All Questions
24
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Find a monic irreducible polynomial equivelent to $(x-x_1)(x-x_2)\Phi_m$
Find a monic irreducible polynomial $f(x) = (x - x_1) ... (x - x_n)$， $|x_1| > 1$ and $x_1$ is real, |x_2| < 1 and
$x_2$ is real, $|x_j| = 1$ for all $j > 2$. And First, prove $n > 3$ ...
- 11
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Showing the polynomial has integer coefficients
Show that $\Phi_n(X)$ has integer coefficients.
The proofs here states that $$\Phi_n(X)=\frac{X^n-1}{\prod_{d|n,d\ne n}\Phi_d(X)}.$$
And by long division, they get $\Phi_n(X)\in \Bbb{Q}[X]$. However, ...
- 1,711
3
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2
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Using cyclotomic polynomials to show a polynomial is irreducible over $\mathbb{Q}$
I have been given the polynomial $$f(x)=x^8+x^7-x^5-x^4-x^3+x+1$$ and have been asked to show it is irreducible over $\mathbb{Q}$ by considering the product $(x^2-x+1)f$. (Looking it up, I realise $f$ ...
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Is $g(x,y)= \frac{f(x^{2y+1},y)}{f(x,y)}$ always an integer?
This question is similar to this other question:
Let $$ f(x,y):= \frac{x^y -1}{x+(-1)^y}$$
and $$ g(x,y):= \frac{f(x^{2y+1},y)}{f(x,y)}.$$
Let $y\ge1$ be an integer. Show that $g(x,y)$ is a ...
- 3,716
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For $p$ prime and for $1 ≤ e ∈ Z$ the prime-power $p^e$-th cyclotomic polynomial $Φ_{p^e} (x)$ is irreducible in $Q[x]$
This is part of 'Abstract Algebra, Paul Garrett, 243-244p'
Recall that
$Φ_{p^e}(x) = Φ_p(x^{p^{e-1}}) = \frac{x^{p^{e}}-1}{x^{p^{e-1}}-1} $
First, we check that $p$ divides all but the highest-...
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Proof? of irreducibility of cyclotomic polynomial
Let $\omega = e^{\frac{i2\pi}{n}}$.
I am trying to show that the minimal polynomial of $\omega$ over $\mathbb{Q}$ is the cyclotomic polynomial, that is the polynomial whose roots are the primitve $\...
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Factoring $x^{2020} + x^{2019} + \cdots + x + 1$ on $\mathbb Q[x]$ [duplicate]
An instructor asked me to factor $x^{2020} + x^{2019} + \cdots + x + 1$ on $\mathbb Q[x]$, which he considers to be tricky.
This polynomial is trivial to factor on $\mathbb C[x]$ and $\mathbb R[x]$. ...
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Irreducibility of the polynomial [duplicate]
Prove or disprove.
Let $f_n(x)=x^{n-1}+x^{n-2}+\cdots +x+1$. Then $f_p(x^{p^{e-1}})$ is irreducible in $\mathbb Q[x]$ for all prime $p$.
I know if $p$ is a prime $f_p(x)$ is $p$-th cyclotomic ...
- 612
3
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1
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681
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when is the $n$-th cyclotomic polynomial irreducible over $\mathbb{R}$
It is well known that the cyclotomic polynomials
$\Phi_n(x)$
are irreducible over the field of rationals $\mathbb{Q}$.
I am curious about their reducibility over the real numbers $\mathbb R$.
We have ...
- 325
1
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2
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If a monic rational polynomial of degree $p-1$ has $p$-th root of unity as a root, is it the cyclotomic polynomial?
If a monic rational polynomial of degree $p-1$ has a $p$-th root of unity as a root, where $p$ is prime, does that make it the cyclotomic polynomial $x^{p-1}+...+1$?
I think this is the same as ...
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How to show that a given polynomial is irreducible in a cyclotomic field
I'm beginning to study McCarthy's Algebraic Extensions of Fields, and one of the first problems is to give a factorization of $x^4 + 1$ in $K[x]$, where $K=\mathbb{Q}(a)$ and $a$ is a root of $x^4+1$ (...
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54
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Notation and interpretation of Polynomials in $\mathbb{F}_{p}[x]$
i'm confused with some notation that involves reduction of polynomyals on $\mathbb{Z}[x]$ to $\mathbb{F}_p[x]$. It's part of the proof that Cyclotomic polynomials are irreducible over $\mathbb{Q}[x]$.
...
- 1,403
1
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1
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Unique subfield of order $2$ of $\mathbb{Q(\zeta_{7}})$ over $\mathbb{Q}$
Actualy this subfield is $\mathbb{Q}(i\sqrt{7})$ since $X^{2}+7$ is irreducible over $\mathbb{Q}$ and has $i\sqrt{7}$ as root.
My problem here is to show unicity, I tried something using the tower ...
- 1,403
6
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1
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Proving that $x^{2^n} + 1$ is irreducible in $\mathbb Q[x]$
$x^{2^n} + 1$ is irreducible in $\mathbb Q[x]$.
I've been working on this and this is my process:
I would like to use Eisenstein's criterion so I considered the substitution $y=x-1$. So
$$x^{2^n}+1=(...
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Proof that $a^{n}+b^{n}$ is irreducible over $\mathbb Q$
The sum of fourth powers cannot be factored over $\mathbb Q$, since
$ a^4+b^4 = (a^2+\sqrt{2}ab+b^2)(a^2-\sqrt{2}ab+b^2)$
And these quadratic factors does not have any real rational factors.
How ...
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