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39 views

Disjoint exceptional lines on non-minimal cubic surface

A diagonal cubic surface $\sum_{i=0}^3a_iT_i^3=0$ is not minimal if, for example, $a_1a_2a_3^{-1}a_4^{-1}\in(k^*)^3$. This should be because there is an exceptional line $D$ such that no element in ...
fp1's user avatar
  • 115
4 votes
3 answers
177 views

Prove that $\mathbb Q(\cos\tfrac\pi7)\neq\mathbb Q(\cos\tfrac\pi9)$

Let $\tau=2\pi$ be the full angle. (tau) For any integer $k$ and any angle $\theta$, $\cos(k\theta)$ is a polynomial in $\cos\theta$. In particular, $\cos(2\theta)=2\cos^2\theta-1$, which shows that $\...
mr_e_man's user avatar
  • 5,726
3 votes
4 answers
256 views

Prove irreducible cubic polynomial over $\mathbb{Q}$ with a cyclic galois group has real roots

I want to prove the following: Let $f\in \mathbb{Q}[x]$ be an irreducible cubic polynomial, whose Galois group is cyclic. Prove that all of the roots of $f$ are real. I know that the Galois group $G$...
Ariel Yael's user avatar
10 votes
0 answers
231 views

On the solvable octic $x^8-44x-33 = 0$ and the tribonacci constant

I had discussed the solvable octic trinomial, $$x^8-44x-33=0\tag1$$ way back in this old MSE post, but I revisited this inspired by another solvable octic, $$y^8-y^7+29y^2+29=0\tag2$$ which I also ...
Tito Piezas III's user avatar
5 votes
0 answers
208 views

Factorization of $x^3-x-1$ over $\mathbb{F}_p$

Factoring quadratic polynomials over finite fields can easily be done by determining if the discriminant is a quadratic residue modulo characteristic in question, and if so, apply the quadratic ...
T. Rex's user avatar
  • 405
2 votes
2 answers
101 views

Finding the roots of $x^5+x^2-9x+3$

I have to find all the roots of the polynomial $x^5+x^2-9x+3$ over the complex. To start, I used Wolfram to look for a factorization and it is $$x^5+x^2-9x+3=(x^2 + 3) (x^3 - 3 x + 1)$$ I can take it ...
Maria Camila Espinal Ramirez's user avatar
0 votes
1 answer
154 views

What are the necessary and sufficient conditions for the cubic equation to have at least 1 positive real root?

What are the necessary and sufficient conditions for the cubic equation to have at least 1 positive real root? I'm just dealing with the $2$ simplest cases. Case-1: $$x^3+px+q=0$$ where, $p<0$ and ...
nonuser's user avatar
  • 506
2 votes
1 answer
530 views

Denesting Cardano's Formula

For a depressed cubic equation $x^3 + px + q =0$ having exactly one real root, Cardano's formula gives the real root as $$\sqrt[3]{-\frac{q}{2} +\sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} + \sqrt[3]{-\...
Pranav Bisht's user avatar
2 votes
0 answers
64 views

Galois group of polynomial over $\mathbb{Q}(\sqrt{-3})$

Find the Galois group of the polynomial $f(x) = x^3 -10$ over the field $\mathbb{Q}(\sqrt{-3})$, given that the discriminant of $f$ is $-2700.$ I know that the Galois group of a given polynomial is ...
user867777's user avatar
2 votes
0 answers
240 views

In characteristic 2 the splitting field of a cubic has degree 6

I have been working on the following problem, from an old p-set of a Galois theory course I found online: Let $F = \mathbb{F}_2(t)$, the field of rational functions on $\mathbb{F}_2$, and let $f(x) = ...
horned-sphere's user avatar
3 votes
3 answers
571 views

Lagrange Method of Solving Cubic Equations

Let $K$ be a field (for simplicity, one may assume $K\subseteq\mathbb{C}$), and let $d\in K$. Denote $w=\sqrt[3]{d}$, and $\zeta=(-1+\sqrt{-3})/2$. If $f(x)=x^3+ax^2+bx+c$ is the minimal polynomial of ...
Roy Sht's user avatar
  • 1,371
1 vote
1 answer
115 views

Show that $x^3+3y^3+9z^3-9xyz=1$ has infinitely many integer solutions. [duplicate]

Show that $x^3+3y^3+9z^3-9xyz=1$ has infinitely many integer solutions. I have found that (1,0,0) and (1,-18,12) are two solutions and tried (1,-18+n,12-n). There is a hint saying that I should try ...
LukeG's user avatar
  • 121
1 vote
1 answer
174 views

Solving the sextic equation with 14th root of unity

I am solving the sextic equation $t^6-t^5+t^4-t^3+t^2-t+1=0$ satisfied by the 14th root of unity (a problem from Ian Stewart's book). I was able to get up to the point where you have the polynomial $u^...
Killaspe's user avatar
  • 338
0 votes
0 answers
71 views

Cubic extension adjoining a root of an irreducible cubic polynomial.

Assume $f(x) \in \mathbb{Q}[x]$ is an irreducible cubic polynomial with discriminant $D$, let $x_1, x_2, x_3$ be the three distinct roots of $f(x)=0$. Let $p$ be any odd prime coprime to $D$. Are the ...
J. King's user avatar
  • 19
2 votes
2 answers
101 views

None of the methods I know to factor cubics are working here ....

I have been trying to find all the different methods for factoring cubics and so far in my search I have come across: 1)Using the sum/difference of cubes 2)The grouping method 3)Using the rational ...
excalibirr's user avatar
  • 2,815

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