Galois theory allows one to reduce certain problems in field theory, especially those related to field extensions, to problems in group theory. For questions about field theory and not Galois theory, use the (field-theory) tag instead. For questions about abstractions of Galois theory, use ...

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Galois group of $f$ is cyclic if $n$ is prime

Hello I am learning Galois Theory by myself and got lost in the following exercise: Let $f$ be an irreducible polynomial of degree $n$, and suppose that the splitting field of $f$ is generated by a ...
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28 views

“Breaking the symmetry” in solving algebraic equations

I've heard somewhere a discussion about solving algebraic equations before: When solving a quadratic equation, we are essentially doing the following. Observe that ...
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28 views

Is it possible to have a matrix with eigenvalues that cannot be constructed from a finite number of basic arithmetic operations, and nth roots?

For example, a characteristic polynomial $ p(\lambda) = \lambda^5 - \lambda -1 $ has the root 1.167304..., but this number cannot be written as a finite number of arithmetic operations (addition, ...
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1answer
17 views

prove $[E(a):E] \le [F(a):F]$

Let $F<E<K$ be field extensions, such that $a \in K$, and $[K:F]<\infty$, Is it true that $[E(a):E] \le [F(a):F]$? How can I show this?
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A question about “the general equation of degree n can not be solved by radicals for n$\ge$5”

This is the proposition 40 on Dummit's textbook. In the proof it used Theorem 32: The general poly $x^n-s_1x^{n-1}+...+{(-1)}^ns_n$ over field F($s_1,s_2,...,s_n)$, is separable with Galois group ...
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Axioms for constructability

When we are doing geometric constructions we assume that the only operations we can perform are We can draw a line between to points. We can draw a circle with one point as the centre and the other ...
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Proving that the Galois group of minimal polynomial of constructible number is a power of $2$

I was trying to understand whether a real number has degree which is power of 2 over rationals is constructible. Then I found this. But I am having trouble in understanding why this statement has to ...
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2answers
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What is the $l$ in this group? [on hold]

What is the $l$ in this group? $a^7=e$, $b^3=e$, $b^{-1}ab=a^2$, $ba=a^lb$. $l=?$
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1answer
11 views

Solving the Cubic Equation (using Lagrange Resolvents)

This is from my textbook. I am having trouble working out the calculations that the author skips over. So we start with the polynomial $\ X^3 - aX^2 + bX -c$ with zeros $x_1,x_2,x_3$. Then we define ...
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Suppose K:Q is normal with Gal(K:Q) isomorphic to $\mathbb{Z}_4\times\mathbb{Z}_6$. Prove that K has 3 distinct non-trivial square roots.

Suppose K:Q is normal with Gal(K:Q) isomorphic to $\mathbb{Z}_4\times\mathbb{Z}_6$. Prove that K has 3 distinct non-trivial square roots. Q is the set of rational numbers. The only clue that I've ...
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Decompose a vector space into invariant subspaces?

Consider the following proposition: Suppose $V$ is a finite dimensional vector space over a field $F$, and $K/F$ is a finite Galois extension with Galois group $G$. If $V$ has a $(K,K)$ bimodule ...
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Galois theory about automorphisms of the field of rational functions

Suppose That $F$ is a field and $G=Aut(F(x))$ is the group of field automorphisms of the field of rational functions $F(x)$ and fix $F$, and that $E\subset F(x)$ is the fixed field of G. please prove ...
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26 views

Question on Galois theory polynomials

Galois theory states that we cannot expect to extract roots of general degree $5$ polynomial equation using radical operations. Supposing we have promise that equation contains only real roots, do we ...
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24 views

When is the normality of field extensions transitive?

I want to answer the following question: Given field extensions $K \subset M \subset L $, when is the case that $M:K$ is normal and $L:M$ is normal implies $L:K$ is normal? In my algebra class we have ...
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1answer
15 views

Galois groups of extensions over $\mathbb{Q}$

Let $K=\mathbb{Q}(\sqrt[8]{2},i)$ and let $F=\mathbb{Q}(i)$. I need to show that $\operatorname{Gal}(K/F)\cong Z_8$ where $Z_8$ is the cyclic group of order 8. I have already shown that $[K:F]=8$, ...
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1answer
50 views

When is $\mathbb{Q}(x)$ a finite extension of $\mathbb{Q}$?

Let $x\in\mathbb{C}$, and consider the field $\mathbb{Q}(x)$. If this field is a finite extension of $\mathbb{Q}$, then $x$ is algebraic over $\mathbb{Q}$, so satisfies some polynomial $P$. Any field ...
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2answers
31 views

Definition of a Finite Normal Extension

I'm reading Fraleigh's A First Course in Abstract Algebra, Seventh Edition, and he makes the following definition on page 448: A finite extension $K$ of $F$ is a finite normal extension of $F$ if ...
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1answer
58 views

the Galois closure of $\mathbb{F}_4(x,y) / \mathbb{F}_4(x)$

I want to show that $\mathbb{F}_4(x,y) / \mathbb{F}_4(x)$ is NOT a Galois extension. Consider the field extension $\mathbb{F}_4(x,y) / \mathbb{F}_4(x)$ where $y$ is root of the polynomial $g(T) = ...
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3answers
55 views

Show that the splitting field of $x^8-3$ has degree 32 over $\mathbb{Q}$

I have already determined that a splitting field for $f(x) = x^8 - 3$ over $\mathbb{Q}$ is $K= \mathbb{Q}(i , \sqrt{2}, 3^{\frac{1}{8}})$. I have the following tower relationship: $$[K: \mathbb{Q}] = ...
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$i \notin \mathbb{Q}[\sqrt[4]{2}]$ without using topological properties of $\mathbb{R}$

I can think of two related ways to prove that $i \notin K = Q[\sqrt[4]{2}]$: $K$ is a subset of the real numbers and $i$ is not a real number. $K$ is orderable and no ordered field can contain ...
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2answers
49 views

Galois group of $x^4-2$

I am trying to explicitly compute the Galois group of $x^4-2$ over $\mathbb{Q}$. I found that the resolvent polynomial is reducible and the order of the Galois group is $8$ using the splitting field ...
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1answer
59 views

$\mathbb{F}_4(x,y) / \mathbb{F}_4(x)$ is a Galois extension, with $\mathbb{F}_4(x,y)$ a function field of an algebraic curve

Consider the field extension $\mathbb{F}_4(x,y) / \mathbb{F}_4(x)$ where $y$ is root of the polynomial $$ f(T)= x^4 + x^2T^2 + x^2T + x^2 + xT^2 + xT + T^4 + T^2 + 1 \in \mathbb{F}_4(x)[T]. $$ I ...
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Determine the elements of the Galois group [duplicate]

I want to determine the elements of the Galois group of $x^p-2$. I have never seen anything like this before and been struggling with some of the Galois problem. Thank you for any input!
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Way to check the Galois groups of given polynomials online?

Is there some way (for instance a way of typing into Wolfram Alpha) that will give me the Galois group of a given polynomial over Q.
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2answers
34 views

Generator of the cyclic Galois group $\operatorname{Gal}(\mathbb{Q}[\xi_{n}]/\mathbb{Q})$

I would like to know what is a generator of a $\operatorname{Gal}(\mathbb{Q}[\xi_{n}]/\mathbb{Q})$ cyclic group if we know that the elements of the group are automorphisms such as ...
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2answers
40 views

Example of Field Extension $E/F$ with $Char(F)=2$ and $[E:F]=2$, but is not Galois

I understand that for a field extension $E/F$, if $Char(F)\neq 2$ and $[E:F]=2$ then it must be a Galois Extension. I have proved this, but I am having trouble finding a counterexample when the ...
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1answer
26 views

General polynomial of degree $n$ is irreducible from Gauss' Lemma

I'm studying "old-fashioned" Galois theory and the following is an elementary and fundamental problem to keep proceed with my studies. I'm really stuck at that question. Can someone help me please? ...
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3answers
90 views

proving that $\mathbb{Q}(\sqrt{5}, \sqrt{6}) = \mathbb{Q}(\sqrt{5}+ \sqrt{6}) $

Here is an extract from my Galois Theory notes proving that $\mathbb{Q}(\sqrt{5}, \sqrt{6}) = \mathbb{Q}(\sqrt{5}+ \sqrt{6}) $ My question is after rearranging equation (1) has my lecturer omitted an ...
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All profinite groups are Galois groups (Thm 3.3.2 in J. Wilson's Profinite groups)

I am reading about infinite Galois theory in Wilson's Profinite groups, and I have a problem in understanding the proof of Lemma 3.3.1 and Theorem 3.3.2 (here you can see them). In particular, I don't ...
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1answer
22 views

Automorphisms of a field extension (proof verification)

I am asked to compute the automorphisms of the field extension $\mathbb{Q}(\sqrt[4]{2})/\mathbb{Q}(\sqrt{2})$. I know that $[\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}]=4$ since $$ ...
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45 views

A question related to Galois theory

I'm working with this problem: Let L/K be a Galois extension with Galois group $S_4.$ Then L is the splitting field of a monic degree 4 irreducible polynomial over K. Char(K)=0. My method is since ...
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49 views

Why does this specific Quintic Equation have a closed form and this similar one does not?

I read on Wikipedia that x^5 -x -1 = 0 has a real root, but that you can't express it in radicals. So I thought maybe all of the x^5 -x -A =0 don't have a real root that can be expressed as a radical ...
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1answer
29 views

Cyclotomic Fields - Showing that the fixed field of $G(\mathbb Q(\xi)/\mathbb Q)$ is $\mathbb Q$.

If $p$ is a prime and $\xi$ is a primitive $p$th root of unity, I know that $G(\mathbb Q(\xi)/\mathbb Q) = \{\psi_{\xi,\xi^k}\}_{1\leq k<p}$, where for each $k$, $\psi_{\xi,\xi^k}(\xi) = \xi^k$. I ...
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26 views

Normal extension and action of automorphisms on factors

Let $N/K$ be a normal extension of fields. Let $f\in K[X]$ be an irreducible polynomial with monic irreducible factors $g,h\in N[X]$. Show that there exists an automorphism $\varphi$ on $N$ which ...
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3answers
35 views

How are all the roots of unity of cyclotomic extension are of this form? [closed]

Suppose $x \in Q(\zeta_n) $ which satisfy $x^t =1, t \in \mathbb{N}$. Then show that $x$ is of the form $\zeta_n^k$ for some $k$ where $1 \leq k \leq n-1$ ?
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14 views

Approximation of a trisecting an angle

I learned a proof that it is impossible to trisect an angle. Is there some research that if we have been given an angle, a ruler and a compass and we are allowed to draw $m$ circles and $n$ lines/line ...
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1answer
47 views

Determine whether the extension is Galois [duplicate]

I am trying to prove that $K=\mathbb{Q}(2^{1/3}, i\sin{2\pi/3})$ is Galois extension over $\mathbb{Q}$. It is easy to see that $K=\mathbb{Q}(2^{1/3},i\sqrt{3})$. I know it is Galois since $K$ is a ...
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3answers
43 views

If field $K/F$ is generated by the $\alpha_1,…,\alpha_n$, then an $\sigma\in $ Aut$(K/F)$ of $K$ uniquely determined

Does this proof seem correct? I'm having second doubts concerning the bolded material. Show that if the field $K$ is generated over $F$ by the elements $\alpha_1,...,\alpha_n$, then an automorphism ...
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41 views

Understanding algebraic closure [duplicate]

I am trying to understand what it means to have an extension that is an algebraic closure of the base field. I'm looking for someone who can help conceptually. I understand how $C/R$ looks. The basis ...
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What is the automorphism group of the field of all constructible numbers?

Let $\Omega\subseteq \mathbb{C}$ be the field of all constructible numbers (i.e. $\Omega$ is the smallest subfield of $\mathbb{C}$ which is closed under taking square roots). What is known about the ...
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1answer
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Is there closed form solution for this infinite polynomial or high-order polymonial?

The equation is as follows \begin{align} \sum_{N=1}^{\infty}P(N)x^N=Z, \end{align} where $P(N)$'s are real number satisfying $0\leq P(N)\leq 1$. Another equation is \begin{align} \sum_{N=1}^{\bar ...
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1answer
46 views

Determine $n$, $m$ for which $\Bbb Q_n \subseteq \Bbb Q_m$.

Determine $n$, $m$ for which $\Bbb Q_n \subseteq \Bbb Q_m$. How to derive this. I am having no clue. I guess $n|m$. But what is the proof?
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A sequence of rational polynomials whose splitting fields over $\mathbf{Q}$ have dihedral Galois groups.

It is well known that the splitting fields of $x^3-2$ and $x^4-2$ over $\mathbf{Q}$ have Galois groups $D_6$ and $D_8$, Dihedral groups of $6$ and $8$ elements respectively. However, this pattern ...
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1answer
42 views

Galois theory, resolvent, Frobenius group

I need to prove that the Galois group of the polynomial $x^5+15x+12\in \mathbb{Q}[x]$ is the Frobenius group of order 20. The discriminant of that polynomial is $D=2^{10}\cdot 3^4\cdot 5^5$, i.e. it ...
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1answer
33 views

Let $G$ be the finite abelian group show that there is a galois extension $K/\Bbb Q$ with $Gal(K/\Bbb Q) \equiv G$ [duplicate]

Let $G$ be the finite abelian group show that there is a galois extension $K/ \Bbb Q$ with $Gal(K/\Bbb Q) \cong G$. I have seen one proof using For a fixed positive integer $n$, there are infinitely ...
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20 views

If $d \in \Bbb Q$ , show that $\Bbb Q( \sqrt d)$ lies in some cyclotomic extension of $\Bbb Q$ [duplicate]

If $d \in \Bbb Q$ , show that $\Bbb Q( \sqrt d)$ lies in some cyclotomic extension of $\Bbb Q$. I have thaught that for $\Bbb Q(\sqrt 3)$ is in $\Bbb Q(w)$ where $w$ is in the $3$rd root of unity. ...
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51 views

Cyclotomic polynomial, after adjoining a radical

Suppose that $p>2$ is prime and that $a$ is a rational number for which $\sqrt[p]{a}$ is in $\mathbb C\backslash\mathbb Q$. The cyclotomic polynomial $\Phi_p$ is well-known to be irreducible over ...
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17 views

Splitting field of a cubic polynomial understanding

The cubic polynomial $f(x) = x^3+px+q\in K[x]$ has 3 roots $a_1,a_2,a_3\in \mathbb C$ Hence, the splitting field extension $L=K(a_1,a_2,a_3)$ $\delta=(a_1-a_2)(a_1-a_3)(a_2-a_3)\in L$ since ...
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1answer
47 views

Galois group of $x^6+1$

$x^6+1$ has $6$ roots: $i,i\xi,i\xi^2,i\xi^3,i\xi^4,i\xi^5$ where $\xi=e^{\tfrac{2\pi i}{6}}$. Since $x^{12}-1=(x^6-1)(x^6+1)$ the splitting field of $x^{12}-1$ contains the splitting field of $x^6+1$ ...
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20 views

If $S=\sum (\frac{n}{p})\zeta^n$ then how to prove that $S^2=(\frac{-1}{p})p$? [duplicate]

Here $\zeta$ is a primitive $n$-th root of unity and ($\frac{n}{p}$) denotes the Legendre symbol. Can someone please give a proof of this fact? I tried writing $S^2$ as the product of two sums $S=\sum ...