2
votes
0answers
142 views

$e=1$ in Theorem 30 from Marcus book “number fields”

Theorem 30 in Marcus book states that, if $p\in\mathbb Z$ is an odd prime and $q$ is a prime $\neq p$, then, fixing $d$ as a divisor of $p-1$ we have that $q$ is a $d$-th power $\operatorname{mod}q$ ...
1
vote
1answer
28 views

Showing the center of an endomorphism ring is a direct summand

I am reading A. Fröhlich's Formal Groups, and I am working on the proof that if $F$ is a formal group defined over a separably closed field $k$ of characteristic $p$, then the endomorphism ring $E$ of ...
7
votes
1answer
51 views

An extension of an algebraic number field which makes an integral ideal $I$, a principal ideal

I want to show that, given an ideal $I \subseteq \mathcal O_K$ (where $K/\mathbb Q$ is an algebraic number field), there is a finite extension $K'/K$ such that, $I\mathcal O_{K'}$ becomes a principal ...
0
votes
0answers
30 views

$A\alpha_1+\dots +A\alpha_n$ is a $B$-module

Let $A$ be an integrally closed domain with quotient field $K$ and $L/K$ a finite separable field extension. We denote with $B$ the integral closure of $A$ in $L$. Now let $\alpha_1,\dots,\alpha_n\in ...
1
vote
1answer
35 views

Localization of the Integer Ring

Let $\mathbb{Z}$ be the ring of integers and let $p$ be a prime, then the $p$-localization of $\mathbb{Z}$ is defined as $\mathbb{Z}_{(p)}=\{\displaystyle\frac{a}{b}|a,b\in\mathbb{Z},p\nmid b\}$. I ...
1
vote
1answer
46 views

When Is there no Local Power Integral Basis?

Let $A$ be a Dedekind domain, $K, L, B$ the usual designations of $A$'s quotient field, a finite separable extension of $K$, and integral closure of $A$ in $L$ respectively. If $\alpha \in B$ ...
1
vote
0answers
58 views

Can a ring of integers be free over a non-PID?

Let $K \subseteq L$ be an extension of number fields, and $A \subseteq B$ the corresponding rings of integers. $B$ is an $A$-module, generated by $[L : K]$ elements. If $K$ has class number one, ...
1
vote
1answer
22 views

Extensions of nonarchimedean valuations

This is a question from Janusz 'Algebraic Number Theory'. Let $R$ be a DVR with maximal ideal $\mathfrak p=\pi R$. Let $K$ be the quotient field of $R$ and $\mid\cdot\mid_{\mathfrak p}$ the ...
3
votes
1answer
39 views

Showing two numbers are relatively prime in number fields

Solve $x^3-2=y^2$ in integers. The standard way to solve this problem is to consider the arithmetic of the ring of algebraic integers $\mathcal{O}_{\mathbb{Q}(\sqrt{-2})}$ and to show that ...
23
votes
1answer
249 views

Does there exist two non-constant polynomials $f(x),g(x)\in\mathbb Z[x]$ such that for all integers $m,n$, gcd$(f(m),g(n))=1$?

Does there exist two non-constant polynomials $f(x),g(x)\in\mathbb Z[x]$ such that for all integers $m,n$, gcd$(f(m),g(n))=1$? I think there are no such polynomials, but how to prove?
3
votes
1answer
75 views

How to prove this innocent looking isomorphism

I've got a Dedekind domain $R$ with quotient field $K$, a non-zero prime ideal $P$ of $R$. I form the completion $\widehat{K}$ of $K$ wrt the valuation $v_P$ associated to $P$. Let $\widehat{R}$ be ...
5
votes
1answer
99 views

Units of $\mathbb{Z}[\sqrt[4]2]$

How would one compute the units in $\mathbb{Z}[\sqrt[4]2]$? According to one source, it can be shown that the fundamental units are $1 + \sqrt[4]2$ and $1 + \sqrt{2}$, but it does not specify the ...
3
votes
1answer
38 views

Examples where there is no power integral basis

Let $A$ be a completion discrete valuation ring with quotient field $K$, $L/K$ finite and separable, and $B$ the integral closure of $A$ in $L$. Let $P, \mathfrak P$ be the unique maximal ideals of ...
10
votes
2answers
275 views

Ring of integers is a PID but not a Euclidean domain

I have noticed that to prove fields like $\mathbb{Q}(i)$ and $\mathbb{Q}(e^{\frac{2\pi i}{3}})$ have class number one, we show they are Euclidean domains by tessalating the complex plane with the ...
1
vote
1answer
71 views

Why is the Ideal Norm Multiplicative?

I had asked this question before and got a partial answer. Let $A$ be a Dedekind domain with quotient field $K$, $L$ a finite separable extension of $K$ of degree $n$, and $B$ the integral closure of ...
3
votes
1answer
68 views

Determine Units of a Ring $\mathbb{Z}[\alpha]$

I am trying to determine the units of $Z[\alpha]$ where $\alpha$ satisfies the monic polynomial $\alpha^4+\alpha^3+\alpha^2+\alpha+1$. I found $Z[\alpha] := \lbrace a+b\alpha+c\alpha^2+d\alpha^3\;|\; ...
2
votes
1answer
78 views

Localization of lattices over Dedekind domains

I'm trying to understand the proof of lemma 4.12 on modules over Dedekind domains from Frohlich and Taylor's book 'Algebraic Number Theory' page 94. I have a Dedekind domain $\mathcal o$, a non-zero ...
1
vote
1answer
193 views

Is $\mathbb{Z}[\sqrt{2} + \sqrt{3}]$ closed under multiplication?

Bonus question: if it's not, is it a subdomain of some ring of algebraic integers? This is just something I was thinking about a few weeks ago. I forgot about the concept of algebraic degrees, which ...
2
votes
1answer
33 views

Expressing $\mathcal O_L$ as a certain free module of rank 1

I have a finite Galois extension of number fields $L/K$ with group $G$ and respective rings of integers $\mathcal O_L$ and $\mathcal O_K$. If $\Gamma$ is an $\mathcal O_K$-order in $K[G]$ and ...
0
votes
0answers
21 views

A finite Galois extension $L/K$ of number fields.

I have a finite Galois extension $L/K$ of number fields with group $G$. Let the respective rings of integers be $\mathcal O_L$ and $\mathcal O_K.$ Suppose that $\Gamma$ is an $\mathcal O_K$-order in ...
2
votes
0answers
30 views

A certain $\mathcal O_K$-lattice where $K$ is a number field

I have a finite Galois extension of number fields $L/K$ with group $G$. Let $\mathcal O_L$ and $\mathcal O_K$ be the respective rings of algebraic integers. I want to show that $\mathcal O_L$ is ...
3
votes
1answer
40 views

Rings of algebraic integers

A basic question on algebraic numbers. If $L/K$ is a finite extension of number fields with respective rings of integers $\mathcal O_L$ and $\mathcal O_K$ then is it true that $\mathcal O_L$ is ...
4
votes
2answers
98 views

The prime elements of the ring $ \mathbb{Z} \! \left[ \dfrac{i \sqrt{3} - 1}{2} \right] $.

I have to study the prime elements of the ring $ \mathbb{Z} \! \left[ \dfrac{i \sqrt{3} - 1}{2} \right] $. For the moment, I cannot find the general form of such elements. Can you help me? Thanks! :) ...
3
votes
1answer
27 views

Associated order of a Galois extension of number fields

I have a finite Galois extension $L/K$ of number fields with group $G$. Let $\mathcal O_L$ be the ring of algebraic integers of $L$. We let $\mathcal A_{L/K}$ be the subring $\{x\in K[G]:x\mathcal ...
0
votes
1answer
41 views

Polynomial Diophantine Equations

So in general how does one decide if: $$ a_0 + a_1x + a_2x^2 ... a_{n_1}x^{n_1} = b_1y + b_2y^2 ... + b_{n_2}y^{n_k}$$ Has solutions for integers $x,y$ given real numbers $a_0, a_1. .. a_{n_1}, b_1 , ...
1
vote
2answers
48 views

Show that it doesn't exist any of natural number $ n = 4m + 3$ that $ n= x^2+y^2 $ for any natural x and y [duplicate]

Show that it doesn't exist any of natural number $ n = 4m + 3$ that $ n= x^2+y^2 $ for any natural x and y Show that every prime number in form $ p=4m+1 $ could be showed as $ p = x^2+y^2$ (x and y ...
2
votes
1answer
36 views

ideal calculations: $2\mathcal{O}_K=\mathfrak{B}^4$ in the ring of integers of $K=\mathbb{Q}(i,\sqrt{2m})$

Let $K=\mathbb{Q}(i,\sqrt{2m})$ where $m \in \mathbb{Z}$ is odd and squarefree. Let $\alpha = (1+i)\sqrt{2m}/2$. Then $\alpha^2=im$, such that $\alpha$ is part of the ring of integers $\mathcal{O}_K$. ...
4
votes
1answer
90 views

Question on extensions of discrete valuation fields

Let $F$ be a discrete valuation field. Let $L$ be a finite extension of $F$. Let $L=F(\alpha)$ where $\alpha$ belongs to ring of integers of $L$, denoted by $O_L$. Is it always true that ...
2
votes
0answers
33 views

$\{1,\sqrt[3]{3},(\sqrt[3]{3})^2 \}$ is integral basis for $\Bbb{Q}(\sqrt{3^3})$

prove that $\{1,\sqrt[3]{3},(\sqrt[3]{3})^2 \}$ is integral basis for $\Bbb{Q}(\sqrt{3^3})$ my aim is to show that $\Bbb{Q}_K=\Bbb{Z}+\sqrt[3]{3}\Bbb{Z}+(\sqrt[3]{3})^2\Bbb{Z}$ $\supseteq$ : ...
3
votes
2answers
60 views

How can I prove an ideal is a product of two irreducible ones

I'm trying to solve this question: I have a guess that $(6+\sqrt{11})=(2,4+\sqrt{11})(2,-3\sqrt{11})$ using some formulas in this book page 48. However I couldn't verify if the multiplication of ...
2
votes
2answers
126 views

Silly mistake in this number theory book

My question is very easily to be solved (at least I hope so) I think this book has a mistake: When I calculate I get $b_3\equiv -2 (\mod{2})$ which implies $b_3=0$, am I right? Another question, ...
0
votes
1answer
59 views

Is this ring an integral domain?

I'm starting to study Algebraic number theory and I'm having problems with the first examples of this book. I'm trying to prove this is a quadratic domain, i.e., an integral domain: I'm sorry I ...
1
vote
1answer
24 views

Normalisation of $\mathbb{Z}$ in $\mathbb{Q}(\sqrt{d})$ is UFD…

We know that the normalisation of $\mathbb{Z}$ in $\mathbb{Q}(\sqrt{d})$ where $d\in \mathbb{Z}$ is $$O=\mathbb{Z}[\beta], \text{$\beta=\sqrt{d}$ if $d\equiv2,3 \pmod 4$}; \ \frac{1+\sqrt{d}}{2} ...
6
votes
1answer
76 views

Is the extension Galois if $\mathrm{Aut}(K)$ acts transitively on the non-ramified prime ideals?

Let $K/\mathbb Q$ be a finite extension such that $\mathrm{Aut}(K)$ acts transitively on the prime ideals that are not ramified above the same prime $p\in\mathbb N$. Is $K$ Galois? Thanks in advance. ...
2
votes
0answers
38 views

The number of abolute value on $\mathbb{Q}(\sqrt{2})$

Let $|.|$ be the usual absolute value on $\mathbb{Q}$. The number of absolute value on $Q(\sqrt{2})$ extending |.| is 2 since $x^2-2=(x-\sqrt{2})(x+\sqrt{2})$ in $\mathbb{R}[x]$. Let ...
0
votes
1answer
31 views

Free abelian group of fractional ideals

This question is from Ch.2 of Frohlich and Taylor's Algebraic Number Theory, page 42. Let $R$ be a Dedekind domain, $I_R$ the multiplicative group of fractional $R$-ideals. There is an isomorphism of ...
5
votes
0answers
60 views

Literature to the ring $\mathbb{Z}[\phi]$ where $\phi=\frac{1+\sqrt{5}}{2}$ is the golden ratio

I know few about algebraic number theory but recently I stumbled upon the ring $\mathbb{Z}[\phi]$ where $\phi = \frac{1+\sqrt{5}}{2}$ is the golden ratio. It seems to be a very interesting object to ...
4
votes
1answer
56 views

generalized ideal class group for infinitely many moduli (Cox 8.4)

I am given the following definition (without the proof or technical details). and I need to understand that I tried the following: Since $P_{K,1}(\mathfrak{m}) \subseteq ...
4
votes
1answer
102 views

Localization in formal power series

I saw in a textbook the following assertion: Let $R$ be a commutative ring with unity, and $R[[X]]$ be the ring of power series in one indeterminate $X$. If the homomorphism $\phi∶ R[[X]] \to R$ ...
1
vote
1answer
54 views

What is known about the ramification index of ramified primes in an arbitrary cyclotomic extension of $\mathbb{Q}$

Let $\zeta$ be a primitive $m$th root of unity, and $L = \mathbb{Q}(\zeta)$. Then $B = \mathbb{Z}[\zeta]$ is the integral closure of $\mathbb{Z}$ in $L$. If $P$ is a prime ideal of $B$ and ...
3
votes
2answers
97 views

Cyclotomic polynomial in $\mathbb{Z}/(p)$

Let $p$ be prime, $n \in \mathbb{N}$ and $p \nmid n$. $\Phi_n$ is the $n$-th cyclotomic polynomial. How can I find the maximum $n \in \mathbb{N}$ (with $p \nmid n)$ so that $\Phi_n$ splits into ...
1
vote
1answer
28 views

$ K(\sqrt{a})$ is unramified if and only if $a \mid d_K$ and $a \equiv 1 \mod{4}$.

Let $K$ be an imaginary quadratic field of discriminant $d_K$ and let $K(\sqrt{a})$ be a quadratic extension where $a \in \mathbb{Z}$. Then $K \subset K(\sqrt{a})$ is unramified if and only if $a$ can ...
1
vote
1answer
28 views

Divisibility in the ring of integers.

For example, let $R=\Bbb Z [\sqrt{-5}]$, and I want to explain $3$ does not divide $2-\sqrt{-5}$. I think the following proof will be right: Suppose $3(a+b\sqrt{-5})=2-\sqrt{-5}$, then taking ...
0
votes
0answers
24 views

Methods for computing subextensions for a n-th cyclotomic field.

So the problem is 1)find all quadratic and cubic subextensions of $\mathbb{Q}[\zeta^{527}]$ and 2)describe how it's primes split completely in the cubic subextensions. Can you give me some ...
1
vote
1answer
54 views

Is $f(x)=x^{4}-2x^2 +3$ Eiseinstein in 2-adic $\mathbb{Q}_{2}$?

I think it is because $|1|_{2}=1$, $|2|_{2}=2^{-1}\leq 1$ and $|3|$,where 3 is prime I am using the following Eisenstein criterion for $f(x)=a_{n}x^{n}+...+a_{0}$: $|a_{n}|=1$, $|a_{0}|=prime$ and ...
0
votes
0answers
16 views

Discriminant of p-adic $\mathbb{Q}_{p}[\phi]$, where $0=f(\phi)=\phi^{p}-\phi-1$

Any suggestions using the minimal polynomial? How about $D_{\mathbb{Q}_{p}[\phi]/\mathbb{Q}_{p}}=(-1)N_{\mathbb{Q}_{p}[\phi],\mathbb{Q}_{p}}(f')$? But foremost I prefer you suggest me a correct ...
0
votes
2answers
46 views

Finding number of roots of a polynomial in p-adic integers $\mathbb{Z}_{p}$

The problem is to find the number roots of $x^3+25x^2+x-9 $ in $\mathbb{Z}_{p}$ for p=2,3,5,7. I read this equivalent to having a root mod $p^{k}~\forall k\geq 1$. By Newton's lemma I can get whether ...
0
votes
1answer
23 views

Possible Quadratic extensions of $\mathbb{Q}_{2}$ are 1-1 with $\mathbb{Q}_{2}^{*}/(\mathbb{Q}_{2}^{*})^{2}$

Why do they have to be units? $\mathbb{Q}_{2}[\sqrt{2}]$ is a quadratic extension but $|2|_{2}=\frac{1}{2}\neq 1$. Where do I need them to be units below? ...
1
vote
1answer
26 views

What does “A mod P generates the residue class field extension” mean?

We have K and finite algebraic extension L. P is a prime ideal in $O_{L}$ over prime $p\in O_{K}$ and $A\in O_{L}$. Then the problem says $\bar{A}:=A ~mod ~P$ generates the residue class field ...
0
votes
0answers
22 views

ANT Frohlich Proposition 3, (v). Induced map of dual modules has the same determinant

$R$ is a Dedekind domain, $V$ is an $n$ dimensional vector space over its quotient field $K$, $B(-,-)$ is a $K$-bilinear form on $V$, and $M, N \subseteq V$ are free $R$-modules of rank $n$. Also ...