ can be written as a polynomial in $e_1=x+y$ and $e_2=xy$, say $P_k(e_1,e_2)$. Then we have

$$x^k+x^{-k}=P_k(x+x^{-1},1)\in\Bbb Z[x+x^{-1}].$$

We can of course interchange the formal variable $x$ with a specific nonzero complex number as we desire. In fact, this is the power sum $p_k(x,x^{-1})$, and the relationship between the power sums and elementary symmetric polynomials is given recursively by Newton's identities.

For a quick inductive proof of the fundamental theorem,

If $x_n|f$ then $x_1\cdots x_n|f$, and dividing out we are left with a symmetric polynomial of smaller degree than before. Otherwise, write $f(x_1,\cdots,x_{n-1},0)$ as a polynomial $p$ in the elementary symmetric polynomials $\hat{e}_i$ of the first $n-1$ variables, $p(\hat{e}_1,\cdots,\hat{e}_{n-1})$. Now the polynomial $$f(x_1,\cdots,x_n)-p(e_1,\cdots,e_{n-1})$$ is symmetric in all of $x_1,\cdots,x_n$ and evaluates to $0$ at $x_n=0$ ie is divisible by $x_n$. Induct.

which I wrote down here.

share|cite|improve this answer
Can we avoid the theorem of symmetric polynomials? – Makoto Kato Jul 25 '12 at 4:55
@MakotoKato yes, as in Arturo's more straightforward answer. I have an affinity for symmetric polynomials though, so I wanted to grab at the chance to mention them. – anon Jul 25 '12 at 4:56
Symmetric polynomials are pretty great though. – Dylan Moreland Jul 25 '12 at 4:58
Great answer! +1 – Belgi Jul 25 '12 at 5:37

Yes. It holds for $k=1$; it also holds for $k=2$, since $$z^2+z^{-2} = (z+z^{-1})^2 - 2\in\mathbb{Z}[z+z^{-1}].$$ Assume that $z^k+z^{-k}$ lie in $\mathbb{Z}[z+z^{-1}]$ for $1\leq k\lt n$. Then, if $n$ is odd, we have: $$\begin{align*} z^n+z^{-n} &= (z+z^{-1})^n - \sum_{i=1}^{\lfloor n/2\rfloor}\binom{n}{i}(z^{n-i}z^{-i} + z^{i-n}z^{i})\\ &= (z+z^{-1})^n - \sum_{i=1}^{\lfloor n/2\rfloor}\binom{n}{i}(z^{n-2i}+z^{2i-n}). \end{align*}$$ and if $n$ is even, we have: $$\begin{align*} z^n+z^{-n} &= (z+z^{-1})^n - \binom{n}{n/2} - \sum_{i=1}^{(n/2)-1}\binom{n}{i}(z^{n-i}z^{-i} + z^{i-n}z^{i})\\ &= (z+z^{-1})^n - \binom{n}{n/2} - \sum_{i=1}^{(n/2)-1}\binom{n}{i}(z^{n-2i}+z^{2i-n}). \end{align*}$$ If $1\leq i\leq \lfloor \frac{n}{2}\rfloor$, then $0\leq n-2i \lt n$, so $z^{n-2i}+z^{2i-n}$ lies in $\mathbb{Z}[z+z^{-1}]$ by the induction hypothesis. Thus, $z^n+z^{-n}$ is a sum of terms in $\mathbb{Z}[z+z^{-1}]$.

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.