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 ...

learn more… | top users | synonyms

2
votes
1answer
26 views

Field at which $f(x)$ splits

Let $f(x) \in \mathbb{Z}_p[x]$. Show that there is a finite field $\mathbb{F}_{p^n}$ at which $f(x)$ splits. And if $f(x)$ is also separable, show that $f(x) \mid x^{p^n}-x$. Could you give me some ...
0
votes
1answer
26 views

The union of finite field extensions is a finite field extension

Assume that all elements under discussion are algebraic over $F$. Let the notation "$K=F(A)$" mean that $A\subseteq K$ and there is an injective homomorphism $\sigma:F\to K$, and every element of $K$ ...
2
votes
4answers
49 views

Roots of different irreducible polynomials are algebraically independent

Let $F$ be a field, and let $f$ be a monic irreducible polynomial over $F$. Let $\alpha$ be a root of some other monic irreducible $g\ne f$. Then is $f$ still irreducible in $F(\alpha)$? Is it true ...
1
vote
0answers
16 views

Field extensions and quotient fields

STATEMENT: Suppose that $F\subseteq E$ is a field extension of $F$. And assume $u\in E$ is transcendental over $E$. Then it readily follows that $F(u)\cong F(x)$, where $F(x)$ is the quotient field of ...
1
vote
2answers
17 views

Find the splitting field of a polynomial

The extension $\mathbb{Z}_p \leq \mathbb{F}_{p^n}$ is normal, as the splitting field of the polynomial $f(x)=x^{p^n}-x$ ($\mathbb{Z}_p$ is a perfect field therefore each polynomial is separable). So, ...
0
votes
1answer
21 views

Show sentences-Field of rational functions

Let $K=\mathbb{Z}_p(x,y)$ be the field of rational functions of variables $x,y$ with coefficients in the field $\mathbb{Z}_p$, where $p$ is prime. Let $g(t)=t^p-x, h(t)=t^p-y \in K[t]$ and $E$ is the ...
0
votes
0answers
29 views

Show that the fields are equal

I have to show that $\mathbb{F}_{2^2}=\mathbb{Z}_2(a)$, where $a \in \mathbb{F}_{2^2}$ is of degree $2$ over $\mathbb{Z}_2$. $$$$ To show this do I have to take first an element of ...
1
vote
1answer
19 views

Basis for field extension $\mathbb{Q}(\zeta , \sqrt[3]{2})/\mathbb{Q}$

I'm trying to find a basis for the field extension $\mathbb{Q}(\zeta , \sqrt[3]{2})/\mathbb{Q}$, where $\zeta$ is the cube root of unity. I attempted this with starting with a set of elements I know ...
2
votes
3answers
172 views

Degree of field extension is infinite

If we have the field extension $\mathbb{Q}\leq \mathbb{R}$, could you explain me why it stands that $[\mathbb{R}:\mathbb{Q}]=+\infty$ ??
2
votes
1answer
24 views

Multiplicative group of a field contains maximal n-1 elements with order n

Let $F$ be a field and $n\in \mathbb N,n>1$. I want to show that the multiplicative group $K$\ $\{0\}$ contains maximal $n-1$ elements with order $n$. I actually don't have any ideas how to solve ...
0
votes
1answer
23 views

Galois group of a polynomial and subfields

Suppose $f(x)\in \mathbb{Z}[x]$ is an irreducible quartic whose splitting field has Galois group $S_4$ over $\mathbb{Q}$. Let $\theta$ be a root of $f(x)$ and set $K=\mathbb{Q}(\theta)$. a) Prove ...
0
votes
1answer
18 views

This is only a subspace if $b=0$ - Axler - LADR p13

I have written here in Axler - Linear Algebra Done Right, page $13$. If $b\in \mathbb{F}$, then $\{(x_1,x_2,x_3,x_4)\in \mathbb{F}^4: x_3 = 5x_4 + b\}$ is a subspace of $\mathbb{F}^4$ if and only ...
1
vote
2answers
30 views

Show that an extension is separable

Let $K$ be a field with $\operatorname{char} K=p$, where $p$ is a prime, and let the degree of the extension $K \leq L$ be coprime to $p$. How can I show that the extension is separable?? Could you ...
0
votes
1answer
17 views

Prove that $\Bbb F_p^\times$ is equal to Miller–Rabin primality test for prime number

I want to prove, that $\Bbb F_p^\times = MRP(p)$. I think, that I have to start with this statement: $\{a \in \Bbb F_p^\times | a^2 = 1 \} = \{1; -1\}$ But I do not know how to continue this idea.
1
vote
2answers
28 views

Let $F$ and $K$ be fields. If $F\subseteq K$ and $r\in K$ s.t. $r^2$ is algebraic over $F$. Then $r$ is algebraic over $F$.

Let $F$ and $K$ be fields. If $F\subseteq K$ and $r\in K$ s.t. $r^2$ is algebraic over $F$. Then $r$ is algebraic over $F$. Assume polynomial $p(x)\in F[x]$ s.t. $p(r^2)=0$ If $r\in K$ and $r^2$ is ...
1
vote
3answers
381 views

Each element is a square of some element

I have to show that each element of $\mathbb{F}_{2^n}$ is a square of some element. Could you give me some hints how I could do that??
1
vote
1answer
74 views

Showing $\mathbf Q(\sqrt2,\mathrm i)=\mathbf Q( \sqrt2+\mathrm i)$.

How can we prove that $\mathbf Q(\sqrt2,\mathrm i)=\mathbf Q( \sqrt2+\mathrm i)$? I have never seen the $\mathbf Q(\sqrt2, \mathrm i)$ notation before, so I am confused as to what it means - is it ...
0
votes
0answers
31 views

Automorphisms of the field of real numbers

How many automorphisms of the field of real numbers are there? If we use only algebra not analysis (analysis means every real number is a limit of some sequence of rational numbers and algebra means ...
0
votes
2answers
38 views

If $p$ and $q$ are prime numbers and $m\gt n$ show that $\sqrt[m]{p}\notin \mathbb Q(\sqrt[n]{q})$

If $p$ and $q$ are prime numbers and $m\gt n$ show that $\sqrt[m]{p}\notin \mathbb Q(\sqrt[n]{q})$ I really have no idea how to prove this problem. I started to consider: Assume $\sqrt[m]{p}\in ...
2
votes
2answers
28 views

Dimension Field True/False.

I'm having trouble approaching how to determine truthfulness and falsehood of the following type of problems. $F$ and $K$ are fields. 1) Suppose that $F\subseteq K$ and $r\in K$. If $[F(r):F]=4$ ...
4
votes
0answers
33 views

Subfield Criteria - Proof or Counterexample

I am interested in whether the following claim is true for all fields $F$: Conjecture: A subset $X\subset F$ is a subfield if and only if (1) $1\in X$, (2) $x,y\in X\Rightarrow x-y\in X$; and (3) ...
2
votes
1answer
61 views

Proof or counterexample: If $F\subseteq K$ and $r\in K$. If $[F(r):F]=4$ then $F(r)=F(r^3)$.

$F$ and $K$ are fields. Proof or counterexample: If $F\subseteq K$ and $r\in K$. If $[F(r):F]=4$ then $F(r)=F(r^3)$. I think I need to find a polynomial in $F(r^3)[x]$ that has $r$ as a root. I ...
0
votes
0answers
21 views

Show that it is/is not a normal extension

Let $a \in \mathbb{R}$ with $a^4=5$. Show that: $\mathbb{Q}(ia^2)$ is a normal extension of $\mathbb{Q}$. $\mathbb{Q}(a+ia)$ is a normal extension of $\mathbb{Q}(ia^2)$ $\mathbb{Q}(a+ia)$ is not a ...
0
votes
1answer
58 views

is $K = \mathbb{Z}[x] / (x^4+9x+6)$ a field or not?

Let $p(x)=x^4+9x+6$. it is irreducible over $\mathbb{Z}[x]$ by Eisenstein criterion(Because $3^2$ does not divide 6). My question is whether $K = \mathbb{Z}[x] / (p(x))$ is a field or not ? There is ...
17
votes
2answers
1k views

Are there number systems corresponding to higher cardinalities than the real numbers?

As most of you know, the set $\omega$ with cardinality $\aleph_0$ corresponds to what we normally know as the natural numbers $\mathbb{N}$, and the set $\mathcal{P}(\omega)$ with cardinality ...
2
votes
2answers
68 views

Show that $\mathbb{Q}( \sqrt2) \neq \mathbb{Q}( \sqrt3)$

The way that I'm thinking is by showing that the field extension $\mathbb{Q}( \sqrt2) /( \sqrt3) \neq \mathbb{Q}( \sqrt3)$, but is there a simpler way I'm ignoring?
0
votes
1answer
15 views

Notation question regarding field extensions (What does $K^2 \subseteq k$ mean)

recently I am reading a paper on pfister forms in characteristic 2 and stumbled across a notation I do not know. It can be found here Suppose $k$ is an arbitrary field of characteristic 2. Let ...
2
votes
1answer
19 views

Do two isomorphic finite field extensions have the same dimension?

If $E = F(u_1, \cdots u_n) \cong \bar{E} = F(v_1, \cdots v_m)$ then do the two extensions necessarily have the same dimension over $F$?
2
votes
1answer
36 views

Questions on the field extension $K = \mathbb{Q}[x]/\langle x^2 − 5\rangle$

Given the field extension $K = \mathbb{Q}[x]/\langle x^2 − 5\rangle$ of $\mathbb{Q}$, and letting $a = [x] ∈ K$; 1) Show $K ≃ \mathbb{Q}(\sqrt5) $ and $[K : \mathbb{Q}] = 2.$ 2)Find the ...
2
votes
3answers
58 views

Explain to me the difference between the notation $\mathbb{Q}( \sqrt2) $and$ \mathbb{Q}[ \sqrt2]$

Please explain to me the difference between the notation $\mathbb{Q}( \sqrt2) $and$ \mathbb{Q}[ \sqrt2]$. I know that these two fields are equal. But what difference do the different brackets imply? ...
3
votes
1answer
41 views

Universal property of the algebraic closure of a field

At page 4 of Strom's "Modern Classical Homotopy Theory" there is a universal formulation of the algebraic closure of a field. You can read it here from google books. Exercise 1.2a is then to convince ...
0
votes
0answers
15 views

Normal basis in every subgroup

Let $K$ be finite Galois extension of $k$, $G=Gal(K/k)$ and $k$ infinite. Prove: Exists $\beta\in K$ such that for every $N<G$, we have $\{\sigma(\beta):\ \sigma\in N\}$ is a basis for ...
0
votes
2answers
19 views

The field closure of a countable union of countable fields is countable?

If $ K_1 \subset K_2\cdots \subset K_n \subset \cdots$ is a tower of countable fields then their union $ \bigcup_n K_n$ is a countable field. If $\{K_a\}$ is a countable family, but not a tower, ...
2
votes
0answers
32 views

element in field, not a square [duplicate]

I am doing a specific exercise where the quaternion group is realised as a Galois group of some field extension. It goes like this: let $K = \mathbb Q(\sqrt 2,\sqrt 3)$ and $\alpha = (2 +\sqrt 2)(3 ...
0
votes
1answer
19 views

Are simple extensions of isomorphic fields isomorphic?

If $F(u) \cong F(v)$, and both are subfields of a larger extension, $K$, then for $k \in K$, is $F(u)(k) \cong F(v)(k)$?
0
votes
1answer
30 views

Field, algebraic element

1) Let $E/F$ an extension and let $\alpha,\beta\in E$ be algebraic elements over $F$. If $\alpha\neq 0$, prove that $\alpha+\beta$, $\alpha\beta$ and $\alpha^{-1}$ are all algebraic over $F$. 2) If ...
0
votes
0answers
13 views

Trace Map and Solution to Equation

Consider the equation $x^{31} - 1 =0$. Determine the number of solutions $\gamma \in \mathbb{F}_{1024}$ to the equation that satisfies $Tr(\gamma)=0$. I did some research on properties of field ...
0
votes
1answer
24 views

Splitting field of $x^ {a^n}$ −1 in Z/aZ[x]

What is the splitting field of the polynomial $x^ {a^n}$ −1 in \ the ring Z/aZ[x] with n natural? I´m working in some kind of proof of the best known theorem that says it´s impossible there exist a ...
3
votes
0answers
74 views

What branch of math is this?

In this paper: http://arxiv.org/pdf/hep-th/0505016v1.pdf what are the branch(es) of math being used? The unnumbered eq. on the top of page 3 and eq. (7) are good examples. All I've been able to figure ...
0
votes
2answers
41 views

Let k be a field and let p, q ∈ N be two prime numbers such that p · 1 = q · 1 = 0. Show that p = q.

My current train of though is letting p =/= q then proving that q must be divisible by p, the contradiction then being that q is prime. But I'm not sure how to go about doing this.
-1
votes
1answer
39 views

Help with problem of the book Algebra of T. Hungerford

This is a problem in “Algebra” by T. Hungerford: If $|K|=q$ and $f\in K[x]$ is irreducible, then $f$ divides $x^{q^n}-x$ if and only if $\deg f$ divides $n$. I found it difficult to solve ...
0
votes
1answer
29 views

Show that the Frobenius homomophism is bijective.

Let k be a finite field with characteristic $p ≥ 2$, show that the homomorphism $F:k\rightarrow k$ where $x \mapsto x^p$ is bijective. Could someone please explain the statement $p ≥ 2$ to me, as ...
0
votes
1answer
25 views

Showing two field extensions (of Q) are isomorphic using primitive elements, and why every element is primitive

I'm taking a class on Galois Theory in another language and the prof is saying my answer on this is incorrect and I'm wondering why, particularly since sometimes there's a language barrier. Basically ...
0
votes
2answers
40 views

Is $x^2-2$ irreducible over R and Q?

I'm not sure if it is valid to say that $x^2 - 2$ can be factorised to $2\cdot\left(\frac 12x^2 - 1\right)$ for it to be reducible in Q. Though I know $(x + \sqrt{2})(x - \sqrt{2})$ works in the ...
0
votes
1answer
27 views

Show that $x^3 + 3x^2 + 9x + 3$ and $x^3 + 3x^2 + 3x − 4$ are irreducible in $\mathbb{Z}[x]$

I need to show, as stated in the title, that $x^3 + 3x^2 + 9x + 3$ and $x^3 + 3x^2 + 3x − 4$ are irreducible in $\mathbb{Z}[x]$ I know that in case of second polynomial, if $f(x-1) = x^3 - 5$ which ...
1
vote
1answer
25 views

Algebraic closure.

I was trying to solve this: Let $E$ be a extension field of $K$. If $K$ is a algebraically closed field, then the algebraic closure $A$ of $K$ on $E$ is a algebraically closed field. But for ...
1
vote
1answer
19 views

Ruler and compass constructions and fields

Use the fact that $\alpha=$cos$(2\pi/5)$ satisfies the equation $x^2+x-1=0$ to conclude that the regular $5$-gon is constructible by straightedge and compass. So, the polynomial $x^2+x-1$ is ...
3
votes
1answer
30 views

Showing that a polynomial over subring is reducible

Suppose that $R_1\subseteq R_2$, and both are integral domains. Further suppose that $R_2$ is a field, where each element $r\in R_2$ is a zero of a polynomial in $R_1[x]$ with the leading coefficient ...
0
votes
1answer
21 views

quadratic equation modulo some number

I read a post that $$ax^2+bx+c \equiv 1 \pmod p$$ can be solved in a similar way we solve a simple quadratic equation, just by replacing division by $2a$ by modulo inverse of $2a$ and square root of ...
1
vote
1answer
25 views

How to find the splitting field of $X^4-10X^2+1$?

How to find the splitting field of $X^4-10X^2+1$ ? I found the roots \begin{align*} X^4-10X^2+1=0&\iff (X^2-5)^2-24=0\\ &\iff X^2-5=\pm 2\sqrt 6\\ &\iff X^2=5\pm 2\sqrt 6\\ &\iff ...