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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Why is $\{a + b\sqrt2 + c\sqrt3 : a\in\Bbb{Z}, b, c \in\Bbb{Q}\}$ not closed under multiplication?

The set $R = \{a + b\sqrt{2} + c\sqrt{3}: a \in \Bbb{Z}, c, b \in \Bbb{Q}\}$ is not closed on multiplication, my textbook states. Why is this? And related to that: why then is $S = \{a + b\sqrt{2} : ...
7
votes
2answers
142 views

Galois group of the splitting field of the polynomial $x^5 - 2$ over $\mathbb Q$

We know that the splitting field of $x^5 - 2 $ over $\mathbb Q$ is $\mathbb Q(2^{1/5}, \rho)$, where $\rho$ is a fifth root of unity. Therefore, $\left[\mathbb Q(2^{1/5} , \rho) : \mathbb Q \right] = ...
7
votes
3answers
956 views

Help with proof that $\mathbb Z[i]/\langle 1 - i \rangle$ is a field.

I have been having a lot of trouble teaching myself rings, so much so that even "simple" proofs are really difficult for me. I think I am finally starting to get it, but just to be sure could some one ...
7
votes
2answers
1k views

Construct algebraic closure of $\mathbb{Q}$

In abstract algebra lecture, the lecturer wants to construct an algebraic closure of $\mathbb{Q}$. The construction is as follow: Suppose $\mathbb{Q_1}=\mathbb{Q}$. Let $\mathbb{Q}_2$ be the set ...
7
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3answers
401 views

Do maximal proper subfields of the real numbers exist?

To clarify the problem, consider the field ${\Bbb R}$ as a field extension of ${\Bbb Q}$ using some sort of Hamel basis. Does there exist a field $F\subsetneq{\Bbb R}$ such that $F(\sqrt{2})={\Bbb R}$ ...
7
votes
2answers
218 views

Explicit Galois theory computation in cyclotomic field

Let $p$ be an odd prime. Let $\zeta=e^{\frac{\pi i}{4 p}}$, thus $\zeta$ is a primitive $8p$-th root of unity. There is a unique $\mathbb Q$-automorphism $\tau$ of the number field ${\mathbb ...
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2answers
534 views

Determining subfields of $\mathbb{Q}(\zeta_3,\sqrt[3]{2})$ without Galois theory?

Somehow I had convinced myself recently that one could determine the subfields of $\mathbb{Q}(\zeta_3,\sqrt[3]{2})$ (here $\zeta_3 = \frac{-1+\sqrt{3}i}{2}$ is a primtive $3$rd root of $1$) without ...
7
votes
1answer
365 views

For field extensions $F\subsetneq K \subset F(x)$, $x$ is algebraic over $K$

Let $x$ be an element not algebraic over $F$, and $K \subset F(x)$ a subfield that differs from $F$. Why is $x$ algebraic over $K$? Thanks a lot!
7
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1answer
823 views

Need help understanding separable polynomials

I have the following definition of what it means for a polynomial to be separable: Let $E$ be a field and $P \in E[x]$ be irreducible. Then we say $P$ is separable if it has no repeated root in ...
7
votes
1answer
103 views

Short method to prove irreduicibility of $x^7+21x^5+35x^2+34x-8$ over $\Bbb Q$?

I am given a task to prove that polynomial $f=x^7+21x^5+35x^2+34x-8$ is irreducible over $\Bbb Q$? In my algebra course we learnt reduction and Eisenstein criterion. Eisenstein doesn't seem to work ...
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votes
2answers
480 views

Elementary proof of $\mathbb{Q}(\zeta_n)\cap \mathbb{Q}(\zeta_m)=\mathbb{Q}$ when $\gcd(n,m)=1$.

In an answer to another question I used the fact that $\mathbb{Q}(\zeta_m)\subseteq \mathbb{Q}(\zeta_n)$ if and only if $m$ divides $n$ (here $\zeta_n$ stands for a primitive $n$th root of unity, ...
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2answers
921 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 ...
7
votes
1answer
122 views

Infinite $p$-extension contains $\mathbb{Z}_p$-extension

Does the Galois group of every infinite $p$-extension $K$ of a number field $k$ contain a (closed) subgroup such that the quotient group is isomorphic to $\mathbb{Z}_p$? My feeling is "yes", but I'm ...
7
votes
1answer
139 views

Extension degree of residue field.

Let $k$ be a field, and $A$ be a finitely generated $k$-algebra with $\text{dim}(A)\leq 1$. Then for any maximal ideal $\mathfrak{m}$ of $A$, does this inequality $[A/\mathfrak{m}:k]<\infty$ hold? ...
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3answers
478 views

Subextension of a finitely generated extension of fields

If $E/K$ is a finitely generated field extension and $F$ is an intermediate field how can I prove that $F/K$ is finitely generated?
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514 views

Derivations in a ring. What applications do they have outside algebra?

INTRODUCTION Let $R$ be any ring, possibly without unity. We call a function $d:R\longrightarrow R$ a derivation on $R$ if it satisfies the following conditions. $(1)$ It is an endomorphism of the ...
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1answer
220 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 ...
7
votes
1answer
214 views

A shortcut in Galois theory

How could we prove Galois correspondence without using Dedekind’s Lemma on group characters, Artin’s lemma and the primitive element theorem ? I just came across Meinolf Geck's article On the ...
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3answers
324 views

Why are the surreals considered “recreational” mathematics?

One of my college math professors once remarked to me that it was interesting that John Conway's two "biggest" contributions to math were both recreational: the Game of Life and the Surreals. No one, ...
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3answers
199 views

Proving that $X^6-15X^4-6X^3+75X^2-90X-116$ is irreducible over $\mathbb Q$

When asked to find the minimal polynomial of $\sqrt[3]{3}+\sqrt[2]{5}$ over $ \mathbb Q$, I easily found out that $X^6-15X^4-6X^3+75X^2-90X-116$ has $\sqrt[3]{3}+\sqrt[2]{5}$ as a root. It's very ...
7
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1answer
143 views

A first order theory whose finite models are exactly the $\Bbb F_p$

Is there a first order theory $T$ in the language of rings such that its finite models are exactly the fields $\Bbb F_p$ with $p$ prime (but no $\Bbb F_q$ with $q$ a proper prime power is a model of ...
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2answers
130 views

Equivalent definition of algebraically closed

In Hungerford's Algebra text, it is stated that a field $K$ is algebraically closed iff there exists a subfield $F$ such that $K$ is algebraic over $F$ and all polynomials in $F[x]$ split in $K[x]$. ...
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1answer
146 views

prove that the field extension is cyclic

Let's define the sequence $x_0=0$ and $ x_{i+1} = \sqrt{x_i+2}$ taking always the positive root. Prove that the field extension $\Bbb Q \subset \Bbb Q(x_i) $ is cyclic with degree $2^i$ Well.. at ...
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votes
2answers
344 views

Is a completion of an algebraically closed field with respect to a norm also algebraically closed?

Assume we have an algebraically closed field $F$ with a norm (where $F$ is considered as a vector space over itself), so that $F$ is not complete as a normed space. Let $\overline F$ be its completion ...
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votes
2answers
163 views

What's the theoretical basis for integration using partial fractions?

Exercises involving integration using partial fractions depend on expressing a rational function $\frac{P(x)}{Q(x)}$ (where the degree of $P$ is less than the degree of $Q$) as a sum of ...
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votes
2answers
101 views

Euclid and finite fields

In 300 BC or so Euclid pointed out that if $S$ is any finite set of prime numbers then the prime factors of $1+\prod S$ are not in $S$, so that $S$ can always be extended to a larger finite set. Much ...
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2answers
182 views

If $E/F$ is algebraic and every $f\in F[X]$ has a root in $E$, why is $E$ algebraically closed?

Suppose $E/F$ is an algebraic extension, where every polynomial over $F$ has a root in $E$. It's not clear to me why $E$ is actually algebraically closed. I attempted the following, but I don't think ...
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3answers
174 views

Every element is radical in a field extension.

Let $L/K$ be an algebraic field extension. Suppose for each $x\in L$, there exists an integer $n>0$ such that $x^n\in K$, where $n$ may depend on $x$. If the characteristic of $K$ is zero, does it ...
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2answers
182 views

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 ...
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2answers
130 views

Showing $\mathbb{Q}(2^{1/3}+2^{1/5})=\mathbb{Q}(2^{1/3},2^{1/5})$

In order to prove $[\mathbb{Q}(2^{1/3}+2^{1/5}):\mathbb{Q}]=15$, I want to show $\mathbb{Q}(2^{1/3}+2^{1/5})=\mathbb{Q}(2^{1/3},2^{1/5})$. Any suggestions?
7
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1answer
528 views

An exercise on cyclic extensions of Hungerford's book, Algebra.

I'm trying to show the following exercise of the book, it is in the section of cyclic extensions and says the following: Let $\overline{\mathbb Q}$ be a fixed algebraic closure of $\mathbb Q$, let ...
7
votes
1answer
169 views

What does this notation mean: $\displaystyle\lim_{\leftarrow} \,\mathbb{Z}/n\mathbb{Z}$?

Just a small notation question from this Wikipedia page: The absolute Galois group of a finite field $K$ is isomorphic to the group $$\hat{\mathbb{Z}}=\lim_{\leftarrow} \mathbb{Z}/n\mathbb{Z}.$$ ...
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1answer
319 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 ...
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1answer
239 views

How to show that $(\mathbb{Q}(\sqrt{2}))^{\times} \cong (\mathbb{Q}(\sqrt{3}))^{\times}$

How to show that $$(\mathbb{Q}(\sqrt{2}))^{\times} \cong (\mathbb{Q}(\sqrt{3}))^{\times}$$ where $(\mathbb{Q}(\sqrt{2}))^{\times}$ is multiplicative group of $\mathbb{Q}(\sqrt{2})$. Mapping ...
7
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3answers
408 views

Intersection of Cyclotomic Fields

How would I prove that $\mathbb{Q_m} \cap \mathbb{Q_n} = \mathbb{Q_{(m, n)}}$ (here $\mathbb{Q_n}$ denotes the $n$th cyclotomic field)? I already know of a solution involving the fact that given two ...
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2answers
273 views

Splitting field of $x^{13}+1$ over $\mathbb{Q}$

I want to know how to calculate the degree of the field extension $[K:Q]$ where $K$ is the splitting field of $x^{13}+1$ over $\mathbb{Q}$. I'm new to this area and this is not really covered in my ...
7
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1answer
252 views

Algebraic Extensions and Separability

I have been wrestling with the following problem for the past few hours, but I have made no progress whatsoever, namely: Let $K/k$ be an algebraic extension with characteristic $p>0$ and let ...
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2answers
230 views

Application of the Artin-Schreier Theorem

This is exercise $6.29$ out of Lang's book: Let $K$ be a cyclic extension of a field $F$, with Galois group $G$ generated by $\sigma$. Assume that the characteristic is $p$, and that ...
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1answer
140 views

Lower Bound on Number of Wildly Ramified Extensions of $\mathbb{Q}_p$

Suppose $p$ is prime and $p$ divides $e$ divides $n$. Typically up to isomorphism there are a lot of wildly ramified extensions of $\mathbb{Q}_p$ which have degree $n$ and ramification index $e$. ...
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386 views

Compositum of fields with trivial intersection

Let $E/F$ be a finite extension. Let $L,K$ be two intermediate fields with $L\cap K = F$, and also $$[L : F] [K:F] = [E:F].$$ Must it hold that the compositum $LK$ equals $E$? If we assume that $E/F$ ...
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1answer
131 views

“Real part” of a number field

Let ${\mathbb K} \subseteq {\mathbb C}$ be a finite extension of $\mathbb Q$, and let $n=[{\mathbb K} : {\mathbb Q}]$. Let $X_{\mathbb K}$ denote the set of all “components” (i.e., real and imaginary ...
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1answer
159 views

Exercise on finite intermediate extensions

Let $E/K$ be a field extension, and let $L_1$ and $L_2$ be intermediate fields of finite degree over $K$. Prove that $[L_1L_2:K] = [L_1 : K][L_2 : K]$ implies $L_1\cap L_2 = K$. My thinking ...
7
votes
1answer
636 views

$K/E$, $E/F$ are separable field extensions $\implies$ $K/F $ is separable

I wish to prove that if $K/E$, $E/F$ are algebraic separable field extensions then $K/F$ is separable. I tried taking $a\in K$ and said that if $a\in E$ it is clear, otherwise I looked at the minimal ...
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2answers
684 views

How to find irreducible polynomials over $\mathbb{Q}(i)$ with prescribed Galois group?

Here is my recent homework question: For each of the following five fields $F$ and five groups $G$, find an irreducible polynomial in $F[x]$ whose Galois group is isomorphic to $G$. If no example ...
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1answer
211 views

Calculating Separable Closures

In my study of fields, the notion of the separability of an algebraic field extension is one of the more slippery concepts I have encountered thusfar. What is particularly vexing to me is the notion ...
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1answer
97 views

Counting diagonalizable matrices in $\mathcal{M}_{n}(\mathbb{Z}/p\mathbb{Z})$

How many diagonalizable matrices are there in $\mathcal{M}_{n}(\mathbb{Z}/p\mathbb{Z})$ ? Where $p$ is a prime number. Attempt : By definition a matrix is called diagonalizable if there exists an ...
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1answer
137 views

Multivariate coprime polynomials in field extensions

Suppose $f$ and $g$ are polynomials in $n$ variables, $n\ge 2$, over a field $E$. Suppose further that $f$ and $g$ are relatively prime over $E$. If $F$ is a field extension of $E$, are $f$ and $g$ ...
7
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1answer
64 views

why similarity over $\bar{\mathbb{F}}$ of $A,B\in M_n(\mathbb{F})$ implies similarity over $\mathbb{F}$?

A clasic problem in linear algebra is to determine if two matrices $A,B\in M_n(\mathbb{F})$ are similar one to another. When $\mathbb{F}=\bar{\mathbb{F}}$, we know that $A,B$ are similar if and only ...
7
votes
1answer
123 views

Computing a uniformizer in a totally ramified extension of $\mathbb{Q}_p$.

Do you know how to compute a uniformizer of $\mathbb{Q}_p(\zeta_{p^n},p^\frac{1}{p})$? Where $\zeta_{p^n}$ is a primitive $p^n$-th root of 1 and $p$ is an odd prime.
7
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1answer
129 views

Should this be viewed as a serious issue with the meadow-theoretic approach?

Meadow theory (see here) allows us to apply the results and concepts of universal algebra to the study of fields. Obviously, this is very, very nice. However, I have the following issue with the ...