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
1answer
3k views

How to prove that the Frobenius homomorphism is surjective?

$R$ is a domain with characteristic $p$ ($p$ is prime).There is a homomorphism $f : R \to R$, $f(a)=a^p$. $f$ is called the Frobenius homomorphism. And I have known this. When $R$ which is mentioned ...
7
votes
2answers
481 views

How does $\cos(2\pi/257)$ look like in real radicals?

We know $\cos(2\pi/p)$ for p a Fermat prime can be expressed in real radicals. The case $p=17$ is a root of an 8th deg eqn, but can be also given as a sequence of nested radicals, $$\begin{aligned} ...
5
votes
2answers
508 views

Is $\mathbf{Q}(\sqrt{2},\sqrt{3}) = \mathbf{Q}(\sqrt{6})$?

Is $\mathbf{Q}(\sqrt{2},\sqrt{3}) = \mathbf{Q}(\sqrt{6})$? If $\mathbf{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}\mid a,b,c,d\in\mathbf{Q}\}$ and $\mathbf{Q}(\sqrt{6})= \{a+b\sqrt{6} | ...
5
votes
2answers
901 views

Galois Field Fourier Transform

there are two definitions for Reed-Solomon codes, as transmitting points and as BCH code ( http://en.wikipedia.org/wiki/Reed%E2%80%93Solomon_error_correction ). On Wikipedia there is written that we ...
2
votes
2answers
820 views

Necessary and Sufficient Condition for a sub-field

Is there any necessary and sufficient condition to determine whether a subset $H$ of a given field $K$ is a subfield? In some paper I have found something like that: $H$ is a field if for all $a, ...
-1
votes
2answers
636 views

What is $\operatorname{Gal}(\mathbb Q(\zeta_n)/\mathbb Q(\sin(2\pi k/n))$?

Let $\zeta_n$ be the $n$-th primitive root of unity and $4 \mid n$. Consider the field extensions $\mathbb Q \subset \mathbb Q(\sin(2\pi k/n) \subset \mathbb Q(\zeta_n)$. What is the degree of the ...
8
votes
4answers
814 views

Why is the product of all units of a finite field equal to $-1$?

Suppose $F=\{0,a_1,\dots,a_{q-1}\}$ is a finite field with $q=p^n$ elements. I'm curious, why is the product of all elements of $F^\ast$ equal to $-1$? I know that $F^\ast$ is cyclic, say generated by ...
8
votes
1answer
2k views

Existence of irreducible polynomials over finite field

Let $F$ be a finite field. How do we prove that for each $n \in \mathbb{N}$ there is an irreducible polynomial of degree $n$? One can assume that $F = \mathbb{F}_{p^m}$ where $p$ is prime. If $n \ge ...
3
votes
2answers
91 views

Field that contains $(\mathbb{Z}/p^n\mathbb{Z})^*$

Let $p$ be an odd prime. There exists a field $F$ that contains an isomorphic copy of $(\mathbb{Z}/p^n\mathbb{Z})^*$ as a multiplicative subgroup? Clearly if $n=1$ we can take $F=\mathbb{F}_p$, but ...
1
vote
1answer
117 views

Show that $9+9x+3x^3+6x^4+3x^5+x^6$ is irreducible given one of its roots

Given a polynomial $f(x)=9+9x+3x^3+6x^4+3x^5+x^6$ and one of its roots $\alpha=2^{1/3}+e^{2\pi i/3}$. Show that $f(x)$ is irreducible in $\mathbb Q[x]$. Eisenstein's criterion fails, it also didn't ...
1
vote
1answer
170 views

Minimal polynomial of intermediate extensions under Galois extensions.

Let $K$ be a Galois extension of $F$, and let $a\in K$. Let $n=[K:F]$, $r=[F(a):F]$, and $H=\mathrm{Gal}(K/F(a))$. Let $z_1, z_2,\ldots,z_r$ be left coset representatives of $H$ in $G$. Show that ...
5
votes
1answer
553 views

field of algebraic numbers

Would anyone be able to give me an outline or a hint towards the proof that the field of algebraic numbers is an infinite extension of the field of rationals? Many Thanks
5
votes
2answers
165 views

Closure of a number field with respect to roots of a cubic

Consider the following property of subfields ${\mathbb K}$ of ${\mathbb C}$ : $$ \text{Any polynomial of degree 3 with coefficients in} \ {\mathbb K} \ \text{has a root in } {\mathbb K} \ \ \ \ ...
5
votes
1answer
1k views

Sums and products of algebraic numbers

How does one go about proving that the sums and products of two algebraic numbers over a field $F$ (say $a,b\in K$, where $K/F$ is a field extension) is also algebraic? If we call $f_a$ and $f_b$ ...
2
votes
2answers
102 views

The set of algebraic numbers is a field.

I need help proving this stunning result. Let me state it this way. Let $\mathbb L$ be a field and $\mathbb K$ be a field with $\mathbb K \subset \mathbb L$ Let $A$ be the set of algebraic ...
2
votes
3answers
147 views

Explain why $\mathbb{Z \times Z}$ and $\mathbb{R \times R}$ is not a field [duplicate]

Explain why $\mathbb{Z \times Z}$ and $\mathbb{R \times R}$ is not a field and that why any external direct sum of two fields cannot be a field. I believe it has much to do with the lack of every ...
1
vote
1answer
61 views

Convert from a field extension to an elementary field extension

I have a basic question about algebraic field extensions: How can I convert a multiple extension like $\mathbb{Q}(\sqrt{2},\sqrt{3})$ to a single (elementary) field extension (like ...
0
votes
3answers
335 views

is it possible for two fields with the same characteristic to not be isomorphic?

I know there is a long list of invariants, but is there a proof or theorem that explains two fields with the same characteristic are isomorphic?
26
votes
6answers
1k views

Is $\{0\}$ a field?

Consider the set $F$ consisting of the single element $I$. Define addition and multiplication such that $I+I=I$ and $I \times I=I$ . This ring satisfies the field axioms: Closure under addition. ...
14
votes
4answers
5k views

Finding the minimal polynomial of $\sqrt 2 + \sqrt[3] 2$ over $\mathbb Q$.

I have to find the minimal polynomial of $\sqrt 2 + \sqrt[3] 2$ over $\mathbb Q$. The suggested way of doing it is to prove that $\mathbb Q[\sqrt 2 + \sqrt[3] 2]=\mathbb Q[\sqrt 2,\sqrt[3] 2]$ first. ...
26
votes
3answers
2k views

Why isn't the inverse of the function $x\mapsto x+\sin(x)$ expressible in terms of “the functions one finds on a calculator”?

The function $f(x)=x+\sin(x)$ is easily checked to be a bijection from the reals to itself, and so it has a unique inverse $y\mapsto g(y)$ such that $f\circ g=g\circ f$ are both the identity map. ...
11
votes
1answer
281 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 ...
10
votes
2answers
876 views

What exactly is the fixed field of the map $t\mapsto t+1$ in $k(t)$?

Suppose $k$ is a field, and $k(t)$ is the rational function field. If $f(t)=P(t)/Q(t)$ for some polynomials $P(t)$ and $Q(t)\neq 0$, then the map $t\mapsto t+1$ sends $f(t)$ to $f(t+1)$. So the ...
8
votes
5answers
461 views

Prove $\sqrt[3]{3} \notin \mathbb{Q}(\sqrt[3]{2})$

I've tried solving $\sqrt[3]{3} =a + b* \sqrt[3]{2}+c* \sqrt[3]{4}$, but there is no obvious contradiction, even when taking the norms/traces of both sides. I can't think of another approach. This is ...
16
votes
3answers
2k views

A finite field cannot be an ordered field.

I am reading baby Rudin and it says all ordered fields with supremum property are isomorphic to $\mathbb R$. Since all ordered finite fields would have supremum property that must mean none exist. ...
8
votes
2answers
1k views

Problem in Jacobson's Basic Algebra (Vol. I)

It is section $4.4$, exercise number $5$. It says the following: Let $F$ be a field of characteristic $p$. Show that $f(x)=x^p-x-a$ has no multiple roots and that $f(x)$ is irreducible in $F[x]$ if ...
6
votes
1answer
384 views

Degree of field extension

If $p$ is odd prime and $c=\cos(\frac{2\pi}{p})$, $s=\sin(\frac{2\pi}{p})$ then for which values of $p$ does $\mathbb{Q}(s,c)=\mathbb{Q}(c)$?
6
votes
5answers
2k views

Infinite Degree Algebraic Field Extensions

In I. Martin Isaacs Algebra: A Graduate Course, Isaacs uses the field of algebraic numbers $$\mathbb{A}=\{\alpha \in \mathbb{C} \; | \; \alpha \; \text{algebraic over} \; \mathbb{Q}\}$$ as an example ...
5
votes
1answer
99 views

Why is the extension $k(x,\sqrt{1-x^2})/k$ purely transcendental?

Consider the function field $k(x,\sqrt{1-x^2})$ of the circle over an algebraically closed field $k$. Is $k(x,\sqrt{1-x^2})$ a purely transcendental extension over $k$? I'm curious because I was ...
4
votes
1answer
80 views

Existence of a group isomorphism between $(\mathbb K,+)$ and $(\mathbb K^\times,\cdot)$

Let $(\mathbb K,+,\cdot)$ be a field. Is there a group isomorphism between $(\mathbb K,+)$ and $(\mathbb K^\times,\cdot) $ ? The answer should clearly be negative. I tried to proceed via ...
3
votes
2answers
700 views

isomorphism of Dedekind complete ordered fields

In the last chapter of Spivak's Calculus, there is a proof that complete ordered fields are unique up to isomorphism. I find the first steps in it somewhat suspicious. Specifically, I believe he is ...
3
votes
2answers
467 views

A question about a weak form of Hilbert's Nullstellensatz

Corollary 5.24 on page 67 in Atiyah-Macdonald reads as follows: Let $k$ be a field and $B$ a finitely generated $k$-algebra. If $B$ is a field then it is a finite algebraic extension of $k$. We know ...
10
votes
3answers
265 views

Does $\mathbb{F}_p((X))$ has only finitely many extension of a given degree?

We know that $\mathbb{Q}_p$ has only finitely many extensions of a given degree in its algebraic closure. Is it the same for $\mathbb{F}_p((X))$?
6
votes
2answers
561 views

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 ...
6
votes
1answer
432 views

Galois Groups of Finite Extensions of Fixed Fields

I am trying to prove the following proposition: Let $L$ be an algebraically closed field, $g \in Aut(L)$ and $K=\{x \in L \; | \; g(x)=x\}$. Show that every finite extensions $E/K$ is a cyclic ...
6
votes
2answers
468 views

Why aren't there any coproducts in the category of $\bf{Fields}$?

Well the question is stated in the title. I dont know much about field theory and i was suprised when i read it on wikipedia please provide some examples thanks in advance
6
votes
5answers
6k views

Finding inverse of polynomial in a field

I'm having trouble with the procedure to find an inverse of a polynomial in a field. For example, take: In $\frac{\mathbb{Z}_3[x]}{m(x)}$, where $m(x) = x^3 + 2x +1$, find the inverse of $x^2 + ...
5
votes
2answers
103 views

The degree of $\sqrt{2} + \sqrt[3]{5}$ over $\mathbb Q$

I know that the degree is at most $6$, since $\sqrt{2} + \sqrt[3]{5} \in \mathbb Q(\sqrt{2}, \sqrt[3]{5})$, which has degree $6$ over $\mathbb Q$. I'm trying to construct a polynomial with root ...
5
votes
3answers
1k views

Quadratic subfield of cyclotomic field

Let $p$ be prime and let $\zeta_p$ be a primitive $p$th root of unity. Consider the quadratic subfield of $\mathbb{Q}(\zeta_p)$. For instance, for $p=5$ we get the quadratic subfield to be ...
4
votes
1answer
297 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 ...
4
votes
2answers
299 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 ...
3
votes
1answer
138 views

When is a divisible group a power of the multiplicative group of an algebraically closed field?

It is known that for any algebraically closed field $\mathbb{F}$ its multiplicative group $\mathbb{F}^*$ is a divisible group, and consequently any power $\mathbb{F}^*\times\cdots\times \mathbb{F}^*.$ ...
2
votes
2answers
875 views

Prove that every nonzero prime ideal is maximal in $\mathbb{Z}[\sqrt{d}]$

$d \in \mathbb{Z}$ is a square-free integer ($d \ne 1$, and $d$ has no factors of the form $c^2$ except $c = \pm 1$), and let $R=\mathbb{Z}[\sqrt{d}]= \{ a+b\sqrt{d} \mid a,b \in \mathbb{Z} \}$. ...
2
votes
1answer
352 views

a property of the radical closure of a field

Definition 1: Let $K$ be a field. If $\alpha$ is an algebraic element over K, such that $\alpha^n \in K $ and such that $x^n-\alpha^n \in K[x] $is also irreducible over K. Then we call $ K(\alpha)$ a ...
2
votes
1answer
602 views

If a normal $K/F$ has no intermediate extensions, then $[K : F]$ is prime

Let $K$ be a finite normal extension of $F$ such that there are no proper intermediate extensions of $K/F$. Show that $[K:F]$ is prime. Give a conterexample if $K$ is not normal over $F$.
1
vote
2answers
98 views

Show that $F[x] / (x^2-1) \cong F \times F$ iff $char(F) \neq 2$

Let $F$ be a field. Show that $F[x] / (x^2-1) \cong F \times F$ iff $char(F) \neq 2$. Here is my work this far: If $char(F) = 2$, then $x^2-1 = (x-1)^2$, and hence $F[x]/(x^2-1)$ has a non-zero ...
0
votes
3answers
271 views

Proving that $-a=(-1)\cdot a$

As the title reveals, I want to prove (based on the axioms of field) that $$-a=(-1)\cdot a$$ I've been trying for a while now, but couldn't think of a way to do it and it got me thinking that maybe ...
8
votes
3answers
599 views

Showing a homomorphism of a field algebraic over $\mathbb{Q}$ to itself is an isomorphism.

Suppose $F$ is algebraic over $\mathbb{Q}$ and $\varphi : F\to F$ is a homomorphism. Prove $\varphi$ is an isomorphism. Showing injectivity follows from the fact that the only ideals in a field ...
7
votes
1answer
329 views

Generating Elements of Galois Group

I am trying to prove the following: Let $K=\mathbb{Q}(\sqrt{p_1},\ldots,\sqrt{p_n})$. Show that $K/\mathbb{Q}$ is Galois with Galois group $(\mathbb{Z}/2\mathbb{Z})^n$. I have attached my proof ...
7
votes
1answer
222 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 ...