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15 votes
Accepted

Is there a nice general formula for $\int \frac{dx}{x^n-1}$ and/or $\int \frac{dx}{\Phi_n(x)}$?

You can get a pretty simple expression using partial fractions over $\mathbb{C}$. In general, if $f(x)=(x-a_1)\dots(x-a_n)$ is a monic polynomial with distinct roots over $\mathbb{C}$, then we have ...
Eric Wofsey's user avatar
14 votes
Accepted

About the number of real roots

Let $p(t)$ denote your polynomial. Then it is not hard to see that $$(1+t)p(t)=t^{11}+1,$$ which clearly has $-1$ as its only real root. But $p(-1)=11$, so $p(t)$ has no real roots. This also shows ...
Servaes's user avatar
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14 votes
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Is the "cyclotomic diagonalization" always squarefree?

We want to show that the problem is equivalent to For any number $n$ and any prime $q \mid n^n-1$, then $ord_{q^2}(n) \neq n$ We will see at the end why we expect this to be false, but also that ...
Andrea Marino's user avatar
13 votes

Is the "cyclotomic diagonalization" always squarefree?

Checking the pairs $n,p\le 10^6$ , I found one counterexample , namely : $$283411^2\mid \Phi_{28341}(28341)$$
Peter's user avatar
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12 votes
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Does there exist a formula for product of the primitive $ n $th roots of unity.

If $g$ is a primitive $n$-th root of unity, then so is $g^{-1}$. Unless $n\le2$, we have $g \ne g^{-1}$. When $n=2$, there is only one primitive $n$-th root of unity: $-1$. Therefore, the product of ...
lhf's user avatar
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8 votes

Determining whether $\Phi_7(x)$ is irreducible over $\mathbb{F}_{11}$

You are actually done as soon as you observe that $11^3-1$ is divisible by $7$. As you observed, this implies that there is a primitive $7$th root of unity $\alpha$ in $\mathbb{F}_{11^3}$, whose ...
Eric Wofsey's user avatar
7 votes
Accepted

An equality about cyclotomic polynomials

$\Phi_n(X)$ is the product of all $(X-\zeta)$ where $\zeta$ is an $n$-th root of unity but not a $d$-th root of unity for any proper divisor $d$ of $n$. Hence the degree of the $n$-th cyclotomic ...
ancient mathematician's user avatar
7 votes
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If $K/\mathbb{Q}$ is finite, then $K$ contains finitely many $n$th roots of unity

Every $n$-th root of unity is a primitive $m$-th root of unity for some $m$ (which divides $n$). There are only finitely many primitive $m$-th roots of unity for any given $m$. So if you have ...
Angina Seng's user avatar
7 votes
Accepted

Is the image of $\Phi_n(x) \in \mathbb{Z}[x]$ in $\mathbb{F}_q[x]$ still a cyclotomic polynomial?

You need $\gcd(p,n)=1$ for otherwise there are no roots of unity of order $n$ in any field $\Bbb{F}_q,q=p^n$. Basically this is because $1$ is the only root of $x^p-1=(x-1)^p$. But, assuming $\gcd(n,p)...
Jyrki Lahtonen's user avatar
7 votes

How to prove $X^{4}+X^{3}+X^{2}+X+1$ is irreductible in $\mathbb{F}_{2}$

You can observe that this is the fifth cyclotomic polynomial, ${x^5-1\over x-1}$, so any field element $\alpha$ making it $0$ is a primitive fifth root of unity. That is, $\alpha^5=1$ and no smaller ...
paul garrett's user avatar
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7 votes
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"Easy" proof that $\Phi_n$ has degree $\phi_n$

When you "simply" consider the map given by $\zeta_n\mapsto\zeta_n^k$ how do you know this is a field automorphism? Can you check directly that it is additive and multiplicative? With fields ...
Alex Mathers's user avatar
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6 votes

Finite-order elements of $\text{GL}_4(\mathbb{Q})$

There is something wrong in the first sentence of the accepted answer. The problem is that the minimal polynomial for a matrix may not be irreducible. For example, the matrix $$\begin{bmatrix}1& 1 ...
TH Wang's user avatar
  • 369
6 votes

Proving that cyclotomic polynomials have integer coefficients

This proof seems to be more elementary. We begin by contending that if $$(x^n -1) =({\sum}^p_{i=1}{a_i{x^i}})({\sum}^q_{j=1}{b_j{x^i}}),$$ where ${\sum}^p_{i=1}{a_i{x^i}}\in{\bf{Z}[x]},$ then every ...
student's user avatar
  • 1,344
6 votes

Cyclotomic Polynomials and GCD

Note that $(x^m-1,x^n-1)=x^{(n,m)} - 1$ in $\mathbb{Z}[x]$. i.e. $\exists p(x),q(x)\in \mathbb{Z}[x]$ such that $$(x^m-1)p(x)+(x^n-1)q(x)=x^{(n,m)}-1$$. Now since $x^{(n,m)}-1\mid x^n -1$ and $x^{(n,m)...
gaurav patil's user avatar
6 votes

Discriminant of cyclotomic polynomial $\Phi_p(x)$

Thought I would add my solution, which is less elegant, but doesn't require too much creativity. It suffices to find the discriminant of the number field $L = \mathbb Q(\zeta_p)/\mathbb Q$. We will ...
FlipTack's user avatar
  • 694
6 votes

showing that $n$th cyclotomic polynomial $\Phi_n(x)$ is irreducible over $\mathbb{Q}$

There is also a non-elementary proof that perhaps is more explanatory than the "elementary" argument, using some (but not too much) algebraic number theory, and primes in arithmetic progressions. To ...
paul garrett's user avatar
  • 52.7k
6 votes

What comes after $\cos\left(\tfrac{2\pi}{7}\right)^{1/3}+\cos\left(\tfrac{4\pi}{7}\right)^{1/3}+\cos\left(\tfrac{6\pi}{7}\right)^{1/3}$?

As some people said, the 2nd formula is easy to derive. In Maple, there are commands to get the minimal polynomial of LHS of the 2nd. The following method works for the 2nd formula, but not for the ...
River Li's user avatar
  • 38.9k
6 votes
Accepted

Prove that $X^4+X^3+X^2+X+1$ is irreducible in $\mathbb{Q}[X]$, but that it has two different irreducible factors in $\mathbb{R}[X]$

Finding the complex roots of the polynomial is easy: if $\varphi=2\pi/5$, the roots are $$ r_1=e^{i\varphi},\quad r_2=e^{2i\varphi},\quad r_3=e^{3i\varphi}=\bar{r}_2\quad r_4=e^{4i\varphi}=\bar{r}_1 $$...
egreg's user avatar
  • 239k
6 votes
Accepted

The discriminant of cyclotomic polynomial $\Phi_n(x)$

The computation can be reduced to prime powers, using the multiplicativity of Euler's totient function and "exploiting the fact that cyclotomic fields of relatively prime order are linearly disjoint", ...
Dietrich Burde's user avatar
6 votes

How to prove $X^{4}+X^{3}+X^{2}+X+1$ is irreductible in $\mathbb{F}_{2}$

You can simply try to factor it; if it is reducible, it must have either a linear or quadratic factor. There aren't many linear and (irreducible) quadratic polynomials in $\Bbb{F}_2[X]$.
Servaes's user avatar
  • 63.4k
5 votes
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Structure of $Gal(\mathbb{Q}(\zeta_{15})/\mathbb{Q})$?

Let $G$ be the Galois group of $\Bbb Q(\zeta_n)$ over $\Bbb Q$. An element $f \in G$ is entirely determined by its image on $\zeta_n$. Since $f(\zeta_n)^k=f(\zeta_n^k)=1 \iff \zeta_n^k=1$ (recall ...
Watson's user avatar
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5 votes
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Irreducibility of special cyclotomic polynomial.

We have $f(x)(x^p-1)=x^{p^2}-1$. Reduce mod $p$ and apply $x \mapsto x+1$. We get $$f(x+1)((x+1)^p-1)=(x+1)^{p^2}-1.$$ Using $(x+1)^p=x^p+1$ and $(x+1)^{p^2}=x^{p^2}+1$ in $\mathbb F_p[x]$, we get $...
MooS's user avatar
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5 votes
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Intermediate fields of cyclotomic field $\mathbb{Q}(\zeta_8)$ - Dummit Foote $14.5.2$

To clarify, looking at the fields themselves: the three quadratic fields in $\Bbb Q(\zeta_8)$ are $\Bbb Q(i),\Bbb Q(\sqrt 2)$ and the less obvious $\Bbb Q(i\sqrt 2)$. two are generated with periods, ...
Emmanuel Amiot's user avatar
5 votes
Accepted

Spotting that $\,x^8 + x^7 + 1\,$ is reducible.

The trick is that $8 \equiv 2 \pmod 3$ and $7 \equiv 1 \pmod 3.$ That means that, if we take a cube root of unity, say $$ \omega = \frac{-1 + i \sqrt 3}{2}, $$ we get $$ \omega^8 + \omega^7 + 1 = \...
Will Jagy's user avatar
  • 140k
5 votes

Determining whether $\Phi_7(x)$ is irreducible over $\mathbb{F}_{11}$

More generally, the irreducible factorization of any polynomial $f(x) \in \mathbb{F}_q[x]$ can be obtained by considering the orbits of the Frobenius map $x \mapsto x^q$ acting on the roots of $f$ ...
Qiaochu Yuan's user avatar
5 votes
Accepted

Polynomials with roots of unity as roots

Part 1 is still kind of nice; depending on what you need this for this may or may not be useful: For a real polynomial, its roots are either real or form conjugate pairs. If you have two conjugate ...
Carl Schildkraut's user avatar
5 votes

Is there a nice general formula for $\int \frac{dx}{x^n-1}$ and/or $\int \frac{dx}{\Phi_n(x)}$?

In terms of a general formula $$ \displaystyle \int \frac{dx}{x^n-1} = -x\;_2F_1\left(1,\frac{1}{n};1+\frac{1}{n},x^n\right) $$ which contains a hypergeometric function. I have played around with the ...
Benedict W. J. Irwin's user avatar
5 votes
Accepted

Galois group of weird polynomial

Yes. $f(x)$ is a factor of $x^{4036}-1$ so $f$ splits over $\Bbb{Q}(\zeta_{4036})$. Also, as you observed, $\Phi_{4036}(x)\mid f(x)$, so the splitting field $\Phi_{4036}$ is contained in the splitting ...
Jyrki Lahtonen's user avatar
5 votes
Accepted

minimal polynomial of $\sin (2\pi/7)$ over $\mathbb Q(\sqrt 7)$

Everything happens inside the field $L=\Bbb{Q}(\zeta,i)$, where $\zeta=e^{2\pi i/7}$. The field $K=\Bbb{Q}(\sqrt{7})$ is a subfield of $L$. This implies that the question can be answered by applying ...
Jyrki Lahtonen's user avatar
5 votes

About the number of real roots

Hint: Use the high-school identity $$t^{2n+1}+1=(t+1)(t^{2n}-t^{2n-1}+t^{2n-2}-\dots+t^2-t+1).$$ What can you conclude for the roots of your polynomial?
Bernard's user avatar
  • 176k

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