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 topic....

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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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Galois Group of $(x^3-5)(x^2-3)$

I am having some trouble calculating the Galois group (over $\mathbb{Q}$) of $(x^3-5)(x^2-3)$. I can see the splitting field is $F:=\mathbb{Q}(\sqrt[3]{5},\omega,\sqrt{3})=\mathbb{Q}(\sqrt[3]{5},i,\...
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3answers
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When is $\mathbb{F}_p[x]/(x^2-2)\simeq\mathbb{F}_p[x]/(x^2-3)$ for small primes?

I've been considering the rings $R_1=\mathbb{F}_p[x]/(x^2-2)$ and $R_2=\mathbb{F}_p[x]/(x^2-3)$, where $\mathbb{F}_p=\mathbb{Z}/(p)$. I'm trying to figure out if they're isomorphic (as rings I ...
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3answers
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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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1answer
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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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2answers
261 views

Can all polynomials of a given degree be reducible?

Let $n > 1$ be a fixed integer. Does there exist a field $F$ with the following properties? $F$ is not algebraically closed. Every polynomial $f(x) \in F[X]$ of degree $n$ is reducible. I ...
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1answer
224 views

Is $\operatorname{Gal}(\mathbb{Q}_p^{un})\cong \hat{\mathbb{Z}}$?

Is the absolute Galois group of $\mathbb{Q}_p^{un}$ the profinite completion of $\mathbb{Z}$? I was never quite sure... In similar cases, it is true. Namely, $\mathbb{C}((t))$ does have absolute ...
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1answer
204 views

Are there nonisomorphic fields with isomorphic multiplicative groups?

This is false for finite fields as the multiplicative groups of finite fields are cyclic, and different cardinalities yield cyclic groups of different cardinalities. But I'm unsure how to proceed for ...
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1answer
160 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 (F^...
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2answers
272 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 $[\mathbf{C}:\mathbf{R}]=2$...
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3answers
358 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 $\mathbb{F}_p((X))$?
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2answers
498 views

Necessarily a field between a field and its algebraic extension

This is an exercise in some textbooks. Let $E$ be an algebraic extension of $F$. Suppose $R$ is ring that contains $F$ and is contained in $E$. Prove that $R$ is a field. The trouble is really with ...
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1answer
484 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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2answers
990 views

Is there an explicit embedding from the various fields of p-adic numbers $\mathbb{Q}_p$ into $\mathbb{C}$?

For any field of p-adic numbers $\mathbb{Q}_p$, one can construct the field $\mathbb{C}_p$, the metric completion of one of its algebraic completions. By the axiom of choice, we can prove this to be ...
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1answer
402 views

Faulty definition of a field in Curtis' Abstract Linear Algebra?

On pp.2-3 of Curtis, Abstract Linear Algebra, he gives a definition of a field which seems to fail to exclude a pathological example. He says a field is a set k with two operations (a+b) and (ab) such ...
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2answers
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Splitting field of a separable polynomial is separable

Probably a stupid question, but.. Why is the splitting field of a separable polynomial necessarily separable? Thanks. Follow up question Show that if $F$ is a splitting field over $K$ for $P \in K[...
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votes
1answer
426 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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1answer
159 views

Set of elements of degree $2^n$ over a base field is itself a field

Let $F \subset L$ be two fields, and define $K = \{\alpha \in L\mid [F(\alpha): F] \text{ is a power of 2} \}$. Our problem is to prove that $K$ is a field. Closure under reciprocation is easy (...
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1answer
611 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 $K$, $K(\...
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1answer
571 views

Perfect closure is perfect

I've been self-studying inseparable extensions and there's something that seems obvious to everybody but not to me. Let's clear out some definitions that are not so universal: Let $K$ be a field ...
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2answers
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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} 4\...
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1answer
219 views

Other Euler characteristics?

At the end of V.3.4 in Algebra: Chapter 0, Aluffi describes the construction of a Grothendieck group over the category of finite dimensional $\operatorname{k}$-vector spaces, $K(\operatorname{k-vect^f}...
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2answers
414 views

Analog to the primitive element theorem for transcendental extensions?

The Primitive Element Theorem states that if $E/F$ is a finite separable field extension, then there exists an element $a$ such that $E=F(a)$. There's a similar result I found, that I don't quite ...
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1answer
303 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
580 views

Proof that a finite separable extension has only finite many intermediate fields

Let $E/F$ be finite separable extension. Is there any proof of the fact that there are only finitely many intermediate fields without using primitive element theorem or fundamental theorem of Galois ...
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4answers
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Is $\mathbb{Q}[2^{1/3}]$ a field?

Is $\mathbb{Q}[2^{1/3}]=\{a+b2^{1/3}+c2^{2/3};a,b,c \in \mathbb{Q}\}$ a field? I have checked that $b2^{1/3}$ and $c2^{2/3}$ both have inverses, $\frac{2^{2/3}}{2b}$ and $\frac{2^{1/3}}{2c}$, ...
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4answers
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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 ...
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3answers
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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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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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3answers
874 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 are $...
9
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1answer
814 views

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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2answers
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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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5answers
139 views

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 $\sqrt{3}$...
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2answers
222 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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2answers
551 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 $\sqrt[8]{3},\sqrt[8]{3}...
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5answers
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What is the main difference between a vector space and a field?

In my opinion both are almost same. However there should be some differenes like any two elements can be multiplied in a field but it is not allowed in vector space as only scalar multiplication is ...
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1answer
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Galois Group of $\sqrt{2+\sqrt{2}}$ over $\mathbb{Q}$

So I want to show that $\mathbb{Q}(\sqrt{2+\sqrt{2}})$ is Galois over $\mathbb{Q}$ and determine its Galois group. My thoughts are as follows: Define $\alpha := \sqrt{2+\sqrt{2}}$. Then it is ...
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1answer
472 views

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

Let $x$ be an element not algebraic over $F$, and $K \subset F(x)$ a subfield that strictly contains $F$. Why is $x$ algebraic over $K$? Thanks a lot!
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1answer
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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 |...
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1answer
280 views

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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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
458 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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2answers
267 views

On irreducible factors of $x^{2^n}+x+1$ in $\mathbb Z_2[x]$

Prove that each irreducible factor of $f(x)=x^{2^n}+x+1$ in $\mathbb Z_2[x]$ has degree $k$, where $k\mid 2n$. Edit. I know I should somehow relate the question to an extension of $\mathbb Z_2$ of ...
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1answer
258 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
224 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
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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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4answers
404 views

Showing that $R(x)$ is a proper subset of $R((x))$ if $R$ is a field

I would like to show that if $R$ is a field, then $R(x)$ is a proper subset of $R((x))$, where $R(x)$ is the ring of rational functions, and $R((x))$ is the ring of formal Laurent series. If $f \in ...
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2answers
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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
584 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 $\mathbb{Q}(\sqrt[...
9
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4answers
114 views

Subgroup generated by $1 - \sqrt{2}$, $2 - \sqrt{3}$, $\sqrt{3} - \sqrt{2}$

For a number field $K$, Dirichlet's unit theorem says that$$(O_K)^\times = \mathbb{Z}^{r - 1} \oplus (\text{a finite cyclic group}),$$where $r$ is the number of all infinite places of $K$. An infinite ...