27
votes

Accepted

### Evaluating a remarkable term for primes p = 5 (mod. 8)

According to Theorem 5.1 of
http://math.mit.edu/~rstan/pubs/pubfiles/35.pdf, the number is
$$ \frac 1p\left[ 2^{p-1}+\frac{p-1}{2}(\epsilon^{4h} +
\epsilon^{-4h})\right], $$
where $h$ is the ...

- 45.9k

19
votes

### How small can a sum of a few roots of unity be?

This question grabbed my attention a couple of years ago and I've just put a paper on the arXiv with new upper bounds for $k=5$. I began by computing lots of data, then teased out the structure of ...

- 4,509

15
votes

Accepted

### Vanishing of a sum of roots of unity

For general $N$, we can reason by induction on the $2$-adic valuation of $N$. If $N$ is odd, GH from MO's answer shows that $S_N :=\sum_{k=0}^{N-1} \zeta_{2N}^{2k^2+k} \neq 0$, where $\zeta_{2N} = z = ...

- 19.9k

14
votes

Accepted

### Primes $p=x^2+27y^2$ and Ramanujan's $x_1^{1/3} + x_2^{1/3} + x_3^{1/3}$

Let $\zeta = e^{2\pi i/p}$ be a primitive $p$th root of unity. Then
$2 \cos (2\pi k/p) = \zeta^k + \zeta^{-k}$. The Galois group of $\mathbb Q(\zeta)$
is isomorphic to $(\mathbb Z/p \mathbb Z)^\times$ ...

- 10.7k

12
votes

### Vanishing of a sum of roots of unity

The sum does not vanish when $N$ is odd. I will elaborate on François Brunault's idea.
Assume that $N$ is odd, and let $S$ be the sum in question. Working in the ring $\mathbb{Z}[z]$, we see that $S^{...

- 88.3k

11
votes

### An algebraic number is not a root of unity?

Just noticed this question.
I think the following is an even more elementary/self-contained proof.
First, let $\eta = \xi u$ (so $u$ is a root of unity iff $\eta$ is),
and divide by $\xi^2$ to get
$$
...

- 73k

11
votes

Accepted

### Uniqueness of sums of roots of unity

You can prove this by using a result of H. B. Mann to reduce to the case that $n$ is squarefree, and then applying the Lam-Leung result mentioned in one of the comments above.
Theorem 1 of Mann's ...

- 6,279

9
votes

Accepted

### Summation formulas involving roots of unity to various powers

Your first sum is a special Gauss sum. For its value in general, see Corollary 9.16 in Montgomery-Vaughan: Multiplicative number theory I. Your second sum can also be expressed in terms of Gauss sums (...

- 88.3k

9
votes

### A conjecture on binomial coefficients and roots of unity

Here is an elementary and explicit way to see this:
Suppose we have a set of $p$ integers $A=\{a_1, a_2,\dots, a_p\}$ which forms a complete set of residues modulo p. Then we have
$$\prod_{a\in A}(x-a)...

- 82.7k

8
votes

### $q$ as a prime power and a root of unity

This is very not rigorous, but it's a way of thinking about this topic which I find personally helpful. Several $\mathbb F_1$ papers contains remarks about the idea going back to Weil and Iwasawa that ...

- 82.7k

8
votes

Accepted

### Why are most coefficients of these minimal polynomials divisible by $p$?

So $\zeta=-\xi$, where $\xi$ is a primitive $p$'th root of unity. Let $\pi=1-\xi$. The minimal polynomial of $(1+\zeta^n)a$ is a product of terms of the form $X-(1+\zeta^{nj})\alpha_{ij}$, where $j$ ...

- 43k

8
votes

### Power sums of p-th roots of unity

I think I can answer my own question (which is really my colleague's question).
Let $p>2$, and let $j\in\{1,\dots,p-1\}$ be fixed. Then
$$ \sum_{k=1}^{p-1}\left|z_1^k+\dots+z_j^k\right|^2 = \sum_{...

- 88.3k

7
votes

Accepted

### When is possible to factor a field extension into one which adds no roots of unity, followed by one which adds only roots of unity?

No. Let $k = \mathbb{Q}(\sqrt{-17})$. The class group of $k$ is $\mathbb{Z}/4 \mathbb{Z}$, generated by $\langle 3, 1+\sqrt{-17} \rangle$. (I checked this table for a field of class number $4$ and ...

- 142k

7
votes

Accepted

### Power sums of p-th roots of unity

This is closely related to the problems surrounding Turán's power sum method. For example see the chapter in Montgomery's Ten Lectures book, or this paper of Gonek. Lemma 1 (attributed to Cassels) ...

- 42.7k

6
votes

### Summation formulas involving roots of unity to various powers

Experiment makes it clear that this sum is $\epsilon\delta\sqrt{p}$ where
$\epsilon=1$ if $p=1\pmod{4}$, and $\epsilon=i$ if $p=-1\pmod{4}$
$\delta=1$ if $\beta$ is a quadratic residue mod $p$, and $\...

- 50.8k

6
votes

### Non-standard Gauss sums

We may assume that $l\not\equiv 0\pmod p$ because otherwise the given sum is simple. The answer is a Kloosterman sum.
Let $$\delta_q(x)=\begin{cases}
1,& \text{if } x\equiv 0\pmod{q};\\
0,& \...

- 10.7k

5
votes

### Why are most coefficients of these minimal polynomials divisible by $p$?

The second statement is false as stated. Let $p=3$, $n=1$, $a=1/\sqrt{3}$. The minimal polynomial of $(1+\zeta)a$ is $x^4-x^2+1$. Joe Silverman's argument is inapplicable since the $p$-adic value of $(...

- 45.9k

5
votes

Accepted

### Does the Lehmer quintic parameterize certain minimal polynomials of the $p$th root of unity for infinitely many $p$?

For question 1, the answer is yes, as shown by Emma Lehmer herself. (See the paper here, in particular, equation (5.8) on page 539.) In particular, Lehmer states that one can take
$$
a = \frac{\...

- 18.3k

5
votes

Accepted

### A curious Gauss-Sum type identity

Perhaps, this can be simplified; but here is some (more or less general) computation. One can easily see how far it can be generalized.
Every term of the sum reads
$$
q^n\frac{\prod_{k=0}^{j-2}(q^{...

- 19.8k

5
votes

### $q$ as a prime power and a root of unity

One explicit connection is to look in non-describing characteristic, i.e. over an algebraically closed field $k$ of characteristic $p$ not dividing q, so $q$ can be simultaneously a root of unity and ...

- 1,849

5
votes

Accepted

### Simplification of a sum with roots of unity

For fixed $k$, the product of $(1+\zeta^{jk}x)$ over $p$ consecutive values of $j$ equals $1+x^p$. Your claim follows.

- 91k

5
votes

### Integer eigenvalues of a class of matrices inspired by Prof. Zhi-Wei Sun's conjecture

Fourier transform does it.
Denote by $u_j$ $(j=0,1,\ldots,n-1$) the column-vector with coordinates $(x_{ji})_{1\leqslant i\leqslant n-1}$. Note that $u_0+u_1+\ldots+u_{n-1}=0$ and any $n-1$ vectors $...

- 91k

4
votes

Accepted

3
votes

### A conjecture involving roots of unity

Motivated by Nemo's solution in the case $\delta=0$, here I provide a proof for the case $\delta=1$. Let $\zeta$ be a primitive $m(n-1)+1$-th root of unity, and consider
$$S=\sum_{k=1}^{n-1}\left(\...

- 13.3k

3
votes

### Trace 0 and Norm 1 elements in finite fields

No, not for every $q,\ell,\zeta$. In fact in general there does not even exist one solution.
Trivial example: $\zeta \in \mathbb F_q^\times$ implies $\zeta^{1- q^i} = 1$ and hence $Tr( \zeta^{1-q^i})=...

- 122k

3
votes

### How small can a sum of a few roots of unity be?

This is not an answer to the question, but I think it is somewhat related, and while the question deals with cyclic groups, this result deals with generalized characters of arbitrary finite groups. ...

- 40.1k

3
votes

### q-th powers and roots of polynomials

A family of counterexamples is defined as follows:
$r=2$,
$p_1\ge 2, \quad p_2\ge 3, \quad d_1 \perp p_1, \quad d_2 \perp p_2$,
$a=\dfrac{\tan(\pi d_1/p_1)-\tan(\pi d_2/p_2)}{\tan(\pi d_1/p_1)+\...

- 4,420

3
votes

### Non-standard Gauss sums

In the ring $\mathbb{Z}[\omega_p]$, the OP's second sum $\sum_{k=1}^{p-1} \left(\frac{k^2+k}{p}\right) \omega_p^{kl}$ raised to the $p$-th power is congruent to
$\sum_{k=1}^{p-2} \left(\frac{k^2+k}{p}...

- 88.3k

3
votes

Accepted

### How to calculate $N_{L/k}$(roots of unity)?

The answer should be: if $n = m$, then the image is the roots of unity of order $p^{n-1}$. If $n = m - 1$ then the norm map should be surjective on the roots of unity.
In the first case, since all ...

- 1,413

2
votes

### Unit in cyclotomic field

This is only a partial answer: it reduces the question to another problem.
Taking up Ilya Bogdanov's comment, let $a$ be a root of $2x^2 + x + 2$.
Then (as he says) we are interested in the set of ...

- 10.7k

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