0
votes
2answers
33 views

every irreducible polynomial has a root in some field extension

We know the following fact from field theory. Let $F$ be a field and $p(X)$ an irreducible polynomial in $F[X]$. Then we can find a field extension $L$ of $F$ such that $p(X)$ has a root in $L$. ...
0
votes
2answers
47 views

dimension of $\mathbb{Q}(\sqrt2)$

How can I prove that the splitting Field $\mathbb{Q}(\sqrt2)$ over the rational numbers $\mathbb{Q}$ is two dimensional vector space over $\mathbb{Q}$ ?
4
votes
3answers
58 views

$[F(t):F(t^n)]=n$ where $t$ is trascendental

Let $F$ be a field and let $t$ be trascendental over $F$. Prove that $[F(t):F(t^n)]=n$. Obviously $[F(t):F(t^n)]\le n$ since the polynomial $f(x)=x^n-t^n \in F(t^n)$ has $t$ as a root. But I don't ...
1
vote
1answer
44 views

Calculating Splitting field

Find the splitting field of the polynomial and degree over $\mathbb{Q}$ $P(X)=X^4+2$. The roots of $P(X)$ are $\sqrt[4]{2}\sqrt{i},\ -\sqrt{i}\sqrt[4]{2}, \ i\sqrt{i}\sqrt[4]{2},\ ...
2
votes
1answer
30 views

Galois group of order 2^4

Find the galois group the polynomial $f(X)=(X^2-2)(X^2-3)(X^2-5)(X^2-7)$ over $\mathbb{Q}$. A splitting field for $f(X)$ is $K=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7})$. We must have ...
0
votes
2answers
27 views

Separable extension [closed]

Let $\alpha$ algebraic over $k$ of characteristic $p>0$ Prove that $\alpha$ is separable over $k$ if and only if $k(\alpha)=k(\alpha^p)$. Any suggestion, please.
6
votes
4answers
87 views

Galois group of $X^{5}-2X+7$ over $\mathbb{Q}$

Is there any way to determine the Galois group of $X^{5}-2X+7$ over $\mathbb{Q}$ not using the discriminant? Thanks!
4
votes
1answer
39 views

Finding a cubic polynomial whose splitting field over $\mathbb{Q}$ equals $\mathbb{Q}(a)$ if $a$ is any of its roots

Question: Let $\alpha$, $\beta$ and $\gamma$ be the roots of a rational cubic polynomial $q$. Can we find a (non-trivial) example where the splitting field of $q$ over $\mathbb{Q}$ equals ...
1
vote
0answers
24 views

determine the degree of an extension, check normality

I came across an old exam problem and I wonder if my solution is correct. Let $L=\mathbb{Q}(\omega)$, where $\omega=e^{\frac{2\pi i}{6}}$ is a primitive sixth root of unity: a) determine ...
0
votes
0answers
39 views

Why do nth roots (radicals) have closed forms whilst other polynomial roots do not?

The nth root function - $\sqrt[n]{a_{0}}$ - may be seen as an arithmetic operation (the inverse of the $pow(x, n)$ function) but it can also be interpreted as computing the roots of a specific class ...
4
votes
2answers
45 views

Non-isomorphic field extensions of $\mathbb{Q}$

I'm having a little bit of a problem with the following question: Show that there do not exist two irreducible polynomials $a(x)$ and $b(x)$ in $\mathbb{Q}[x]$ of degrees 6 and 7 respectively ...
2
votes
2answers
33 views

Transitivity Property of Separable Extensions

I was looking for some proof for the transitivity property of separable field extensions. Although this might sound like a very well-known fact and is referred to frequently, I do not seem to find a ...
2
votes
3answers
141 views

Minimal polynomial for $\zeta+\zeta^5$ for a primitive seventh root of unity $\zeta$

Minimal polynomial for $\zeta+\zeta^5$ for a primitive seventh root of unity $\zeta$ I have asked a similar problem Minimal Polynomial of $\zeta+\zeta^{-1}$ and i tried to repeat similar idea ...
2
votes
1answer
41 views

Places of this extension

I'm reading this book. I'm trying to find the degree of the places of the extension $\mathbb C(X)\mid\mathbb R$. I know the places of the extension $\mathbb R(X)\mid\mathbb R$ and I've already ...
2
votes
1answer
33 views

Isomorphic algebraic closures.

I've just stared learning the Galois Theory so my question might be trivial, but could someone give me an example of two different algebraic closures of the same field? Cause I don't get how they can ...
2
votes
1answer
24 views

Sources about transcendence degree

I asked this question: Characterization of the transcendentals over a field I realized I need some knowledge about transcendence degree to prove some facts in the book I'm reading. I would like to ...
1
vote
3answers
44 views

Extensions of degree $1$.

My doubt is very simple: Let $F|K$ be a field extension, if $[F:K]=1$, what can we say about $F$ and $K$? can I say $F=K$? I'm trying to prove the equality without success. Thanks in advance
4
votes
1answer
68 views

Milne's Galois Theory Example

The following example is drawn from Milne's Galois Theory notes, p.42 (http://www.jmilne.org/math/CourseNotes/FT.pdf) We study the extension $\mathbb{Q}[\zeta]/\mathbb{Q}$ where $\zeta=e^{2\pi i/7}.$ ...
1
vote
1answer
26 views

Irreducibility in Galois/non Galois Extensions

Let $k$ be a field and $\alpha$ algebraic over $k$. Let $K$ be the Galois closure of $k(\alpha)$ (obtained by adding all conjugates of $\alpha$). If $f(x) \in k[x]$ is irreducible over $k[\alpha]$ is ...
2
votes
1answer
53 views

A Fixed Field of $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$

I was working on a review problem from Dummit and Foote and came across the following issue. It is clear that the Galois group of the splitting field for the polynomial $(x^2-2)(x^2-3)(x^2-5)$ has ...
0
votes
0answers
24 views

Galois group of a splitting field

$f=x^4-5x^2+6 \in \operatorname Q[x]$ $f=(x^2-2)(x^2-3)$ $\operatorname F=\operatorname Q(\sqrt[2]{2},\sqrt[2]{3})$ $f$ is irreducible for Eisenstein's criterion, $\operatorname{char} \operatorname ...
1
vote
1answer
34 views

Galois group of the splitting field of $ x^6 - 5$

$f = x^6 - 5$ $\in \operatorname Q [x]$ I want to find a splitting field $\operatorname F$ of $f$ over $ \operatorname Q$ $\sqrt[6]{5}$ is a real root of $f$ $u$ is the 6-th root of unity then $ ...
1
vote
2answers
38 views

$K \le B\le F$, If $F$ Galois over $K$ $\Rightarrow$ $F$ a Galois over $B$

Let $F$ a Galois over $K$, and let $B$ be a subfield of $F$ such that : $K \le B\le F$ $\Rightarrow$ $F$ a Galois over $B$ PROOF: $F$ is a splitting field of $f \in K[x]$ separable over $K$. ...
1
vote
1answer
70 views

Solvability of polynomials over fields of characteristic zero

1) Let $K$ be a field, $\operatorname{char}(K)= 0$, and $f ∈ K [x]$ with $\deg(f)\le4$. Then $f$ is solvable by radicals. Proof: $\operatorname{Gal} (F/K) \cong S_4$ then $\operatorname{Gal}(F/K)$ ...
1
vote
1answer
54 views

How to show this is the minimal polynomial

I'm trying to the following problem. But I can't show some irreducibility of the polynomials. Put $L=\mathbb{C}(X,Y,Z)$, $\omega=\frac{-1+\sqrt{-3}}{2}$. Define two automorphism $\sigma, \tau$ of ...
2
votes
1answer
95 views

Radical Solvable quintic polynomial.

I am trying to solve the following exercise: Let $k$ be a field of characteristic 0, and let $f(x)\in k[x]$ be a polynomial of degree 5 with splitting field $E/k$. Prove that $f(x)$ is ...
4
votes
1answer
85 views

What is the difference between field theory and Galois theory

I am about to finish the book Galois theory by Harold Edwards. I am planning to study Galois theory at a more advanced level or field theory. I am unable to decide because I don't know the difference ...
2
votes
1answer
46 views

$(u+1)(u+2)\cdots(u+p)=u^p-u$

Let $E$ be a field with characteristic $p>0$. How should I prove that $$(u+1)(u+2)\cdots(u+p)=u^p-u$$ I can verify this equality for small $p=2,3,5$. Is there a way to prove this result in general? ...
3
votes
0answers
39 views

Proving that a certain maximal subfield has countably infinite index

Let $K$ be a maximal subfield of $\mathbb C$ which does not contain $\sqrt{2}$. I've shown that $\mathbb C$ is an algebraic extension of $K$, and that the Galois group of any finite extension of $K$ ...
2
votes
1answer
29 views

If $L_1/K$ and $L_2/K$ are not Galois (solvable), then $L_1L_2/K$ is not Galois (solvable)

This is part of an exam preparation: Prove/contradict: If $L_1/K$ and $L_2/K$ are not Galois, then $L_1L_2/K$ is not Galois. If $L_1/K$ and $L_2/K$ are not solvable Galois extensions, ...
1
vote
1answer
28 views

Normalizer of a subgroup of a Galois group

I wanted to check whether my solution for this problem was correct. Let $k \subseteq L \subseteq K$ be a finite extension of fields, with $K/k$ Galois $H$ the normalizer of $Aut(K/L)$ in $Aut(K/k)$. ...
0
votes
2answers
58 views

Find the galois group of the polynomial when a root is given

If $\alpha$ is a root of a polynomial $f(x)=x^3 +x^2-4x+1$ then show that $2 - 2\alpha - \alpha^2$ is also a root of $f(x)$. Use this fact to compute the Galois group of the splitting field of $f(x)$ ...
4
votes
2answers
153 views

Is the order of $\operatorname{Gal}(\bar{\Bbb Q}/\Bbb Q)$ known?

Is the order of $\operatorname{Gal}(\bar{\Bbb Q}/\Bbb Q)$ known? And if so is there a description on how the order can be found? My initial thoughts is that because $\bar{\Bbb Q}$ is countable and ...
2
votes
0answers
35 views

Characterizing quadratic number fields that are subfields of cyclic quartic number fields [duplicate]

Given a quadratic number field $F = \mathbb{Q}(\sqrt{d})$, is there a way to determine whether or not $F \subset K$ for some quartic numberfield $K$ with $\operatorname{Gal}(K/\mathbb{Q}) \cong ...
1
vote
0answers
33 views

Splitting field and intermediate fields

Find splitting field $K$ of polynomial $x^3-7$ over $\mathbb{Q} (\sqrt{3} )$ and find $(K: \mathbb{Q} )$ $G(K/ \mathbb{Q} )$ Find intermediate fields between $\mathbb{Q}$ and $K$. So the ...
1
vote
1answer
51 views

Show that a sequence of fields exists

I do not have a clue how to solve the following problem: Let $K\subseteq L$ be Galois extension of degree $p^n$, where $p$ is prime and $n$ is natural. Show that there exists a sequence of subfields ...
1
vote
0answers
35 views

Galois extension of two fields

I am preparing myself to the exam in algebra and I have found the following exercise, which is quite hard for me. Do you have any suggestions for solving it? For fields $K\subseteq L,M\subseteq \bar ...
-1
votes
1answer
37 views

Prove that $u$ is algebraic over $\mathbb{Q}$ [closed]

Let $u$ be a root of the following equation: $$x^{3}+{\displaystyle \dfrac{1+i}{\sqrt{2}}x^{2}+\dfrac{-1+i\sqrt{3}}{2}x+1=0}$$ Prove that $u$ is algebraic over $\mathbb{Q}$ . Thanks in ...
2
votes
1answer
61 views

If $|\operatorname{Aut}_KF|=3$, must we have cube roots of unity?

Let $K$ be a field of zero characteristic. Let $F$ be a finite dimensional extension field of $K$ such that $|\operatorname{Aut}_K F|=3$. Must the equation $x^2+x+1=0$ have a root in $F$ ? Thank you
6
votes
2answers
64 views

Does the data of Galois group, ramified places, and inertia groups, determine a Galois number field?

Suppose I tell you that $K/\mathbb{Q}$ is a finite Galois extension, and I specify the Galois group $G$, and suppose further that I give you a finite list $S$ of places of $\mathbb{Q}$ and for each ...
2
votes
2answers
30 views

Splitting field of $f$ as smallest field extension containing all BUT ONE zero of $f$

I'm just working with splitting fields and I have to prove something which I don't understand. Let $L$ be a splitting field of the polynomial $f$ over $K$ and $f = \prod_{i=1}^n(X-\alpha_i)$. ...
2
votes
1answer
36 views

Separability of a polynomial

I have a non zero polynomial $f\in F[X]$ where $F$ is a field. Let $L$ be a field extension of $F$ so that $f$ splits completely in $L[X]$, so $f(X)=c\prod_{i=1}^n (X-a_i)$ with $c,a_i\in L$. If ...
1
vote
0answers
24 views

Did I Do This Galois Theory Problem Right? Subfields of $\mathbb{Q}(\zeta_{12})$.

Let $\omega$ be a primitive $12$th root of unity. (i) What is $[ \mathbb{Q}(\omega) : \mathbb{Q}]?$ (ii) List the distinct conjugates of $\omega + \omega^{-1}$. (iii) What is $Aut(\mathbb{Q}(\omega ...
1
vote
2answers
42 views

Is my answer correct on this Galois Theory problem? Find the lattice of subfields of $\mathbb{Q}(\zeta_9)$

Problem: let $\zeta$ be a primitive $9$th root of unity, and $K = \mathbb{Q}(\zeta)$. Describe the lattice of subfields of $K$, give generators for each subfield and list its degree over ...
0
votes
0answers
31 views

Splitting fields and isomorphic

Please check these statements whether those are true Let $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ be extension fields of $\mathbb{Q}$, the rational numbers. Since they are not isomorphic as ...
4
votes
1answer
79 views

Find all subfields in extension $\mathbb{Q}\subseteq \mathbb{Q}(\sqrt[4]{2})$

I want to find all intermediate subfields of extension $\mathbb{Q}\subseteq \mathbb{Q}(\sqrt[4]{2})$. I guess that $\mathbb{Q}(\sqrt[4]{2})$ is not a splitting field, since we would have polynomial ...
4
votes
1answer
49 views

A Galois Extension over a field of characteristic p

Hi! I've been having a bit of trouble with this question, namely the very last part. It's from a past paper for a Galois Theory course, in which I have an exam on Monday. Basically, I can show every ...
3
votes
1answer
64 views

Help with computing Galois group of $x^4 - 3$.

Let $f(x) = x^4 - 3$. I believe $Gal(f(x)) = Gal(\mathbb{Q}(\sqrt[4]{3}, i)/\mathbb{Q})$, and then we have $$\sigma_1 = \begin{cases} \sqrt[4]{3} \rightarrow \zeta^n\sqrt[4]{3}, ...
0
votes
1answer
61 views

Why is $x^2-2ux+1$, where $u = \cos(\frac{2\pi}{n})$, irreducible in $\mathbb Q(u)$?

My textbook states that $x^2-2ux+1$, where $u = \cos(\frac{2\pi}{n})$ for some $n \in \mathbb N$ is clearly irreducible in $\mathbb Q(u)$. Is this obvious? I tried to write it as a product of ...
2
votes
1answer
43 views

Fixed Field of automorphisms (of $k(x)$ with $k$ a field) Induced by $I(x)=x$, $\varphi_1(x) = \frac{1}{1-x}$, $\varphi_2 (x)=\frac{x-1}{x}$?

Since $I(x)=x$, $\varphi_1(x)=\frac{1}{1-x}$, $\varphi_2 (x)=\frac{x-1}{x}$ form a group of order 3 the group is cyclic so it is generated by $\varphi_1$ then I have to find the fixed field of ...