# Tag Info

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### 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 ...
• 332k
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 ...
• 63.4k
Accepted

### 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 ...
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### Is the "cyclotomic diagonalization" always squarefree?

Checking the pairs $n,p\le 10^6$ , I found one counterexample , namely : $$283411^2\mid \Phi_{28341}(28341)$$
• 84.7k
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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 ...
• 217k

### 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 ...
• 332k
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 ...
Accepted

### 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 ...
• 159k
Accepted

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 $\gcd(n,p)... • 134k 7 votes ### How to prove$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 ... • 52.7k 7 votes Accepted ### "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 ... • 18.6k 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 $$\begin{bmatrix}1& 1 ... • 369 6 votes ### Proving that cyclotomic polynomials have integer coefficients This proof seems to be more elementary. We begin by contending that if$$(x^n -1) =({\sum}^p_{i=1}{a_i{x^i}})({\sum}^q_{j=1}{b_j{x^i}}),$$where {\sum}^p_{i=1}{a_i{x^i}}\in{\bf{Z}[x]}, then every ... • 1,344 6 votes ### Cyclotomic Polynomials and GCD Note that (x^m-1,x^n-1)=x^{(n,m)} - 1 in \mathbb{Z}[x]. i.e. \exists p(x),q(x)\in \mathbb{Z}[x] such that$$(x^m-1)p(x)+(x^n-1)q(x)=x^{(n,m)}-1$$. Now since x^{(n,m)}-1\mid x^n -1 and x^{(n,m)... 6 votes ### Discriminant of cyclotomic polynomial \Phi_p(x) Thought I would add my solution, which is less elegant, but doesn't require too much creativity. It suffices to find the discriminant of the number field L = \mathbb Q(\zeta_p)/\mathbb Q. We will ... • 694 6 votes ### showing that nth cyclotomic polynomial \Phi_n(x) is irreducible over \mathbb{Q} There is also a non-elementary proof that perhaps is more explanatory than the "elementary" argument, using some (but not too much) algebraic number theory, and primes in arithmetic progressions. To ... • 52.7k 6 votes ### What comes after \cos\left(\tfrac{2\pi}{7}\right)^{1/3}+\cos\left(\tfrac{4\pi}{7}\right)^{1/3}+\cos\left(\tfrac{6\pi}{7}\right)^{1/3}? As some people said, the 2nd formula is easy to derive. In Maple, there are commands to get the minimal polynomial of LHS of the 2nd. The following method works for the 2nd formula, but not for the ... • 38.9k 6 votes Accepted ### Prove that X^4+X^3+X^2+X+1 is irreducible in \mathbb{Q}[X], but that it has two different irreducible factors in \mathbb{R}[X] Finding the complex roots of the polynomial is easy: if \varphi=2\pi/5, the roots are$$ 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 $$... • 239k 6 votes Accepted ### The discriminant of cyclotomic polynomial \Phi_n(x) The computation can be reduced to prime powers, using the multiplicativity of Euler's totient function and "exploiting the fact that cyclotomic fields of relatively prime order are linearly disjoint", ... • 132k 6 votes ### How to prove X^{4}+X^{3}+X^{2}+X+1 is irreductible in \mathbb{F}_{2} You can simply try to factor it; if it is reducible, it must have either a linear or quadratic factor. There aren't many linear and (irreducible) quadratic polynomials in \Bbb{F}_2[X]. • 63.4k 5 votes Accepted ### Structure of Gal(\mathbb{Q}(\zeta_{15})/\mathbb{Q})? Let G be the Galois group of \Bbb Q(\zeta_n) over \Bbb Q. An element f \in G is entirely determined by its image on \zeta_n. Since f(\zeta_n)^k=f(\zeta_n^k)=1 \iff \zeta_n^k=1 (recall ... • 23.8k 5 votes Accepted ### Irreducibility of special cyclotomic polynomial. We have f(x)(x^p-1)=x^{p^2}-1. Reduce mod p and apply x \mapsto x+1. We get$$f(x+1)((x+1)^p-1)=(x+1)^{p^2}-1.$$Using (x+1)^p=x^p+1 and (x+1)^{p^2}=x^{p^2}+1 in \mathbb F_p[x], we get ... • 31.6k 5 votes Accepted ### Intermediate fields of cyclotomic field \mathbb{Q}(\zeta_8) - Dummit Foote 14.5.2 To clarify, looking at the fields themselves: the three quadratic fields in \Bbb Q(\zeta_8) are \Bbb Q(i),\Bbb Q(\sqrt 2) and the less obvious \Bbb Q(i\sqrt 2). two are generated with periods, ... 5 votes Accepted ### Spotting that \,x^8 + x^7 + 1\, is reducible. The trick is that 8 \equiv 2 \pmod 3 and 7 \equiv 1 \pmod 3. That means that, if we take a cube root of unity, say$$ \omega = \frac{-1 + i \sqrt 3}{2}, $$we get$$ \omega^8 + \omega^7 + 1 = \... • 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$f$... • 421k 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 ... • 33.5k 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 ... • 4,331 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 ... • 134k 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 ...
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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?