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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Splitting field implies Galois extension

I hope this isn't too elementary of a question, but I'm not sure I understand Artin's proof that if $K/F$ is a finite extension, then $K/F$ Galois is equivalent to $K$ being a splitting field over $F$ ...
LCL's user avatar
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How to prove polynomials with degree $n$ does not form a vector space?

This is one of my linear algebra problems: Prove that polynomials of degree $n$ does not (The professor made these words bold intentionally) form a vector space. From what I read, the set of ...
Jay Wang's user avatar
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Set of elements in $K$ that are purely inseparable over $F$ is a subfield

Let $F\subset K$ be an algebraic field extension. Is the set of all elements of $K$ that are purely inseparable over $F$ necessarily a subfield of $K$?
ashpool's user avatar
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6 answers
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What is the difference between an axiom and a definition?

Now I know that this question has been asked before here, but the reason I'm asking this again is because the example given in the question there, namely one of Peano's Axioms is very clearly an axiom ...
Saad Haider's user avatar
10 votes
1 answer
860 views

Intermediate field, normal closure and Galois group

Let $K/F$ be Galois with $G=Gal(K/F)$ and let $L$ be an intermediate field. Let $N\subseteq K$ be the normal closure of $L/F$. If $H=Gal(K/L)$ show that $Gal(K/N)=\cap_{\sigma\in G}\sigma H\sigma^{-1}$...
user109584's user avatar
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Galois group of $x^5-x-1$ over $\Bbb Q$

I am trying to compute the Galois group of $x^5-x-1$ over $ \Bbb Q$. I've shown that this polynomial is irreducible over $\Bbb Q$, by showing that it is irreducible over $\Bbb Z_5$. Let $F$ be the ...
blancket's user avatar
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4 answers
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Galois group of $x^6+3$ over $\mathbb Q$

I'm having some difficulties finding the Galois group of the polynomial $g(x)=x^6+3$ over $\mathbb Q$. Here's what I did : I observed that the roots of the given polynomial are $\sqrt[6]3 \xi_{12}^{...
Lorenzo Rossi's user avatar
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2 answers
756 views

Dimension of an algebraic closure as a vector space over its base field.

Let $k$ be an infinite field and $\bar{k}$ its algebraic closure. The Artin-Schreier Theorem tells us (among other things) that $[\bar{k}:k]=1,2,\infty$. There's a natural interpretation of $[\bar{k}:...
JSchlather's user avatar
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Rational function field over uncountable field is uncountably dimensional

Here is the problem and I'm a bit stuck in my proof. We have an uncountable field $k$ and a transcendental element $t$ over $k$ so we consider the field $k(t)$ and want to prove that $\dim(k(t)/k)$...
WrabbitW's user avatar
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Polynomial rings -- Inherited properties from coefficient ring

To avoid mixing up things, I wanted to collect properties a polynomial ring is inheriting from the coefficient ring and what property implies another. Let $R$ be a ring (what else do I need at which ...
Frank H's user avatar
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Abelian Galois group of $f$ implies splitting is simple extensions by a root of $f$.

Given an irreducible polynomial $f\in \mathbb{Q}[x]$ with Abelian Galois group, I would like to show that the splitting field $K$ over the rationals can be written as a simple extension $\mathbb{Q}(\...
MAM's user avatar
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5 answers
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Is $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ the same as $\mathbb{Q}(\sqrt{2},\sqrt{3})$?

Is $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ the same as $\mathbb{Q}(\sqrt{2},\sqrt{3})$? I mean, $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ can be viewed as: $\mathbb{Q}[\sqrt{2}][\sqrt{3}]$, as polynomials of $\sqrt{3}$...
user119615's user avatar
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Can we turn $\mathbb{R}^n$ into a field by changing the multiplication?

Of course $\mathbb{R}$ is a field with usual addition and multiplication. When we move up a dimension into $\mathbb{R}^2$, however, there is not a clear way to multiply two vectors together to get ...
tylerc0816's user avatar
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A field that is an ordered field in two distinct ways

Question: Explain the construction below (taken directly from Counter Examples in Analysis): An ordered field is a field $F$ that contains a subset $P$ such that $P$ is closed with respect to ...
J C's user avatar
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Unramified primes of splitting field

I would like to show the following: Theorem: Let $K$ be a number field and and $L$ be the splitting field of a polynomial $f$ over $K$. If $f$ is separable modulo a prime $\lambda$ of $K$, then $L$ ...
Steven's user avatar
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An element is integral iff its minimal polynomial has integral coefficients.

This is from Algebraic Number Theory by Neukirch Let $A$ be an integral domain which is integrally closed, K its field of fractions, $L|K$ a finite field extension, and $B$ the ...
Mohan's user avatar
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factorization of the cyclotomic $\Phi_n(x)$ over $\Bbb F_p$

One can prove that the $\Phi_n(x)$ are irreducible over $\Bbb Z$. Where $\Phi_n(x)=\prod _{(a,n)=1}\zeta_n^a$ (i.e the product of the primitive n-rooth of unity). I want to find a factorization of $\...
Daniel's user avatar
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Source to learn Galois Theory

What kind of recommendations do you have for a very good source to learn Galois Theory? Is there any Atiyah-MacDonald-type book on Galois theory? What is your opinion on the chapters from Lang, and ...
9 votes
2 answers
693 views

Is $\Bbb Z_p^2$ a Galois group over $\Bbb Q$?

$\newcommand{\Q}{\Bbb Q} \newcommand{\N}{\Bbb N} \newcommand{\R}{\Bbb R} \newcommand{\Z}{\Bbb Z} \newcommand{\C}{\Bbb C} \newcommand{\A}{\Bbb A} \newcommand{\ab}{\mathrm{ab}} \newcommand{\Gal}{\mathrm{...
Watson's user avatar
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Let $K$ be a subfield of $\mathbb{C}$ not contained in $\mathbb{R}$. Show that $K$ is dense in $\mathbb{C}$.

Let $K$ be a subfield of $\mathbb{C}$ not contained in $\mathbb{R}$. Show that $K$ is dense in $\mathbb{C}$. completely stuck on it. can I get some help please.
tuitokari's user avatar
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1 answer
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Example of a nontransitive action of $\operatorname{Aut}(K/\mathbb Q)$ on the roots in $K$ of an irreducible polynomial.

I want to find an irreducible polynomial $f(x)$ over $\mathbb Q$ and a finite nonnormal extension $K/\mathbb Q$ which contains at least two roots of $f(x)$ such that $\operatorname{Aut}(K/\mathbb Q)$ ...
Pratyush Sarkar's user avatar
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1 answer
1k views

Capelli Lemma for polynomials

I have seen this lemma given without proof in some articles (see example here), and I guess it is well known, but I couldn't find an online reference for a proof. It states like this: Let $K$ be a ...
T.C.'s user avatar
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$\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})=\mathbb{Q}$?

Is there an easy way to see that $$\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})=\mathbb{Q}?$$ I know that $\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})$ is a subfield of $\mathbb{Q}(\sqrt[...
Spenser's user avatar
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Is it true that every element of $\mathbb{F}_p$ has an $n$-th root in $\mathbb{F}_{p^n}$?

It is not hard to prove that every element of $\mathbb{F}_p$ has a square root in $\mathbb{F}_{p^2}$: take any $a \in \mathbb{F}_p$ and consider the polynomial $f = X^2 - a$. If $f$ has a root in $\...
Matías's user avatar
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The prime number theorem over a finite field - Lang's *Algebra*, Chapter V, Exercise 23(b)

This is Exercise 23(b) of Chapter V (Algebraic Extensions) from Lang's Algebra. Let $k$ be finite field with $q$ elements, and let $\pi_q(n)$ be the number of monic irreducible polynomials $p \in k[...
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Determine the Galois Group of $(x^2-2)(x^2-3)(x^2-5)$

Determine all the subfields of the splitting fields of this polynomial. I chose this problem because I think to complete it in great detail will be a great study tool for all of the last chapter, as ...
Jared's user avatar
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2 answers
301 views

Are $\mathbb{R}$ and $\mathbb{Q}$ the only subfields of $\mathbb{C}$ with natural structure as ordered fields?

We know that $\mathbb{R}$ and $\mathbb{Q}$ have a unique structure as ordered fields with the usual order, and that $\mathbb{C}$ cannot be realised as an ordered field. Various non-trivial subfields ...
user829347's user avatar
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8 votes
3 answers
469 views

$\mathrm{Aut}_\mathbb{Q}(\overline{\mathbb{Q}})$ is uncountable

How do you show that $\mathrm{Aut}_\mathbb{Q}(\overline{\mathbb{Q}})$ is uncountable ? Thanks in advance
Zorba le Grec's user avatar
8 votes
1 answer
3k views

Degree of composite field

Let $E/K$ and $F/K$ be finite subextensions of $L/K$, denote $EF/K$ the composite subextension. Then $[EF:F]\leqslant[E:K]$ and $[EF:K]\leqslant[E:K]\cdot[F:K]$. If we assume $E\cap F=K$, then will ...
Lao-tzu's user avatar
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8 votes
2 answers
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Find all irreducible polynomials of degrees 1,2 and 4 over $\mathbb{F_2}$.

Find all irreducible polynomials of degrees 1,2 and 4 over $\mathbb{F_2}$. Attempt: Suppose $f(x) = x^4 + a_3x^3 + a_2x^2 + a_1x + a_0 \in \mathbb{F_2}[x]$. Then since $\mathbb{F_2} =${$0,1$}, then ...
Mahidevran's user avatar
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8 votes
4 answers
2k views

$x^4+x^3+x^2+x+1$ irreducible over $\mathbb F_7$

This question came from the answer here. The answer claims that $x^4+x^3+x^2+x+1$ is irreducible over $\mathbb F_7$. I can check that it has no roots in the field, but why can't it be written as a ...
user557's user avatar
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8 votes
4 answers
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Prove that a polynomial of degree $d$ has at most $d$ roots (without induction)

Let $p(x)$ be a non-zero polynomial in $F[x]$, $F$ a field, of degree $d$. Then $p(x)$ has at most $d$ distinct roots in $F$. Is it possible to prove this without using induction on degree? If so, ...
John Smith's user avatar
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8 votes
3 answers
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$\mathbb{R}$ is not isomorphic to a proper subfield of itself

let $\mathbb{R}$ be the field of real numbers. I found stated in this pretty work On Groups that Are Isomorphic to a Proper Subgroup, that there is no proper subfield $K$ of $\mathbb{R}$ which is ...
Maurizio Barbato's user avatar
8 votes
1 answer
3k views

Minimal polynomial over an extension field divides the minimal polynomial over the base field

I need help proving this theorem: Given the field extension: $\mathbf{K} \subseteq \mathbf{L}$, for $\alpha \in \mathbf{L}$ and $g(x) \in \mathbf{K}[x]$, $\alpha$'s minimal polynomial over $K$, ...
Daniella's user avatar
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8 votes
2 answers
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Field Norm Surjective for Finite Extensions of $\mathbb{F}_{p^k}$

I'd rather not have the answer, because I feel like this should be a relatively easy question, and I'm just missing some key step, but could anyone give me a hint on showing that the norm (defined as $...
Jonathan Beardsley's user avatar
8 votes
1 answer
2k views

Intersection of field extensions

Let $F$ be a field and $K$ a field extension of $F$. Suppose $a,b\in K$ are algebraic over $F$ with degrees $m$ and $n$, where $m,n$ are relatively prime. Then $F(a) \cap F(b) = F$. I see that the ...
rorty's user avatar
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8 votes
3 answers
3k views

Field extensions and algebraic/transcendental elements

Let $E$ be an extension of field $F$, and let $\alpha, \beta \in E$. Suppose $\alpha$ is transcendental over $F$ but algebraic over $F(\beta)$. Show that $\beta$ is algebraic over $F(\alpha)$. Okay,...
terrible at math's user avatar
8 votes
1 answer
2k views

Quadratic equations that are unsolvable in any successive quadratic extensions of a field of characteristic 2

Show that for a field $L$ of characteristic $2$ there exist quadratic equations which cannot be solved by adjoining square roots of elements in the field $L$. In $\mathbb{Z_2}$ adjoining all square ...
Gaston Burrull's user avatar
8 votes
4 answers
12k views

Homomorphisms between fields are injective.

How would I prove this? I know that I must show $f(a)=f(b) \Rightarrow a = b$ I also know I must use the definition of homomorphism, ie: $f(a+b)=f(a)+f(b)$ $f(ab)=f(a)f(b)$ $f(1)=1$ I am ...
Mark's user avatar
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8 votes
2 answers
1k views

Counterexamples in algebra

I got the feeling that whenever a subject gets so sophisticated that Zorn's lemma is needed, a book of counterexamples in that subject would probably benefit researchers/ students a lot. Zorn's ...
Hui Yu's user avatar
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0 answers
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When exactly is the splitting of a prime given by the factorization of a polynomial?

Let $L/K$ be an extension of number fields with $L=K(\alpha)$, where $\alpha\in \mathcal{O}_L$. Let $f\in \mathcal{O}_K[x]$ be the minimal polynomial of $\alpha$ over $K$. Let $\mathfrak{p}\...
Ben Blum-Smith's user avatar
8 votes
3 answers
5k views

clarification of algebraic closure and algebraically closed field

Definition of Algebraic closure: An extension $K$ of $F$ is called an algebraic closure of $F$ if (a) $F \subset K$ is algebraic; (b) $K$ is algebraically closed. Given the above definition, I have ...
spaceman_spiff's user avatar
8 votes
1 answer
963 views

Finite Field Extensions and the Sum of the Elements in Proper Subextensions (Follow-Up Question)

I recently posted the following question, to which this question is a follow-up. Regardless, my question here will be self-contained. Let $F$ be a finite field, and let $u,v$ be algebraic over $F$. ...
Elchanan Solomon's user avatar
7 votes
1 answer
1k views

Tensor product of inseparable field extensions

Suppose $K$ is a field and $L, L'$ are finite extensions of $K$. It is known that if $L/K$ (or $L'/K$) is separable, then $L \otimes_K L'$ is a product of finitely many fields. Is there a ...
bob's user avatar
  • 247
7 votes
2 answers
819 views

What is the condition for a field to make the degree of its algebraic closure over it infinite?

As we all know, the algebraic closure often has an infinite degree. Also, this shows the necessary and sufficient condition for a Galois extension to be a finite extension of fields. However, we may ...
awllower's user avatar
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7 votes
3 answers
3k views

Suppose $\alpha$ is a rational root of a monic polynomial in $\mathbb{Z}[x]$. Prove $\alpha \in \mathbb{Z}$ .

Suppose $\alpha$ is a rational root of a monic polynomial in $\mathbb{Z}[x]$. Prove $\alpha \in \mathbb{Z}$ . attempt: Suppose $p(x)$ is monic. Then the leading coefficient $a_n = 1.$ Thus, $p(x) = x^...
Mahidevran's user avatar
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7 votes
2 answers
15k views

characteristic of a finite field

knowing that the characteristic of an integral domain is $0$ or a prime number, i want to prove that the characteristic of a finite field $F_n$ is a prime number, that is $\operatorname{char}(F_n)\not ...
palio's user avatar
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7 votes
2 answers
936 views

There is exactly one algebraically closed field with prescribed characteristic and cardinality

I need to cite and read the proof of the following: Theorem: For every characteristic $p\geq0$ and uncountable cardinal $k$, there is up to field isomorphism exactly one algebraically closed field of ...
Chilote's user avatar
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7 votes
1 answer
622 views

A formula for the roots of a solvable polynomial

Let $F$ be a field and $p(x)\in F[x]$ a separable polynomial, denote $K$ as the splitting field of $p$ and assume that $K/F$ is Galois with a solvable Galois group. I don't understand if this imply ...
Belgi's user avatar
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7 votes
1 answer
513 views

Tensor product $L \otimes_K L$ has no nilpotent elements iff $I/I^2=0$

Let $L \supset K$ be a finite extension of fields. The diagonal $L \otimes_K L$ we can endow with structure of $L$ algebra via $L \to L \otimes_K L,\ l \mapsto l \otimes 1_L$. Especially $L \otimes_K ...
user267839's user avatar
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