Questions tagged [galois-theory]

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 (galois-connections).

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Proof that any extension that is Galois in the classical sense is also Hopf Galois

The usual action of $G$ on $L$ is by automorphisms that fix $\cal{K}$. Explicitly, for any $g\in G, l,m\in L, k\in \cal{K}$ \begin{align*} g(l+m)&=g(l)+g(l)+g(m)\\ g(lm)&=g(l)g(n)\\ g(k)&=...
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31 views

Extensions of an algebraically closed field

First some definitions: A field $k$ is algebraically closed if every non-constant polynomial $f(x) \in k[x]$ has a zero in $k$.Then if I am adding another restriction that if the field is also perfect,...
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Understanding why some Galois extension over $\mathbb{Q}_2$ has Galois group $\operatorname{GL}_2(\mathbb{F}_3)$

Let $K = \mathbb{Q}_2$ and $d_2 = 3x^4 + 12 x^2 + 4x - 4 \in K[x]$. Let $L_0$ be the splitting field of $d_2$ and let $\alpha \in L_0$ be a root of $d_2$. Let $L/L_0$ be the quadratic extension ...
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$E$ is the splitting field over $F$ for some separable $f(X) \in F[X]$ implies that $f(X)$ is separable over $\Phi(\Gamma(E/F))$?

I'm going over the proof of Theorem 25.1 in Abstract Algebra, and I'm trying to understand a particular detail concerning separable polynomials. Here's the relevant portion of the theorem statement: ...
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Extension to an automorphism of topological fields

Here is what I am trying to prove: Let $L$ be an algebraically closed topological field of characteristic zero and $K$ be a proper algebraically closed and topologically closed subfield of $L$. Let $\...
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action of absolute Galois group to discrete top is continuous [duplicate]

Let $G$ be a absolute Galois group of finite field, and $E$ be discrete topological space. I heard that action $G×E→E$ is continuous. Could you tell me the direct proof of this statement? Reference ...
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Determining that a Galois group is not cyclic

I am given a list of criteria and I am supposed to reason that the Galois group cannot be cyclic. Here is the list: $f$ is a fourth degree polynomial in $\mathbb{Q}[x]$ $f$ is separable the ...
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Determine $Gal_{\mathbb{Q}}(f(x)g(x))$ where $f(x)=x^3-2$, $g(x)=x^2-2$.

Let $f(x)=x^3-2$, $g(x)=x^2-2$ and $h(x)=f(x)g(x)$ as polynomials over $\mathbb{Q}$. Note that $Gal_{\mathbb{Q}}(f(x))=D_3$ and $Gal_{\mathbb{Q}}(g(x))=\mathbb{Z}_2$. Determine $Gal_{\mathbb{Q}}(h)$. ...
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Galois group of the polynomial $x^n+x^{n-1} +⋯+x^2+x−1$

As we already know, the following polynomial is irreducible over $\mathbb Q[x]$: $$x^n + x^{n-1} + \cdots + x^2 + x - 1 = \frac{x^{n+1} - 2x + 1 }{ x-1}.$$ By Descartes' rule of signs, it has only 1 ...
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Prove this $\mathbb F_2 (a)$ is splitting field of $X^4+X+1$

I've been solving problems from my Galois Theory course, and I don't find the way to solve this one: Given the polynomial $f(X)=X^4+X+1\in\mathbb F_2[X]$, prove that $f(X)$ is irreducible in $\mathbb ...
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Extension of a field monomorphism to an automorphism .

$\mathbf {The \ Problem \ is}:$ Let $E$ be the splitting field of a polynomial $f$ over $k.$ Let $k \subset K \subset E,$ then show that any $k-$monomorphism $\phi$ from $K \to E$ can be extended to a ...
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55 views

E/F is Galois ext with $[E:F]=p^2$ and its Galois group is not cyclic, then $\exists K$ a proper subfield of $E$.

Let $E$ be a Galois extension of a field $F$. Suppose that the Galois group $Gal(E/F)$ is an abelian group of order $p^2$ which is not cyclic. Then I want to show $\exists K$ a proper subfield of $E$...
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Cyclic Galois extension $F$ with $[F:\mathbb{Q}]=6$ implies $F$ is not a splitting field of a poly of degree $3$ over $\mathbb{Q}$.

Let $F$ be a cyclic Galois extension of $\mathbb{Q}$, the field of rational numbers, such that $[F : \mathbb{Q}] = 6$. Show that $F$ is not a splitting field of a polynomial of degree $3$ in $\mathbb{...
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$X^8-2$ is irreducible over $\mathbb{Q}(i)$ because $\left[\mathbb{Q}(2^{1/8},i):\mathbb{Q}(i) \right]=8$ , what property is used?

$X^8-2$ is irreducible over $\mathbb{Q}(i)$ because $\left[\mathbb{Q}(2^{1/8},i):\mathbb{Q}(i) \right]=8$ It's an assertion i find in an exercice , i don't understand what property is used (the ...
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Algebraic Independence of Automorphisms

Suppose $K$ is a infinite field. Let $\phi_1,\phi_2 \dots \phi_n$ be a finite set of automorphisms of $K$. Then show that they are algebraically independent. I tried this by assuming a polynomial $f(...
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Proving the irreducibility of a polynomial based on its Galois group

Suppose $f(X) \in \mathbb{Q}[X]$ is a polynomial of degree $n$. Let $K$ be the splitting field of $f(X)$ over $\mathbb{Q}$. Prove that if $Gal(K/\mathbb{Q}) \simeq S_n $, then $f(X)$ is irreducible ...
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Degree of the extension $\mathbb{Q}(\sqrt[5]{7}+\sqrt[5]{49})$

The original question is find the degree of the irreducible polynomial of $3+\sqrt[5]{7}+\sqrt[5]{49}$, but it's equivalent to find $[\mathbb{Q}(3+\sqrt[5]{7}+\sqrt[5]{49}):\mathbb{Q}]$. $\mathbb{Q}(3+...
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finding all automorphism of $\mathbb{Q}(\sqrt[4]{2})/\mathbb{Q}(\sqrt{2})$

If we want to find all the automorphisms of $\mathbb{Q}(\sqrt[4]{2})/\mathbb{Q}(\sqrt{2})$, The generator is $\sqrt[4]{2}$, We look at the minimal polynomial, in this case its $x^2 - \sqrt{2}$ (I'm ...
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Understanding the definition of normal field extension

We let $E/F$ be an algebraic extension of a field $F$, and define $S$ to be the set of all $F$-conjugates of the elements of $E$. Now we define $F(S)$ the be the field generated by $S$ over $F$. rove: ...
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28 views

Automorphism group of compositum of fields

$\newcommand{\Gal}{\operatorname{Gal}} \newcommand{\Aut}{\operatorname{Aut}}$ Suppose we have field extensions $F < K_1, K_2 < K$. It is known that if $K_1/F$ and $K_2/F$ are both Galois and $...
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A Field Contained in All Fields

I've been thinking about subfield lattices. Textbook always ask about a extension $F/\mathbb{Q}$ and ask to draw the lattice and ask questions based on that. But what if we go the other way? Say $F/\...
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1answer
31 views

What does it mean when one talks about splitting field of a multivariable polynomial? And then, Galois group of that splitting field?

I came across the following, while I was reading a recent research article. I do not know how to interpret it; the article does not define this concept (perhaps because it is too elementary). Could ...
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1answer
44 views

Criterion for finite Galois extension

I am trying to relax the condition for checking whether a extension is Galois. I got from a textbook that: $E/F$ is a finite Galois Extension iff $E$ is a spitting field of a separable irreducible ...
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Solvable elements of a field extension

Suppose $K$ over $F$ is a field extension, and $\alpha \in K$. My instructor says that "$\alpha$ is solvable over $F$ if there exists a radical extension $L$ of $F$ containing $\alpha$". My ...
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51 views

Show that the following polynomial in $\mathbb{Q}[t]$ is irreducible and not solvable over $\mathbb{Q}$

The polynomial is $f(t)=t^5-4t+2$. I can show it's irreducible with Eisenstein's Criterion, and I know I need to show that it has exactly two complex (non-real) roots to prove that it's not solvable, ...
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A question related to Galois groups of composite extensions [duplicate]

I'm trying to make sense of a proof of the following claim: Suppose $K$ is Galois over $F$ and $G=Gal(K/F)=G_1 \times G_2$ is the direct product of two subgroups $G_1$ and $G_2$. Then, $K=K_1K_2$ and ...
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A question related to intermediate fields

Consider the polynomial $f(X)=X^{4}+aX^{2}+b \in F[X]$ such that $f$ is irreducible over $F$. Suppose $\alpha$ is a root of $f(X)$ and let $K=F(\alpha)$. Prove that there exists a subfield $L$ of $K$ ...
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How can I prove that $\sqrt{2}$ is not inside Q($\sqrt{5 + \sqrt{5}}$)?

I want to prove it, but although it is quite intuitive I don't know how to prove it mathematically. I have tried using the tower rule but I haven't got anything crear. Thanks in advance.
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50 views

Finding Galois extension whose Galois group is Sn

I want to prove that given any integer n, we can find a finite Galois extension K over $\mathbb Q$ such Gal$({K}: {\mathbb Q })$ = $S_n$ For prime p, I know finding a polynomial with exactly 2 nonreal ...
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A question related to abelian extensions

Let $K/F$ be an abelian extension (a Galois extension $K/F$ such that $Gal(K/F)$ is abelian) of degree at least $2$. Prove that there exists a tower of fields $F=K_{0} \subset K_1 \subset \cdots \...
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1answer
86 views

Existence of a Field not containing $\sqrt{2}$, such that any finite extension is cyclic.

I want to show the existence of a field $E$ not containing $\sqrt{2}$, such that any finite extension of $E$ in $\overline{\mathbb{Q}}$ is cyclic. I think the maximal field not containing $\sqrt{2}$ ...
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An isomorphism related to the class group

Suppose that $L/K$ is a Galois extension with abelian Galois group. Let $\phi : Cl(\mathcal{O}_K) \rightarrow Cl(\mathcal{O}_L)$ be the morphism of class groups given by $\phi(I) = I\mathcal{O}_L.$ My ...
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1answer
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What does 'topologically' generated mean?

I heard ablolute galois group of finite field is 'topologically' generated by frobenius map. I understand this sentence except for 'topologically', but what does 'topologically' exactly mean here? ...
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Suppose $K$ is a Galois extension of $F$. Consider $E/F$ a finite extension such that $K\cap E=F$. Show that $[KE:K]=[E:F]$.

I found the following problem: Suppose $K$ is a Galois extension of $F$. Consider $E/F$ a finite extension such that $K\cap E=F$. Show that $[KE:K]=[E:F]$. Can someone give me a hint? I remember ...
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Algebraic Closures are Splitting Fields [duplicate]

Assuming that $\overline{K}$ is an algebraic closure of the field $K$, how would one go about proving that $\overline{K}$ must be a splitting field? I am trying to do this by finding a set $\mathcal{P}...
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1answer
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Quadratic subfields of the cyclotomic field $\mathbb{Q}(\zeta_{14})$

In a nutshell, my question is: what is degree of the field extension $\mathbb{Q} \, ( \zeta_{14} + \zeta_{14}^9 + \zeta_{14}^{11}) $ over $\mathbb{Q}$? As to why I'm asking this, I was trying to ...
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Finding the number of automorphisms in $G_{\Bbb{Q}(\sqrt[4]{2},i),\Bbb{Q}}$?

I am trying to solve the following problem: Given $K$, the splitting field of $x^4-2$ over $\Bbb{Q}$, consider $\sigma,\tau\in G_{K / \Bbb{Q}}$ such that $$\sigma(\sqrt[4]{2})=\sqrt[4]{2}i\hspace{2cm}...
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Why are these extension fields equal? [duplicate]

I have $\alpha= 2^{1/3}+\sqrt{2}$ and $E(\alpha)=\mathbb{Q}$, I want to prove that $E=\mathbb{Q}(2^{1/3}, \sqrt{2})$. I tried to compute the irreducible polynomial of $\alpha$, but I didn't get ...
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Classical Galois Theory and Groupoids?

Given your favorite field $k$, we can build its groupoid of algebraic closures in the obvious way: Look at the category of all possible algebraic closures with field isomorphisms as the arrows. This ...
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2answers
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If all roots of $f$ generate a splitting field, is $f$ irreducible?

Recently I had to prove the existence of some irreducible polynomial. I wanted to use the following statement, but I do not know if it is true: The Statement: Let $F$ be a field. If $f\in F[X]$ is ...
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1answer
35 views

an example of a field $k$ of characteristic $p$, and an irreducible, inseparable polynomial in $k[x]$

As stated, what would be an example to satisfy the above criterion?, what would be an example of a field $k$ of characteristic $p$, and an irreducible, inseparable polynomial in $k[x]$
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1answer
47 views

Local polynomial of Galois representation restricted to subextension

Let $K$ be an extension of $\mathbb{Q}_p$. Consider an Galois representation $\rho: G_K \to GL_2(\mathbb{C})$ and for any finite extension $M/K$ we call $$P(\rho|_M,T) = \det(1-\operatorname{Frob}_{M}^...
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Theorem related to Trace map of a field extension

Let $ E/F$ be a finite separable extension. Then the map $ T_x:E \rightarrow F$ given by $ T_x(y)=Tr (xy)$ is an $ F$-linear map.Show that the map $ \psi: E \rightarrow \operatorname{Hom}(E,F)$ given ...
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1answer
49 views

$\mathbb{Q}(\sqrt[4]{2}(1+i))/\mathbb{Q}$ isn't a normal extension, tower stuff…

I'm trying to prove that $ \mathbb {Q} (\sqrt[4]{2}(1 + i)) /\mathbb {Q} $ is not normal. My attempt is: Let $\alpha = \sqrt[4]{2}(1 + i) $. The polynomial \begin{equation} p (x) = x ^ 4 + 8 = (x- \...
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1answer
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Reconstructing a polynomial's Galois group from those of its subfield factors

Having worked out that the Galois group of $$p(a)=a^{4} \left(16 t^{4} + 16 t^{2}\right) - 64 a^{3} t^{3} + a^{2} \left(- 8 t^{4} + 24 t^{2}\right) + 16 a t + t^{4} + 5 t^{2} - 4$$ over $\mathbb Q[t=\...
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Why is the extension $L/F$ in the proof is normal$?$

In proof of part $(1)$ of this preposition, $L/F$ is a galois extension. However, I didn't understand why. It is a separable extension. But how can one prove the normality$?$ Here, $E/F$ is a finite ...
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Is $x$ always a primitive element of $\text{GF}(2^m)$?

I checked using MATLAB that $x$ is a primitive element of $\text{GF}(2^m)$ for $m\le 16$. Is the statement true for $m > 16$? EDIT: In this question, we represent an element of $\text{GF}(2^m)$ as ...
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39 views

Elements of Galois group map roots of minimal polynomial to which other roots?

If we have $F \leq E \leq \mathbb{C}$, I know that for any $\alpha \in E$ and $\sigma \in \operatorname{Gal}(E/F)$, we have that $\sigma$ maps roots of $m_{\alpha,F}(x)$ to roots of $m_{\alpha,F}(x)$. ...
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1answer
69 views

How to compute the Galois Group of $(x^3-3)(x^4-2)$ over $\mathbb Q$

So I know the splitting field is $\mathbb Q(\sqrt[4]2, i ,\sqrt[3]3,\zeta_3 )$ where $\zeta_3=-\frac12+\frac{\sqrt3}{2}i$ is the root of unity. And the order of the Galois Group equals the degree of ...
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1answer
101 views

Galois group of $\mathbb{Q}_p(\zeta_4,\sqrt[4]{p})/\mathbb{Q}_p$ for $p-1$ not divisible by $4$

Let $p > 2$ be a prime number such that $p-1$ is not divisible by $4$. Consider $K = \mathbb{Q}_p$ and $L = \mathbb{Q}(\sqrt[4]{p},\zeta_4)$ where $\zeta_4 \in L$ is a primitive forth root of unity ...

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