Galois group of x^6-9x^6-9

f = x^6-9 = (x^3-3)(x^3+3) Let L_f be splitting field therefore L_f = \mathbb{Q}[\sqrt[3]{3},e^{\frac{2\pi i}{3}}], [L_f:\mathbb{Q}] = 9. Also Gal\space x^3±3/\mathbb{Q} = S_3 and Gal\space f/\mathbb{Q}-orbits is roots of x^3-3 and x^3+3 therefore, i think, that G = Gal\space f/\mathbb{Q} = S^3\times S^3. But |G| = 36 \ne [L_f:\mathbb{Q}]. What’s wrong? Answer A few issues: … Read more

If every polynomial of odd degree is reducible in $\mathbb{F}$ then $[E:\mathbb{F}]$ is not odd

$\mathbb{F}$ is a field of characteristic zero, and every odd-degree polynomial over $F$ has a root in $\mathbb{F}$. now I have two questions:- If $E$ is a finite extension of $F$ show that $[E:\mathbb{F}]$ is not odd. If $E$ is a finite Galois extension of $\mathbb{F}$, Show that $[E:\mathbb{F}]$ is a power of two. My … Read more

Finding Galois extension isomorphic to \mathbb{Z}_n\mathbb{Z}_n

Exist a method to find an Galois extension E such that Gal(E/\mathbb{Q})\cong\mathbb{Z}_n? Only for n=6 how can I do? Answer Find m such that \text{Gal}(\mathbb{Q}(\zeta_m)/\mathbb{Q}) \cong \mathbb{Z}_m^{\ast} has a quotient group isomorphic to \mathbb{Z}_n. You can do this by letting m be a prime congruent to 1 \bmod n. By Galois theory, the fixed field … Read more

Prove that the intersection of two intermediate subfields of a Galois extension corresponds to the product of its subgroups of its Galois group

Let $F/K$ be a Galois extension and let $E_1, E_2$ be intermediate subfields of $F/K$ with corresponding subgroups $H_1,H_2$ of $Aut(F/K)$. Prove that the intersection of $E_1$ and $E_2$ corresponds to $<H_1H_2>$,and $E_1E_2$ corresponds to the intersection of $H_1$ and $H_2$. Answer Let $H = \langle H_1H_2\rangle$, then for any $\sigma \in H_1H_2$, and $x\in … Read more

Galois group of a product of irreducible polynomials

I’m working on a problem that asks to determine the Galois group of the polynomial (x3−2)(x2−5) over Q. I know by a Theorem in Hungerford that the Galois group of X3−2 is isomorphic to S3. Additionally, since the splitting field of x2−5 is Q(√5), it’s easy to show the automorphism group group over Q (i.e. … Read more

Why is $\mathbb{Q}(\sqrt[4]{2}) $ is not normal over $\mathbb{Q}$?

I read somewhere that $\mathbb{Q}(\sqrt[4]{2}) $ is not normal over $\mathbb{Q}$, and the reason given is: if it were normal over $\mathbb{Q}$, then $i \in \mathbb{Q}(\sqrt[4]{2})$. Could someone explain to me why this is the case? Answer One definition of normal extension $L/K$ is: Every irreducible polynomial in $K[X]$ that has one root in $L$, … Read more

Degree of the splitting field of x3−5 x^3-5 over Q\mathbb{Q}

I am trying to find the degree of the splitting field for x3−5 over Q. I have so far: The splitting field will be Q(3√5,u) where u is the 3rd root of unity. So I’m looking for the degree of Q(3√5,u):Q. x3−5 is irreducible by Eisenstein’s criterion so the degree of Q(3√5):Q = deg(x3−5) = … Read more

A polynomial with a root in $\mathbb{F}_p \ \forall p$, where $p$ is prime, but no root in $\mathbb{Z}$ [duplicate]

This question already has answers here: Polynomial with a root modulo every prime but not in $\mathbb{Q}$. [duplicate] (1 answer) Prove that for every prime $p$, there exists a solution to $(x^2-2)(x^2-6)(x^2-3) \equiv 0 \mod(p)$ (3 answers) Closed 6 years ago. Give an example of a polynomial $f(x) \in \mathbb{Z}[x]$ which has a root in … Read more