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Questions tagged [field-theory]

Use this tag for questions about fields and field theory in abstract algebra. A field is, roughly speaking, an algebraic structure in which addition, subtraction, multiplication, and division of elements are well-defined. This tag is NOT APPROPRIATE for questions about the fields you encounter in multivariable calculus or physics. Use (vector-fields) for questions on that theme instead.

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Subfields of splitting field of $x^4+25$ over $ℚ$.

Let $F$ be the splitting field of the polynomial $x^4+25$ over $ℚ$. List all subfields in $F$ and the corresponding subgroups in the Galois group. is problem $1$ on this pdf. The solution is: As we ...
hbghlyj's user avatar
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If $\alpha$ is algebraic over $F$, then it is algebraic over any extension field $L$ of $F$.

The photo attached is from the book Abstract Algebra by Dummit&Foote. I didn’t understand it at all. How do we know that $L$ contains the element $\alpha$. In other words, what does $\alpha $ have ...
boyler's user avatar
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1 answer
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Proof that the finite-dimensional extension $K[a]$ coincides with the decomposition field $K(a)$

Let $K\subseteq F$ be an extended field and $a\in F$. Let $K[a]=\left \{ f(a)|f\in K[x] \right \}$. Prove that if $K[a]$ is finite-dimensional as a vector space over $K$, then $K[a]=K(a)$ Consider ...
Dmitry's user avatar
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2 votes
1 answer
73 views

If $F(\alpha) = F(\alpha^2)$, then $\alpha$ is algebraic over $F$

Let $K$ be an extension of a field $F$ and let $\alpha \in K$. If $F(\alpha) = F(\alpha^2)$, then $\alpha$ is algebraic over $F$. I have an idea about how to do this that doesn't trivialize the ...
Grigor Hakobyan's user avatar
0 votes
0 answers
20 views

Proof of equivalent definitions of normal extensions

Let $F \supset k$ be an algebraic extension. We have these two characterizations of a normal extension: a) For every $k$-homomorphism $\sigma: F \to \overline{k}$, we have $\sigma(F) \subset F$ b) If ...
lkksn's user avatar
  • 55
0 votes
1 answer
23 views

Splitting fields and isomorphisms

If $K \subseteq L$, $K \subseteq L'$ are field extensions, $L \cong L'$ and $L$ is the splitting field of $f \in K[x]$, is $L'$ also a splitting field of $f$ ? I think $L'$ is a splitting field $\iff$ ...
ed268's user avatar
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Reducibility of $x^2-7$ over $\mathbb{Q}(\sqrt[5]{3})$

Suppose for a contradiction that $x^2-7$ is reducible over $\mathbb{Q}(\sqrt[5]{3})$. Then $\sqrt{7}\in\mathbb{Q}(\sqrt[5]{3})$. It follows that $\mathbb{Q}\subset\mathbb{Q}(\sqrt{7})\subset\mathbb{Q}(...
spinosarus123's user avatar
1 vote
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39 views

Question in an example of prime decomposition in an extension of DVRs.

Choose any element $\alpha \in \mathbb{Q}[[x]]$ such that it is transcendental over $\mathbb{Q}(x)$(This is possible by comparing cardinalities.), and let $\beta=\alpha^2$. Let $K=\mathbb{Q}(x,\beta)$,...
clgdj's user avatar
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How to prove that two algebraically closed field with same characteristic and cardinal are isomorphic? [duplicate]

Hi I would like to prove that for algebraic closed field $F_1,F_2$ if: They have the same characteristic $p$ They have the same cardinal and is uncountable Then they are isomorphic. I have proved ...
Shore's user avatar
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How to prove expended fields whose base has same cardinal are isomorphic?

I would like to prove that for field $F_1,F_2$ if they : have isomorphic sub-field $F'$ $F_1,F_2$ is the vector space over $F'$, and their bases, doneta $B_1,B_2$, has the same cardinal Then $F_1,...
Shore's user avatar
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2 votes
1 answer
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Find a counterexample: for $a,b \in K \setminus\{0\}$, equation $ax^2 + b y^2 = 1$ has a solution iff $ax^2 + by^2 = z^2$ has a nontrivial solution

It is a well-known fact that if $\text{char}(K) \neq 2$ and $a,b \in K \setminus\{0\}$, then the equation $ax^2 + by^2 = 1$ has a solution over $K$ if and only if the homogeneous equation $ax^2 + by^2 ...
Neckverse Herdman's user avatar
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0 answers
69 views

Why is $\mathbb{Z}_2[i]$ not a field even though $i$ is algebraic over $\mathbb{Z}_2$

I know that it has zero divisors $1+i$ such that $(1+i)(1+i)$ $=$ $0$ Hence, it’s not an Integral Domain and thus not a Field. But what I cant understand is that $\mathbb{Z}_2 [i]$ is also adjunction ...
Maths wizard's user avatar
1 vote
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Order of $\mathbb F _p [x] / (f)$.

I could use some help with the following exercise: Find the number of reducible monic polynomials of degree $2$ over $\mathbb F_p$. Show this implies that for every prime $p$ there exists a field of ...
RatherAmusing's user avatar
2 votes
1 answer
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Artin-Schreier extension is cyclic of degree 1 or $p$

Let $K$ be a field of characteristic $p > 0$ and $K \subset L$ the extension obtained by adjoining the zeros of the Artin–Schreier polynomial $f = x^p − x − a \in K[x]$, where $a\in K^*$, to $K$. ...
math_physics's user avatar
1 vote
1 answer
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Sub division rings of dimension 2 of division rings

Suppose $A$ is a division ring and $B$ is a sub division ring such that $A$, as a left vector space over $B$, has dimension $2$. Is it true that $B$ must be commutative ?
Boccherini's user avatar
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1 answer
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Why is the subfield of $\mathbb{Q}(\zeta_p)$ of index $2$ expressible in terms of the sum of $\zeta_p$ to the power of all quadratic residues mod $p$?

Let $p$ be an odd prime, $\zeta_p = e^{2 \pi i / p}$. I've been playing around with calculating the intermediate fields of $\mathbb{Q}(\zeta_p)$. I know that $\mathrm{Gal}(\mathbb{Q}(\zeta_p) / \...
Robin's user avatar
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1 vote
1 answer
38 views

Clarification on the proof that $[\mathbb F (\alpha) : \mathbb F] = \deg_{\mathbb F} \alpha$

Let $\mathbb E / \mathbb F$ be an extension of fields, $\alpha \in \mathbb E$ algebraic over $\mathbb F$. Then $[\mathbb F (\alpha) : \mathbb F] = \deg_{\mathbb F} \alpha$. I could use some help ...
RatherAmusing's user avatar
-2 votes
0 answers
40 views

Is $\mathbb{R}$-{transcendental numbers} still a field or not? [duplicate]

Is $\mathbb{R}$-{transcendental numbers} still a field or not? i.e, can we prove finite many terms of + and $\times$ using non-transcendental numbers is still non-transcendental numbers or not.(non-...
poker resources's user avatar
6 votes
1 answer
85 views

Can the composite of $\mathbb Q_p$ and $\mathbb Q_\ell$ in $\mathbb C$ be $\mathbb C$?

I will motivate my question with two observations. Consider the standard embedding $\mathbb R\to\mathbb C$. Then, for any embedding $\mathbb Q_p\to \mathbb C$, we must have that the composite of $\...
Andrea B.'s user avatar
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3 votes
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30 views

Is a sphere, whose volume is a sum of volumes of two given constructible spheres, constructible?

If I have two spheres with radius $r_1$and $r_2$ a sphere with the volume as the sum of volumes of these two spheres will have radius $(r_{1}^3 +r_{2}^3)^{1/3}$ now given that $[\mathbb{Q}(r_1):\...
Anurenj ER's user avatar
1 vote
0 answers
42 views

Dimension of division ring over a sub division ring

Let $L$ be a division ring ("skew field") and $K$ a sub division ring. Now suppose that $L$, as a left vector space over $K$, has finite dimension $m$. Does $L$, as a right vector space over ...
Boccherini's user avatar
3 votes
3 answers
77 views

Let $L = \mathbb Q(\sqrt{2+\sqrt{3}})$ What is its degree over $\mathbb Q$

Let $L = \mathbb Q(\sqrt{2+\sqrt{3}})$ What is its degree over $\mathbb Q$. So what I did is this, let $a = \sqrt{2+\sqrt{3}}$. consider $x = \sqrt{2+\sqrt{3}}$ then $x^2 = 2 + \sqrt{3}$ then $x^2 -2 \...
Donlans Donlans's user avatar
3 votes
0 answers
49 views

Detecting isomorphism of simple field extensions

Let $k$ be a field and $f, g \in k[x]$ irreducible. How can I tell whether the field extensions $k[x] / (f)$ and $k[x] / (g)$ are isomorphic as extensions of $k$? Is there any necessary and sufficient ...
Tobias Fritz's user avatar
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0 answers
93 views

Exercise 4, Section 5.3 of Hungerford’s Algebra

Hungerford, Algebra, page 257, gives the following as Definition 3.1: Let $S$ be a set of polynomials of positive degree in $K[x]$. an extension field $F$ of $K$ is said to be a splitting field over $...
user264745's user avatar
-2 votes
1 answer
62 views

Exercise 1, Section 5.3 of Hungerford’s Algebra [duplicate]

Definition: Let $S$ be a set of polynomials of positive degree in $K[x]$. An extension field $F$ of $K$ is said to be a splitting field over $K$ of the set $S$ of polynomials if every polynomial in $S$...
user264745's user avatar
3 votes
1 answer
26 views

Homomorphisms between $\mathbb Q[x]/(x^3 - 2)$ and $\mathbb C$? and solvabity of an equation

I am required to find the field homomorphisms between $K = \mathbb Q[x]/(x^3 - 2)$ and $\mathbb C$ and to show that for any $n \geq 1$ the equation $$x_1^2 + \dots + x_n^2 = -1 \qquad x_i \in K $$ has ...
Donlans Donlans's user avatar
8 votes
1 answer
151 views

Assistance with an exercise on field endomorphisms

Working through the problems in a book on field theory (Field Extensions and Galois Theory by Bastida). I came across one which I thought looked like a "routine" exercise, but has been ...
Matt D's user avatar
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1 vote
0 answers
51 views

Automorphism group gives the Galois group over the fixed field.

Question. Let $E$ be a field and let $G=\operatorname{Aut}(E)$ be the group of ring automorphisms. Let $F=\operatorname{Fix}(G)$ be the fixed field of $G$. Is $E$ Galois over $F$? Is $G=\operatorname{...
user108580's user avatar
1 vote
0 answers
40 views

Restrictions of Galois representations

What field embeddings do I need to fix so that if I have a Galois representation $\rho:G_F \to GL(V)$, where $F$ is a number field, then for all finite extensions $L/F$ and places $v$ of F, I can talk ...
user14411's user avatar
  • 403
0 votes
1 answer
42 views

Artin map is the Kronecker symbol

Let $d$ be a squarefree integer. Then the discriminant $\Delta$ of $\mathbb{Q}(\sqrt{d})$ over $\mathbb{Q}$ is either $d$ or $4d$. I know that a prime number $p$ is ramified in $\mathbb{Q}(\sqrt{d})$ ...
Sardines's user avatar
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0 votes
0 answers
57 views

Linear independence of elements in $\mathbb F_5^3$

$\require{enclose}$ $\textbf{Question}$: Is the set ${(2,1,4),(1,3,2),(3,4,1)} \subseteq \mathbb F_5^3 \ $linearly independent? $\textit Attempt$: $$ \begin{align} 2a+b+3c=0 \tag 1\\ a+3b+4c=0 \...
Morten's user avatar
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2 votes
1 answer
47 views

Stationary action principle with a non-local action

Typical presentations of the Euler-Lagrange equations in field theory assume that the Lagrangian density is a function of the fields and their derivatives, so that the action $S$ is the integral over ...
Matt Dickau's user avatar
  • 2,289
0 votes
0 answers
19 views

For $p\geq 2$ show that $f(x)\in \mathbb{F}_p[x]$ is irreducible [duplicate]

For $p\geq 2$ show that $f(x)\in \mathbb{F}_p[x]$ where $f(x)=x^p-x+c$ is irreducible for all nonzero $c$. And separable. My approach: First for irreducibility, notice $f(z)=0 \iff p | f(z)$. And ...
Remu X's user avatar
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2 votes
0 answers
58 views

Patrick Morandi "Field and Galois Theory" - Exercise I.3.12

From the book: Let $K$ be a field, and suppose that $\sigma \in \mathrm{Aut}(K)$ has infinite order. Let $F$ be the fixed field of $\sigma$. If $K / F$ is algebraic, show that $K$ is normal over $F$. ...
Yifan Dai's user avatar
0 votes
0 answers
28 views

Let $F$ be a field and $f(x)\in F[x]$ such that $f(x)|g(x),\forall$ non constant polynomial $g(x)\in F[x].$ Show that $f(x)$ is a constant polynomial.

Let $F$ be a field and $f(x)\in F[x]$ such that $f(x)$ divides $g(x)$ for every non constant polynomial $g(x)\in F[x].$ Show that $f(x)$ is a constant polynomial. My solution goes like this: Let $a\...
Thomas Finley's user avatar
0 votes
2 answers
105 views

Subgroups of the multiplicative group of $\overline{\mathbb{F}_p}=\bigcup_{n\geq 1} \mathbb{F}_{p^n}$

Let $p$ be a prime and $K=\overline{\mathbb{F}_p}=\bigcup_{n\geq 1} \mathbb{F}_{p^n}$ where $\mathbb{F}_{p^m}\subseteq \mathbb{F}_{p^n}$ whenever $m\leq n$. Let $K^{\times}$ be the multiplicative ...
user44312's user avatar
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1 vote
1 answer
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Subgroups and corresponding subfields of Galois group of $x^5-5x^2-3$

I think I have found the correct Galois group of $f(x) = x^5-5x^2-3$ over $\mathbb{Q}$ to be $D_5$. generators are $σ=(1\ 2\ 4\ 5\ 3)$ and $τ=(2\ 3)(4\ 5)$ I can write down the elements of the ...
hbghlyj's user avatar
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-1 votes
1 answer
155 views

For which rings $P, Q, P\times Q$ product is a field? - Abstract Algebra problem from my college exam.

Exactly like in the title, here was my thought process: If $P\times Q$ must be a field, every element (besides $(0, 0)$) must have its inverse. Let's take $(p, q)$ for any $p$ from $P$ and $q$ from $...
Michal Pielka's user avatar
5 votes
1 answer
72 views

Is $\Bbb Q[a^2b+b^2c+c^2d+d^2a]$ the fixed field of the subgroup $\langle(1234)\rangle$ of Galois group $S_4$?

$f$ is an irreducible quartic polynomial in $\Bbb Q[x]$ with Galois group $S_4$. $a,b,c,d$ are four distinct roots of $f$. Let $K$ be the splitting field of $f$. Let $\text{Gal}(K/\Bbb Q)=S_4$. Then $...
hbghlyj's user avatar
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3 votes
1 answer
72 views

Compute $\operatorname{Gal}(\mathbb{Q}(\sqrt{1+2i},\sqrt{1-2i}) / \mathbb{Q})$

Using the primitive element theorem I got that $\mathbb{Q}(\sqrt{1+2i},\sqrt{1-2i}) = \mathbb{Q}(\sqrt{1+2i}+\sqrt{1-2i})$, and by noting that this is and normal extension of a field of characteristic ...
H4z3's user avatar
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0 answers
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Lemma 13, Section 5.2 of Hungerford’s Algebra

Theorem 2.2. Let $F$ be an extension field of $K$ and $f\in K[x]$. If $u\in F$ is a root of $f$ and $\sigma \in \text{Aut}_K F$, then $\sigma (u)\in F$ is also a root of $f$. If $F$ is an extension ...
user264745's user avatar
0 votes
1 answer
73 views

Show that GF(81) is an $x^{26}+x^{8}+x^{2}+1$ decomposition field

I tried decomposing the polynomial, but after taking out $(x^{2}+1)$ you have to break the remainder into polynomials of degree 4, which is manually hard. Perhaps this is solved by using Frobenius ...
mackenzie's user avatar
7 votes
0 answers
53 views

Ring homomorphisms between finite fields

I'm confused by this problem (A-52) from J.S. Milne - Field and Galois Theory Let $E$ and $F$ be finite fields of the same characteristic. Prove or disprove these statements. (a) There is a ring ...
dlvp's user avatar
  • 71
0 votes
1 answer
36 views

If Gal(K,F)=1 then K is a splitting field over F

I found a question asking whether or not this statement is true: “If K extension of F with Gal(K,F)=1 then K is splitting field over F.” My thought was that if Gal(K,F)=1 then K=F but that doesn’t ...
Mathematician's user avatar
2 votes
0 answers
53 views

Lemma 9, Section 5.2 of Hungerford’s Algebra

Let $F$ be an extension field of $K$ and let $H,J$ be subgroups of the Galois group $\text{Aut}_K F$ with $H\lt J$. If $[J : H]$ is finite, then $[H': J']\leq [J : H]$. Proof: Let $[J: H] = n$ and ...
user264745's user avatar
0 votes
0 answers
20 views

When are subgroups of the Galois group Conjugate?

I was working on the following problem: Let $K$ be a field of characteristic $0,$ $f \in K[X]$ an irreducible polynomial and $L/K$ a splitting field for $f.$ Let $\alpha_1, \dots, \alpha_n \in L$ be ...
Charles6Se's user avatar
3 votes
0 answers
107 views

Proof that for all nonzero real numbers $a$, $\frac{1}{a}$ is nonzero

I was wondering if someone could check my proof that "For all $a\in\mathbb{R}$, if $a\neq 0$ then $\frac{1}{a}\neq 0$". The definitions/assumptions I am basing the proof off of come from &...
user1320946's user avatar
0 votes
0 answers
11 views

powers of primitive element form basis of simple extension [duplicate]

Let $E/F$ be a finite extension of fields, $r \in E$ an element whose minimal polynomial is $f \in F[x]$, and $n = [F(r) : F]$. How can I show that $S = \{1, r, \dots, r^{n-1}\}$ is a basis of $F(r)$ ...
Abced Decba's user avatar
1 vote
1 answer
46 views

Inverse problem of the field extension $K/\Bbb{Q}$ of given number of prime above $p$

Let $g$ be an arbitrary fixed positive integer. I want to find an example of $K/\Bbb{Q}$ : field extension and prime number $p$ such that the number of primes of ring of integers of $K$ above $p$ is ...
Pont's user avatar
  • 6,206
0 votes
1 answer
45 views

Algebraic number on the unit circle has norm in the unit circle [closed]

Let $\alpha\in \mathbb{C}$ be algebraic over $\mathbb{Q}$ with modulus 1. For $f$ its minimal polynomial, is it always true that $f(0)$ lies on the unit circle? I've been trying things and apparently ...
Guillaume Berlat's user avatar

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