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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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 ...
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514 views

Polynomial irreducible - maximal ideal

I have a couple of ideals which I wonder if I correctly classify as maximal/prime ideal. $I_1 = \langle 2x^2 + 9x -3\rangle$, $I_2 = \langle x - 1\rangle$ $\mathbf 1)$ Is $I_1$ a maximal ideal in ...
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Why is it called a 'ring', why is it called a 'field'?

The definitions of 'ring' and 'field' are pretty straightforward. For a ring (e.g. integers): addition is commutative $( 1 + 2 = 2 + 1 )$ addition and multiplication are associative $(2 +(2+2)) = ...
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When is a Morphism between Curves a Galois Extension of Function Fields

My apologies if this question has already been answered somewhere on this site: when I searched, I could only find specific examples rather than the general question. It is known that the category of ...
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2answers
537 views

Is $\sqrt{2}\in\mathbb{Q}(\sqrt[8]{3})$ or not?

My hunch is that $\sqrt{2}\not\in\mathbb{Q}(\sqrt[8]{3})$. For practice, I want to compute the splitting field and its degree of $x^8-3$ over $\mathbb{Q}$. I know the roots are ...
9
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1answer
147 views

Does there exist a field $(F,+,*)$ so that $(F,+) \cong (F^*,*)$?

This question occurred to me earlier today. I can see that if the field has a unit, then there is an element of multiplicative order $2$, namely $-1$. Thus if there was an isomorphism $(F,+) \cong ...
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2answers
252 views

This tower of fields is being ridiculous

Suppose $K\subseteq F\subseteq L$ as fields. Then it is a fact that $[L:K]=[L:F][F:K]$. No other hypotheses are needed (I'm looking at you, Hungerford V.1.2). Now obviously ...
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Embedding of a field extension to another

Can $\mathbb{Q}(\sqrt {-2})$ be embedded into a cyclic extension of degree 4 over $\mathbb{Q}$?
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349 views

Reducibility of $P(X^2)$

This question is inspired by a comment discussion in If $K=K^2$ then every automorphism of $\mbox{Aut}_K V$, where $\dim V< \infty$, is the square of some endomorphism.. Let $k$ be a field of ...
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2answers
434 views

Why is a variety over a non-alg. closed field a hypersurface?

Exercise $3$ on page $8$ of Kunz's Introduction to Commutative Algebra and Algebraic Geometry is as follows: If the field $K$ is not algebraically closed, then any $K$-variety $V \subset A^n(K)$ can ...
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1answer
246 views

Why doesn't this hold for $p=2$?

I have a question about the following lemma: Assume that the characteristic of $F$ is $p$ and $p>2$. Then $(t^m-1)/(t^n-1)$ is a square in $F[t, t^{-1}]$ if and only if $(\exists s \in ...
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1answer
340 views

Find intermediate fields of $\mathbb{Q}(\sqrt[3]{2}, \sqrt{3},i) \, | \, \mathbb{Q}(i)$

This is the problem I am facing: Compute the intermediate fields of the extension $K | \mathbb{Q}(i)$ where $K = \mathbb{Q}(\sqrt[3]{2}, \sqrt{3},i)$ and find the intermediate fields $M$ such ...
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1answer
196 views

Complete ordered field is an Archimedean field that cannot be extended to an Archimedean field

As a bonus problem, our professor of real analysis asked us to prove that the real numbers (a complete ordered field) cannot be extended into an Archimedean field, with no definition of what he meant ...
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1answer
222 views

How to compute the Galois group of $x^5+99x-1$ over $\mathbb{Q}$?

I was stuck trying to compute the Galois group of $x^5 + 99x -1$. The problem asks to compute the Galois group over $\mathbb{F}_2, \mathbb{F}_3, \mathbb{F}_5, \mathbb{F}_{11}$ and $\mathbb{Q}$. I was ...
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2answers
270 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 ...
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2answers
549 views

$\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})=\mathbb{Q}$?

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 ...
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526 views

Can a field be isomorphic to its subfield?

Let $K$ be a field and $K(X)$ be the field of its rational functions. Now let $\phi \in K(X)$ be a rational function such that $K(\phi) \neq K(X)$. Now, since $\phi$ is transcendental over $K$, ...
9
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1answer
109 views

Help understanding fields.

Hi guys I have a test this tuesday and I am given practice questions to do , and I have trouble understanding fields. Like I know by definition what they are, but applying them is kind of confusing. ...
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2answers
386 views

The Noether-Deuring Theorem

I have to solve the following exercise taken from the book "Introduction to Representation Theory" by P. Etingof, O. Golberg, S. Hensel, T. Liu, A. Schwendner, D. Vaintrob, E. Yudovina and S. ...
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4answers
280 views

Hyperreal field extension

In non-standard analysis, assuming the continuum hypothesis, the field of hyperreals $\mathbb{R}^*$ is a field extension of $\mathbb{R}$. What can you say about this field extension? Is it ...
9
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1answer
229 views

Question about a property certain algebraic extensions $E/K$ (not necessarily separable) have.

A few days ago I found this question here on math.stackexchange, which gave a sufficient criterion for a separable, algebraic extension $E/K$ to be an algebraic closure of $K$. However it was claimed ...
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333 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 ...
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1answer
340 views

What is the overall idea of Galois theory?

I am a third year undergraduate, doing a course on Field and Galois theory. Now, while I seem to understand most of the concepts locally, I do not seem to get the 'Whole picture' of what is happening ...
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2answers
782 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} ...
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1answer
261 views

Is it possible to do calculus on any field with a topology?

I'll try to make my point clear: when we consider the field of complex numbers $\mathbb{C}$ we can do calculus there because we have properties of a field and in the same time we have a topology to ...
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0answers
204 views

Relationship between intersection and compositum of fields

This issue came up in a number theory lecture today. Let $K$ be a number field and let $L/K$ be an abelian (finite Galois) extension. Then there exists a primitive $m$th root of unity $\zeta_m$ so ...
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3answers
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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 ...
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5answers
506 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 ...
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If $A$ an integral domain contains a field $K$ and $A$ over $K$ is a finite-dimensional vector space, then $A$ is a field. [duplicate]

Possible Duplicate: Proof that an integral domain that is a finite-dimensional $F$-vector space is in fact a field I need to prove this result, but the only starting point I think of is to ...
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632 views

Show that $\pi \notin Q(\pi^3)$

As the title says. I think a proof by contradiction is the most natural thing. Suppose $\pi \in Q(\pi^3)$. Then \begin{equation} \pi = ...
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3answers
261 views

Can you construct a field over every set $M$?

I know there are finite fields like $\mathbb F_2$, $\mathbb F_4$ or the $\mathbb Z/n\mathbb Z$ for prime $n$ with modulo operations. For other special $n$, I've seen fields $\mathbb F_n$ with $n$ ...
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1answer
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Is $\mathbb Q_r$ algebraically isomorphic to $\mathbb Q_s$ while r and s denote different primes?

It is obvious that $\mathbb{Q}_r$ is topologically isomorphic to $\mathbb Q_s$ while $r$ and $s$ denote different primes. But I really don't know whether it is true in the aspect of algebra. As I ...
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Two finite fields with the same number of elements are isomorphic

Fraleigh(7ed) Theorem33.12. Let $p$ be a prime and let $n\in\mathbb{Z}^+$. If $E$ and $E'$ are fields of order $p^n$, then $E \simeq E'$. Proof in the text: Both $E$ and $E'$ have ...
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2answers
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Constructing a Galois extension field with Galois group $S_n$

Constructing a Galois extension field $E$ with $Gal(E/F)= S_n$ How do I construct one?
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Does there exist a field $K$ such that $\mathbb R \subsetneq K \subsetneq \mathbb C$?

I'm thinking of unions of $\mathbb R$ with some subset of $\mathbb C$ but am not sure how to approach this without ending up with all of $\mathbb C$. Doe anyone have any suggestions?
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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 ...
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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] = ...
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Show that an algebraically closed field must be infinite.

Show that an algebraically closed field must be infinite. Answer If F is a finite field with elements $a_1, ... , a_n$ the polynomial $f(X)=1 + \prod_{i=1}^n (X - a_i)$ has no root in F, so F ...
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2answers
947 views

How can one express $\sqrt{2+\sqrt{2}}$ without using the square root of a square root?

I was trying to review some analysis, and came across problem 3 from page 78 of Walter Rudin's Principles of Mathematical Analysis. As part of the problem, I wanted to try to write ...
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608 views

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

Is $\mathbb Q(\pi,i\pi):\mathbb Q$ a simple extension?
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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} : ...
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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 $K(a)=K(b)$. ...
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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 ...
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315 views

Good undergraduate level book on Cyclotomic fields

I have Lang's 2 volume set on "Cyclotomic fields", and Washington's "Introduction to Cyclotomic Fields", but I feel I need something more elementary. Maybe I need to read some more on algebraic number ...
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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 ...
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2answers
206 views

Non-algebraically closed field in which every polynomial of degree $<n$ has a root

My problem is to build, for every prime $p$, a field of characteristic $p$ in which every polynomial of degree $\leq n$ ($n$ a fixed natural number) has a root, but such that the field is not ...
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Unique quadratic subfield of $\mathbb{Q}(\zeta_p)$ is $\mathbb{Q}(\sqrt{p})$ if $p \equiv 1$ $(4)$, and $\mathbb{Q}(\sqrt{-p})$ if $p \equiv 3$ $(4)$

I want to prove the assertion: The unique quadratic subfield of $\mathbb{Q}(\zeta_p)$ is $\mathbb{Q}(\sqrt{p})$ when $p \equiv 1 \pmod{4}$, respectively $\mathbb{Q}(\sqrt{-p})$ when $p \equiv 3 ...
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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 ...
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144 views

Can we make $\mathbb{Z}$ into a field?

This is probably an elementary question about fields, but I think it is a little tricky. Can we make the integers $\mathbb{Z}$ into a field? Let me be more precise. Is it possible to make ...
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3answers
524 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 $F(\sqrt{2})={\Bbb R}$ ...