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. Please use (galois-theory) instead for questions specifically about that ...

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8
votes
3answers
525 views

$\mathbb{Q}(\pi, i\pi)$ over $\mathbb{Q}$

Is $\mathbb Q(\pi,i\pi):\mathbb Q$ a simple extension?
3
votes
2answers
487 views

Purely inseparable extension

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 $K$?
5
votes
3answers
682 views

Irreducibility and Splitting Fields

Show that over any field $F$, the polynomial $x^3-3x+1$ is either irreducible or splits into linear factors. Edited: This is my attempt: Let $f(x)=x^3-3x+1$. Let $a_1,a_2,a_3$ be the roots of ...
2
votes
0answers
178 views

Is there a subfield F of the complex field with [C:F]=3? [duplicate]

Possible Duplicate: What is the condition for a field to make the degree of its algebraic closure over it infinite? More generally, what is known about subfields $\mathbb{F}\subset ...
4
votes
2answers
364 views

Interesting property of a polynomial ring with coefficients algebraic over $\mathbb{Q}$

Minutes ago I read that if $F$ is the field of algebraic numbers over $\mathbb{Q}$, then every polynomial in $F[x]$ splits over $F$. That's awesome! Nevertheless, I don't fully understand why it is ...
8
votes
2answers
254 views

If $a,b\in\mathbb{Z}$, and if $a+b\sqrt{2}$ has a root in $\mathbb{Q}(\sqrt{2})$, then the root is actually in $\mathbb{Z}[\sqrt{2}]$

I'm working my way though a classical geometry book by Hartshorne right now, but this problem popped up in a section I'm reading. It is Problem 13.10 from Hartshorne's Geometry: Euclid and Beyond if ...
9
votes
2answers
311 views

algebraic version of “finite covering of a compact space is compact”

The following statement is an exercise in point set topology: If $E \to X$ is a covering with nonempty finite fibers and $X$ is compact, then also $E$ is compact. Now Grothendieck generalized covering ...
1
vote
1answer
120 views

Does $\forall n,d\!\in\!\mathbb{N}$ $\forall$ field $\mathbb{F}$ exist an irreducible $f\!\in\!\mathbb{F}[x_1,\ldots,x_n]$ of degree $d$?

how can one show (hopefully in an elementary manner) that there exist irreducible polynomials of arbitrary degree and number of variables over arbitrary field? thank you P.S. induction? EDIT: ehm, ...
6
votes
2answers
991 views

Splitting field of $x^{n}-1$ over $\mathbb{Q}$

From I.N.Herstein's Topics in Algebra. Chap 5 Sec 5.3 Page 227 Problem 8 Problem 8: If $n>1$ prove that the splitting field of $x^{n}-1$ over the field of rational numbers is of degree $\Phi(n)$ ...
4
votes
2answers
282 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 ...
2
votes
1answer
1k views

“Surjectivity is stable under base change” and field compositums

If $f:X\rightarrow Y$ is a surjective morphism of schemes and $g:X'\rightarrow Y$ is another morphism of schemes, one can show that $p_{2}:X\times_{Y}X'\rightarrow X'$ is also surjective. ...
7
votes
2answers
748 views

A question regarding normal field extensions and Galois groups

The following is possibly true but I can't find a corresponding theorem: If $E/F$ is the splitting field of some polynomial in $F$ and $F \subset K \subset E$ then: $Gal(E/K)$ normal subgroup of ...
3
votes
2answers
217 views

Extending morphisms with Zorn's lemma

I have stumbled upon a remarkable similarity between the proof of Baer's criterion and an extension theorem in field theory. Here are the statements: Baer's criterion: Let $R$ be a ring. A left ...
12
votes
3answers
603 views

A question regarding the definition of Galois group

In my book, Galois group is defined to mean the set of automorphisms on $E/F$ that "leave alone" the elements in $F$. On Wikipedia it says: "If $E/F$ is a Galois extension, then $Aut(E/F)$ is called ...
1
vote
1answer
124 views

Is there a specific name for this notion of extensions of fields?

As this question suggests, I quite like the notion of permuting the coefficients of polynomials. And, moreover, I have another question on this direction:If L|F is a finite normal field extension, ...
1
vote
1answer
348 views

Question regarding separability of polynomials and perfect fields

I have this definition A field $F$ is perfect if it has characteristic $0$ or if it has characteristic $p$ and $F^p = F$ From Wikipedia, I have this fact about separable polynomials: Irreducible ...
3
votes
2answers
266 views

A question regarding finite field extensions

If I understand correctly, the definition of the degree of a field extension $L/K$ is the dimension of $L$ over $K$ interpreted as a vector space. Now if the degree is $n < \infty$, the basis looks ...
2
votes
2answers
591 views

Minimal polynomials

Can someone tell me if this is right: I would like to find the minimal polynomial of (i) $\sqrt[4]{2}i$ over $\mathbb{Q}$ (ii) $\sqrt[4]{2}i$ over $\mathbb{R}$ (i): $\sqrt[4]{2}i$ is a root of ...
1
vote
6answers
232 views

Proving $|K| = p^n \Rightarrow \operatorname{char}(K) = p$

I would like to show that if $K$ is a field with $p^n$ elements then its characteristic has to be $p$, $p$ prime. I'm not sure where to start. It's clear to me that I can construct a field of order ...
0
votes
3answers
564 views

Conjoining elements to fields which are the roots of irreducible polynomials

I know that if $F$ is a field, and $x,y$ are roots of an irreducible polynomial over $F$ lying in an algebraic extension of $F$, then $F(x) \cong F(y)$. But it seems it is not true (in general) that ...
6
votes
1answer
136 views

Cyclic Extensions of $\mathbb{R}(t)$

Let $\mathbb{R}(t)$ be the field of rational functions over $\mathbb{R}$ (the fraction field of $\mathbb{R}[x]$). I am looking for elements in the Brauer group of the field, and the current idea I ...
4
votes
3answers
393 views

Equivalence of Archimedian Fields Properties

I'm trying to prove that an Archimedian field is a subfield of the real numbers, my plan is to use the fact that the rationals are dense within the field and their Dedekind completion is the real ...
2
votes
1answer
93 views

If p divides the index, then there exists $\beta \in O_K, \beta \notin \mathbb{Z}[\alpha]$ such that $p\beta \in \mathbb{Z}[\alpha]$

Again, this is a homework question, so (for the sake of learning) I'd be happy I get subtle hints only. Anyhow, the setup is as follows: "$K=\mathbb{Q}(\alpha)$ is of degree n over $\mathbb{Q}$, ...
2
votes
2answers
560 views

Field and Sets Proof

Need some clarification on this one. Question: Is the set [0, infinity) with the usual addition and multiplication axioms, a field? Attempted solution: Yes it is a field. let 0 <= r, s < ...
1
vote
3answers
962 views

Addition and Multiplication of Field Proof

I have just started out proofs so I am not fully grasping this concept yet. Question: Is the set $\mathbb{R}^2$, with addition and multiplication defined below a field? Explain. $(a, b) + (c, d) = ...
4
votes
1answer
284 views

Is every finite separable extension of a strictly henselian DVR totally ramified?

Let $R$ be a discrete valuation ring with field of fraction $K$ and residue field $k$ and let $K'$ be a finite and separable extension of $K$. If $R$ is henselian ("Hensel's lemma holds", e.g. if ...
3
votes
2answers
433 views

Construct an isomorphism between fields

The first part of my question asked : State all the irreducible Polynomials in $\mathbb{Z}_2[x]$ of order 3. I was able to do this and get the following polynomials : $x^3 + x^2 + x + 1 ...
4
votes
3answers
1k views

Example of a complete, non-archimedean ordered field

I'm looking for a concrete example of a complete (in the sense that all Cauchy sequences converge) but non-archimedean ordered field, to see that these two properties are independent (an example of ...
7
votes
1answer
205 views

Field Extensions

Let $L/K$ a finite extension and $f(x)\in K[x]$ a non-linear irreducible polynomial. Prove that if $\mathrm{gcd}\left( \mathrm{deg}(f) , \left[ L:K \right] \right)=1$ then $f(x)$ has no roots in ...
5
votes
3answers
152 views

Count of elements in $\Bbb{Z}_7[x]/(3x^2+2x)$

Hi I have some problem how to get count of elements in $\Bbb{Z}_7[x]/(3x^2+2x)$. I think there belong to only polynomials which are indivisible with $3x^2+2x$ ($\gcd=1$). I think it is so as far I ...
4
votes
1answer
108 views

Is an unramified cover of the p-adics determined by its degree?

If $K_1$ and $K_2$ are subfields of a pre-chosen $\overline{\mathbb{Q}_p}$, and if they're both unramified at $p$, and $[K_1:\mathbb{Q}_p]=[K_2:\mathbb{Q}_p]$, does that imply that $K_1=K_2$? My ...
7
votes
2answers
375 views

Can two different roots of an irreducible polynomial generate the same extension?

Let $K$ be a field and $f(x)$ be an irreducible polynomial over $K$. Suppose, $f(x)$ has degree at least $2$. Is it possible that if $a,b$ are two roots of $f(x)$ with $a\neq b$, then $K(a)=K(b)$. ...
2
votes
1answer
159 views

Field extension question

A question I recall from Carl Linderholm's "Mathematics Made Difficult", Chapter 3 Exercise 8. ...
8
votes
1answer
200 views

Is $\mathbb{Q}_p(\zeta_p)$ the same as $\mathbb{Q}_p(p^{\frac{1}{p-1}})$?

It seems so. $\mathbb{Q}_p(\zeta_p)$ is a $p-1^{th}$ extension of $\mathbb{Q}_p$ which doesn't extend the residue field; and so is $\mathbb{Q}_p(p^{\frac{1}{p-1}})$. However I can't see how to express ...
2
votes
0answers
104 views

Field reductions. part three

Follow-up to Field reductions. part two For a field $K$, an element $a \in K$, and $K(\setminus a)$ the set of field reductions (maximal subfields of $K$ not containing $a$) are there any interesting ...
4
votes
1answer
282 views

Field reductions. part two

This is a follow-up to the question on field reductions. (EDIT: Originally this question used the notation $\{\mathbb{R}(\setminus a)\}$, but now uses $\mathbb{R}(\setminus a)$ instead.) There ...
11
votes
1answer
274 views

Field reductions

If there is a field $F$ that is a field reduction of the real numbers, that is $F(a)=\mathbb{R}$ for some $a$, let's also denote this $F=\mathbb{R}(\setminus a)$, then given $x \in \mathbb{R}$ is ...
5
votes
3answers
3k views

Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$?

In this post we saw isomorphism of vector spaces over $\mathbb{Q}$. Just came across this question: Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$? In know these as $\mathbb{Q}$-Vector ...
8
votes
2answers
1k views

Degree of $\sqrt{2}+\sqrt[3]{5}$ over $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt[3]{5})$

I'm self-studying field extensions. I ran over an exercise which I can't completely solve. (I haven't yet started studying Galois theory, and I think this exercise isn't meant to be solved using it, ...
8
votes
1answer
649 views

Roots of unity and field extensions

Can we always break an arbitrary field extension $L/K$ into an extension $F/K$ in which the only roots of unity of $F$ are those in $K$, followed by an extension $L/F$ which is of the form ...
4
votes
2answers
759 views

How do I factorize polynomial over Galois field?

How does the factoring of polynomials over Galois fields work? I cannot seem to understand the basic concept. For example: How do I factorize $x^6 - 1$ over $\operatorname{GF}(3)$? I know that the ...
6
votes
3answers
161 views

almost n-th power in a field

The field of real numbers has a nice property - for a fixed $n\in \mathbb{N}$, every $r\in \mathbb{R}$ is either an n-th power (if n is odd or if r>0) or -r is an n-th power. I want to generalize this ...
2
votes
2answers
971 views

inverse element in a field of sets

A field is an abelian group under addition and a group under multiplication. But for the power set of a set S, under union and intersection operations and with S and the empty set being their ...
2
votes
1answer
763 views

Field of sets and Sigma algebra of sets

(1). According to Wikipedia, a field of subsets of X is defined to be a non-empty subset of the power set of X closed under the intersection and union of pairs of sets and under complements of ...
3
votes
6answers
584 views

Are there broad or powerful theorems of rings that do not involve the familiar numerical operations (+) and (*) in some fundamental way?

I am of, and I would like to retain, a mindset that mathematics does not have to have numbers as the central object of interest. With that in mind, I have done a fair amount of self-study on topics in ...
21
votes
2answers
1k views

Why is $\mathbb{Q}(t,\sqrt{t^3-t})$ not a purely transcendental extension of $\mathbb{Q}$?

This question is taken from Dummit and Foote (14.9 #6). Any help will be appreciated: Show that if $t$ is transcendental over $\mathbb{Q}$, then $\mathbb{Q}(t,\sqrt{t^3-t})$ is not a purely ...
14
votes
5answers
439 views

Why does $K \leadsto K(X)$ preserve the degree of field extensions?

The following is a problem in an algebra textbook, probably a well-known fact, but I just don't know how to Google it. Let $K/k$ be a finite field extension. Then $K(X)/k(X)$ is also finite with ...
4
votes
4answers
1k views

Examples of fields of characteristic 0?

I was preparing for an area exam in analysis and came across a problem in the book Real Analysis by Haaser & Sullivan. From p.34 Q 2.4.3, If the field F is isomorphic to the subset S' of F', show ...
2
votes
2answers
192 views

Powers of $x$ as members of Galois Field and their representation as remainders

first question on math.stackexchange :) I'm studying for a Cryptography - Communication Security exam, and it involves a certain quantity of number theory - finite field theory, so be warned: this is ...
5
votes
1answer
375 views

Algebra structure of tensor product of two Galois extensions

Sorry if this question is too basic. It is from Fröhlich and Taylor's "Algebraic Number Theory". Let $E/F$ be a finite Galois extension of fields, with $G=Gal(E/F)$, and let $K$ and $L$ be two ...