# All questions with tag [math: galois-theory]

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## Field Extension problem beyond $\mathbb C$

There are great deals of areas in between $\mathbb C$ and also Meromorphic Functions on $\mathbb C$. As an example set of "All Even Meromorphic Functions on $\mathbb C$" is a subfield in between $\mathbb C$ and also Meromorphic Functions on $\mathbb C$. Inquiry: How to classify such subfields? I have no suggestion whether someone res...
2022-07-25 17:47:10
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Let $n \geq 5$. I want to show that $\exists p\in \mathbb{Q}[X]$ of degree $n$ with roots which are not possible to find via radicals and rational functions from the coefficients of $p$. I've got the following theorem (Abel-Ruffini) written down in my notes. (1) $\exists$ transcendental $\alpha_1,\dots\alpha_n\in\mathbb{C}$ s.t. $F=\mathbb{Q}(\a... 2022-07-25 17:44:21 0 ## Solubility of a Galois Group going over some past papers with no answers and would like a bit of help if possible.. I've shown that for p a prime number then$x^p-1 \in K[x]$is abelian where K is a subfield of$\mathbb{C}$. I've now been asked to show that the Galois group of$x^p-a$over K is soluble with$a \neq 0 \in K$. I know any abelian group A is soluble, since${...
2022-07-25 17:43:41
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## Find all finite fields k whose subfields form a chain: if $k'$ and $k''$ are subfields of $k$, then either $k' \subseteq k''$ or $k'' \subseteq k'$.

Find all finite fields $k$ whose subfields form a chain: that is, if $k'$ and $k''$ are subfields of $k$, then either $k' \subseteq k''$ or $k'' \subseteq k'$. So, I understand that I'm trying to find the values of $n$ such that the subfields of $\mathbb{F}_{p^n}$ (where $\mathbb{F_{p^n}}$ is the Galois field of order $p^n$) form a chain. Howe...
2022-07-25 17:20:51
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## Two Equivalent Notions of Algebraic Simple Extension (Proof)

When reviewing messages concerning area expansions I've found the adhering to 2 interpretations for the straightforward expansion $K(\alpha)$ where $\alpha$ some algebraic number over $K$. They are (1) $K(\alpha)$ is the tiniest area having both $K$ and also $\alpha$ (2) $K(\alpha)=\{\sum_{0}^{\infty}k_i\alpha^i:k_i\in K\}$ I have not had t...
2022-07-25 16:43:36
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Let $L/K$ be a Galois extension of fields and let $\zeta_n$ be a primitive $n$th root of unity in some field extension of $L$, where $n$ is not divisible by the characteristic. Prove that $\mathrm{Aut}(L(\zeta_n)/K(\zeta_n))$ is a subgroup of $\mathrm{Aut}(L/K)$. I'm unsure because I don't really know what is required. If $\sigma \in \mathrm{Aut... 2022-07-25 16:40:55 1 ## Proof that a polynomial is irreducible in$\mathbb{Q}$Let$p$be a prime number, and let$m,k_1,\ldots,k_{p-2}$be even numbers. Define the polynomial$h(x)=(x^2+m)(x-k_1)\cdots(x-k_{p-2})$and$r=\min \{|h(a)|\mid a\in\mathbb{R},h'(a)=0\}$. Under these condition I've proven that$r&gt;0$. Now let$n$be an odd number large enough so that$\frac{2}{n}&lt;r$and let$f(x)=h(x)-\frac{2}{n}$, ... 2022-07-25 16:39:35 1 ## Why is this extension of Galois? Let$F$be a subextension of$\mathbb{C}$maximal with respect to not containing$\sqrt2$. Let$K/F$be a finite extension with$K\subset\mathbb{C}$. Then$K/F$is of Galois and$[K:F]$is a power of a prime. Could you help me solve this exercise? Following the idea of Olivier: Let$L$be the normal closure of$K/F$. If$K\neq F$, then$\sqrt2\i...
2022-07-25 13:30:47
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Let $K = \mathbb F_p(x)$, and let $H = \left\{\begin{pmatrix} d &amp; a \\ 0 &amp; 1 \end{pmatrix} \ \big| \ a \in \mathbb F_p, d \in \mathbb F_p^\times\right\}$ be a group under multiplication which acts on $K$ via $\begin{pmatrix} d &amp; a \\ 0 &amp; 1 \end{pmatrix} x = dx + a$. How can I find the fixed field of $H$? If $f =... 2022-07-25 13:28:44 2 ## Is the size of the Galois group always$n$factorial? I am researching field theory, and also I simply began the phase on Galois theory. Given that a Galois expansion is the splitting area of some polynomial$p(x)$and also this polynomial have specifically$n$origins in the expansion area I assume that every automorphism of the expansion permutes the origins of$p(x)$[and for every single permu... 2022-07-25 13:15:01 1 ## Why$H\neq N_G(H)$? Let$K$be a field,$f(x)$a separable irreducible polynomial in$K[x]$. Let$E$be the splitting field of$f(x)$over$K$. Let$\alpha,\beta$be distinct roots of$f(x)$. Suppose$K(\alpha)=K(\beta)$. Call$G=\mathrm{Gal}(E/K)$and$H=\mathrm{Gal}(E/K(\alpha))$. How can I prove that$H\neq N_G(H)$? My idea was to take a$\sigma: E\rightarrow\ba...
2022-07-25 12:52:23
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## Example of field $K$ with $\mathrm{char}(K) > 0$, such that $[K(\alpha):K] = [K(\beta):K]$ but $K(\alpha) \not \cong K(\beta)$

I'd like to fine an example of field $K$ and elements $\alpha, \beta$ such that $\mathrm{char}(K) = p&gt; 0$, $[K(\alpha):K] = [K(\beta):K]$ but $K(\alpha) \not \cong K(\beta)$. This obviously can't work if $K$ is a finite field. So I need to find a non-finite $K$. The only ones that pop into my head are $\mathbb F_p(t)$, $\mathbb F_p(t^p)$...
2022-07-25 12:45:29
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## Show the norm map is surjective

Let $K/F$ be an extension of finite field. Show that the norm map $N_{K/F}$ is surjective. Here is what I have so far: Since $F$ is a finite field and $K/F$ is a finite extension of degree $n$, so $\operatorname{Gal}(K/F)=\langle\sigma\rangle$, where $\sigma(a)= a^{q}$ with $q=p^{m}=|F|$. In addition, by primitive element theorem, $K=F(\alpha)$ ...
2022-07-25 12:39:59
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## Galois Group of $(x^3-5)(x^2-3)$

I am having some trouble calculating the Galois group (over $\mathbb{Q}$) of $(x^3-5)(x^2-3)$. I can see the splitting field is $F:=\mathbb{Q}(\sqrt[3]{5},\omega,\sqrt{3})=\mathbb{Q}(\sqrt[3]{5},i,\sqrt{3})$, where $\omega$ is a primitive 3rd root of unity, and it has degree 12 over $\mathbb{Q}$. Since the extension is Galois (it is a splitting ...
2022-07-25 12:37:37
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## What is a maximal abelian extension of a number field and what does its Galois group look like?

How does one know that a number field $K$ has a maximal abelian extension (unique up to isomorphism) $K^{\text{ab}}$? I've read proofs involving Zorn's lemma that it has an algebraic closure (And that algebraic closures are unique up to isomorphism.) $\bar{K}$ All these proofs involved ideals of the polynomial ring in variables $x_f$, $f$ an i...
2022-07-25 12:34:03
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In an assignment I got I have been asked to try to determine when $E/F$ is a Galois Extension, and determine the Galois group of such an extension. $\textbf{Context:}$ $F$ is any field of characteristic zero and $E = F(\sqrt{c},\sqrt{a + b\sqrt{c}})$, where $a,b,c \in F$, $c$ is not a square in $F$ and $a + b\sqrt{c}$ is not a square in $F(\sqr... 2022-07-25 07:54:03 1 ## Endomorphisms and Automorphisms Let$L$be a field extension of$K$. Consider the set$\operatorname{End}_KL$of all functions from$L$to$L$which are linear over$K$. A subset of$\operatorname{End}_KL$is$\Gamma(L:K)$, the group (under composition) of all automorphisms on$L$which fix$K$. So, we can consider$\operatorname{End}_KL$to be a vector space over$K$. We h... 2022-07-25 07:50:14 1 ## Order and generator of the Galois group of an extension of finite fields I'm searching for the order and also define a generator of the team $$\mathrm{Aut}_{\mathrm{GF}(2^3)}(\mathrm{GF}(2^{12}))$$ It is clear that the order is 4, yet just how would certainly you define the generator? Many thanks! 2022-07-25 07:49:09 1 ## An example that the Galois correspondence fails if the extensions is not Galois Let$F/K$be a finite extension of fields.$S$the set of subgroups of$\mathrm{Aut}_K(F)$and$I$the set of intermediate fields of the extension$F/K$. Define the function$\varphi:S\rightarrow I$as$\varphi(G)= F^G$where$F_G$is the subfield of$F$fixed by$G$. Could you help me to find an example of extension$F/K$where$\varphi$is not... 2022-07-24 06:41:19 1 ## When the group of automorphisms of an extension of fields acts transitively Let$F$be a field,$f(x)$a non-constant polynomial,$E$the splitting field of$f$over$F$,$G=\mathrm{Aut}_F\;E$. How can I prove that$G$acts transitively on the roots of$f$if and only if$f$is irreducible? (if we suppose that$f$doesn't have linear factor and has degree at least 2, then we can take 2 different roots that are not in$F...
2022-07-24 06:40:58