Skip to main content
18 votes
Accepted

Galois group of the irreducible cubic equation $x^3-3x+1$

You don't actually need to find the last root. Consider the field $\mathbb{Q}(\alpha)$, where $\alpha$ is any root of $f$. This field not only contains $\alpha$, but also $2- \alpha^2$ per the field ...
Kaj Hansen's user avatar
  • 33.1k
14 votes
Accepted

Galois group of an irreducible , separable polynomial be abelian , then each of the roots of the polynomial generates the splitting field?

The proof goes as follows: Since $\operatorname{Gal}(E/k)$ is abelian, any subgroup is normal, hence any intermediate field of $E/k$ is normal. In particular $k(a)/k$ is normal and by the definition ...
MooS's user avatar
  • 31.6k
13 votes
Accepted

Prove $f(x)=x^8-24 x^6+144 x^4-288 x^2+144$ is irreducible over $\mathbb{Q}$

Your polynomial $f(x)$ takes prime (or $-$ prime) values at $x = \pm 1, \pm 7, \pm 11, \pm 13, \pm 23, \pm 67, \pm 85, \pm 109, \pm 145, \pm 197, \pm 205, \pm 209, \pm 241, \pm 373, \pm 397, \pm 403, \...
Robert Israel's user avatar
13 votes
Accepted

Can I prove that a splitting field is normal without using zorn lemma

The argument doesn't actually require an algebraic closure; you can just replace the algebraic closure by sufficiently large finite extensions in each step of the argument. Specifically, say $F$ is ...
Eric Wofsey's user avatar
11 votes
Accepted

Expressing the roots of a cubic as polynomials in one root

Based on the answers in this question, I'm able to finish the general case. Let $x^3+px+q$ be a cubic with rational coefficients and having real splitting field of degree $3$. Let its roots be $a,b,c$...
lhf's user avatar
  • 217k
10 votes

Galois groups of $x^3-3x+1$ and $(x^3-2)(x^2+3)$ over $\mathbb{Q}$

You do not need to know the roots of the cubic to find its Galois group. You should consult an algebra book about Galois groups and discriminants here. Then the solution is as follows. By the Rational ...
Dietrich Burde's user avatar
10 votes

Find a splitting field for $x^4 - x^2 - 2$ over $\Bbb Z_3$

Notice that $$x^4-x^2-2=x^4+2x^2+1=(x^2+1)^2$$ in $\Bbb Z/3\Bbb Z$. Thus the splitting field has to contain $\alpha $ such that $\alpha ^2+1=0$; and if an extension contains such an $\alpha$, then it ...
Arnaud D.'s user avatar
  • 20.9k
9 votes
Accepted

Why every extension of a characteristic zero field is separable?

An extension $K/F$ is separable iff every element of $K$ has a separable minimal polynomial over $F$. Minimal polynomials are irreducible, and in characteristic $0$ all irreducible polynomials are ...
Arthur's user avatar
  • 200k
8 votes

$\operatorname{char} k=0$, $a,b,c,d,e \in k$ , then is the polynomial $f(x)=ax^8+bx^6+cx^4+dx^2+e$ solvable by radicals over $k$?

Hint: Is $g(x)=ax^4+bx^3+cx^2+dx+e$ solvable by radicals?
Hagen von Eitzen's user avatar
8 votes

Prove $f(x)=x^8-24 x^6+144 x^4-288 x^2+144$ is irreducible over $\mathbb{Q}$

You just have to show that $f(x)$ is the minimal polynomial of $\eta=\sqrt{(2+\sqrt{2})(3+\sqrt{3})}$ over $\mathbb{Q}$, i.e. to show that $\eta$ is an algebraic number over $\mathbb{Q}$ with degree $...
Jack D'Aurizio's user avatar
8 votes

What fields are isomorphic to $\mathbb{Q}[\sqrt{2}]$ other than $\mathbb{Q}[\sqrt{2}]$ itself?

If you define $\Bbb Q[\sqrt 2]$ as the smallest subfield of $\Bbb C$ that contains $\sqrt 2$, then here's a totally different field that is isomorphic to it: $\Bbb Q[X]/(X^2-2)$, one (of two possible) ...
Hagen von Eitzen's user avatar
8 votes
Accepted

Galois number fields that have the imaginary unit.

Items 2 and 4 are incompatible. Assume that $$ \Bbb{Q}\subset \Bbb{Q}(i)\subset K_f, $$ where $K_f$ is Galois over $\Bbb{Q}$, $G=Gal(K_f/\Bbb{Q})\simeq A_4$. By Galois correspondence $\Bbb{Q}(i)$ is ...
Jyrki Lahtonen's user avatar
7 votes

Galois group of the irreducible cubic equation $x^3-3x+1$

I'm fairly sure that Kaj Hansen's answer (+1) is the intended solution. If you want to say more, you can observe that if $\zeta=e^{2\pi i/9}$, and $u=\zeta+\zeta^{-1}=2\cos(2\pi/9)$, then $$ u^3-3u=(\...
Jyrki Lahtonen's user avatar
7 votes
Accepted

Can $\sqrt{3}$ be written as a polynomial expression in $\sqrt[3]{3}$ and $\zeta_3$

No. The field $K=\Bbb Q(\zeta_3,\sqrt[3]3)$ has degree $6$ over $\Bbb Q$ and Galois group $S_3$. It has a unique quadratic subfield, $\Bbb Q(\zeta_3)=\Bbb Q(\sqrt{-3})$.
Angina Seng's user avatar
7 votes
Accepted

Degree of a splitting field is no greater than $n!$

$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$Proceed by induction on $n$. Let $\alpha$ be a root of $P$ in some extension field of the field $F$ you start with. Then $\Size{F(\alpha) : F} \le ...
Andreas Caranti's user avatar
7 votes
Accepted

Galois group of $ X^6-2tX^3+1 $ over $ \mathbb{Q}(t) $

You cannot in general deduce the irreducibility of $f(X^n)$ from the irreducibility of $f(X)$. (Consider $f$ as polynomial over $\mathbb Q[t]$ and reduce modulo the prime element $2t+1$, then we get ...
Lukas Heger's user avatar
  • 21.7k
7 votes
Accepted

Find the splitting field of $x^4+1$ over $\mathbb Q$.

This seems to be OK, excepting $\mathbb Q(i,\sqrt 2)=\mathbb Q(\sqrt {-2})$. (See the comments.) Another, more general approach, is to look at this and noting that $x^4+1$ is the cyclotomic polynomial ...
Nicky Hekster's user avatar
7 votes
Accepted

Can $\mathbb{Q}(u)$ be of degree 2 over $\mathbb{Q} (u^3)$?

Let $u$ be a root of $x^2+x+1$. This polynomial is irreducible over $\Bbb{Q}$. Also note that $$(x-1)\cdot(x^2+x+1)=x^3-1.$$ Since $u$ is a root of $x^2+x+1$, it is also a root of $x^3-1$. So we ...
user729424's user avatar
  • 5,071
6 votes

How to prove that algebraic numbers form a field?

I'm late to the party, but none of these answers have mentioned polynomial resultants. To me this is "the natrual" approach. Note that if $a,b \in \bar{F}$ have minimal polynomials $A(x),B(x) \in F[x]...
Doris's user avatar
  • 646
6 votes
Accepted

Field Extension with Galois group $S_n$ is the splitting field of a polynomial of degree $n$

Let $ H_i $ denote the isomorphic copy of $ S_{n-1} $ in $ S_n $ composed of elements which fix $ i $. For $ L/K $ a Galois extension with Galois group $ S_n $, consider the fixed field $ F_1 $ of $ ...
Ege Erdil's user avatar
  • 17.8k
6 votes
Accepted

Degree of a field extension $K/\mathbb{Q}$

$K$ embeds in the complex numbers, but not in the real numbers. Applying complex conjugation to this embedding thus gives a nontrivial automorphism with order $2$, so $2$ divides the order of the ...
Angina Seng's user avatar
6 votes

Galois group of $x^3+2x+2$

Here's a general theorem which fits your problem perfectly: If $f$ is an irreducible polynomial of prime degree $p$ with rational coefficients and exactly two non-real roots, then the Galois group ...
lhf's user avatar
  • 217k
6 votes
Accepted

Compute the degree of the splitting field of $x^{12} - 2$ over $\mathbb{Q}$ and describe its Galois Group as a semidirect product

$\newcommand{Gal}{\operatorname{Gal}}$Preliminaries: $\operatorname{irr}(\zeta, \Bbb Q) = \Phi_{12}(X) = X^4-X^2+1$ $\zeta = \dfrac{\sqrt3+i}2$ $\sqrt3 = \zeta + \zeta^{-1}$ $i = \zeta + \zeta^5$ $\...
Kenny Lau's user avatar
  • 25.1k
6 votes
Accepted

Degree of splitting field of $X^n-1$ over some finite field

It is not true that $o(\alpha)=n$ for every root $\alpha\in K$ of $X^n-1$; after all $1\in K$ is also a root of $X^n-1$. But because $K$ is the splitting field of $X^n-1$ there exists a root $\alpha\...
Servaes's user avatar
  • 63.5k
6 votes

Galois group of $x^6+1$ over $\mathbb{F}_2$

Note that $x^6+1=(x^3+1)^2=(x-1)^2(x^2+x+1)^2$. The polynomial $x^2+x+1$ is irreducible over $\mathbb{F_2}$ and hence $\mathbb{F_2}/(x^2+x+1)\cong\mathbb{F_4}$ is an extension field of $\mathbb{F_2}$. ...
Mark's user avatar
  • 40.9k
6 votes
Accepted

Finding primitive element of field extension in characteristic 2 corresponding under Galois correspondence to the group $G_f\cap A_n$

Assume $\mbox{char}(F)=2$ and let $\alpha_1,\dots,\alpha_n$ be the distinct roots of $f(X)$. Define $$ \delta_f = \sum_{i<j} \frac{\alpha_i}{\alpha_i + \alpha_j} $$ and $$ D(f) = \sum_{i<j} \...
Albert's user avatar
  • 2,932
6 votes

Treating splitting fields as the same dangerous?

Let's consider the polynomial $f(x) = x^2 + 1$ over $\mathbb{Q}$. Then one way to construct a splitting field for $f$ is to take the subfield of $\mathbb{C}$ generated by its roots, which of course is ...
Qiaochu Yuan's user avatar
5 votes

$\mathbb{C}$ is not the splitting field of any polynomial over $\mathbb{Q}$ (without cardinality)

The existence of a transcendental number provides such a proof.
Noah Schweber's user avatar
5 votes

Separable Extension and Splitting Field

$Q(2^{1/3})$ is separable since it is an extension of a field of characteristic zero, but is not a splitting field, since it does not contain complex roots of $X^3-2$. If a field is not perfect, its ...
Tsemo Aristide's user avatar
5 votes
Accepted

Splitting Field Problem

As $\sqrt[6]{7}$ is a zero of the polynomial and $\sqrt[6]{7} \not \in \mathbb{Q}$ we adjoin it to $\mathbb{Q}$ to obtain $\mathbb{Q}(\sqrt[6]{7})$. Now in $\mathbb{Q}(\sqrt[6]{7})$ the polynomial ...
Stefan4024's user avatar
  • 35.9k

Only top scored, non community-wiki answers of a minimum length are eligible