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Concern about algebraic integer

About algebraic integer and algebraic numberIf $alpha$ and $beta$ are algebraic integers then any solution to $x^2+alpha x + beta = 0$ is also an algebraic integer.Representing an algebraic number in an algebraic number fieldShowing that every polynomial over the Algebraic Numbers has a $0$ in the Algebraic Numbers.The general form of (algebraic) number field?Set of algebraic integer form a ring.minimal polynomial of algebraic integer over $mathbbZ$ and $mathbbQ$Trigonometric expression as an algebraic integerIs $fracsqrt3+sqrt52$ an algebraic integer?Extension with algebraic element is finite

0

$begingroup$

To be honest, I don't figure out how to attack this problem:

Let $alpha$ an algebraic integer i.e. there is a monic polynomial $f(x) in mathbbZ[x]$ s.t. $f(alpha)=0$.

Let $R:=mathbbZ[alpha]$. For some positive integer $m$, prove that $R/mR$ is finite and determine its order.

I'll appreciate any help/hint.

Thanks.

abstract-algebra free-modules

share|cite|improve this question

edited Mar 21 at 17:03

Andrews

1,2812422

asked Mar 21 at 15:09

Charles SeifeCharles Seife

837

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Can you find an expression for elements of $mathbbZ[alpha]$ using powers of $alpha$?
  $endgroup$
  – Arturo Magidin
  Mar 21 at 17:07
  

  
  
  
  

add a comment | 

0

$begingroup$

To be honest, I don't figure out how to attack this problem:

Let $alpha$ an algebraic integer i.e. there is a monic polynomial $f(x) in mathbbZ[x]$ s.t. $f(alpha)=0$.

Let $R:=mathbbZ[alpha]$. For some positive integer $m$, prove that $R/mR$ is finite and determine its order.

I'll appreciate any help/hint.

Thanks.

abstract-algebra free-modules

share|cite|improve this question

edited Mar 21 at 17:03

Andrews

1,2812422

asked Mar 21 at 15:09

Charles SeifeCharles Seife

837

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Can you find an expression for elements of $mathbbZ[alpha]$ using powers of $alpha$?
  $endgroup$
  – Arturo Magidin
  Mar 21 at 17:07
  

  
  
  
  

add a comment | 

$begingroup$

To be honest, I don't figure out how to attack this problem:

Let $alpha$ an algebraic integer i.e. there is a monic polynomial $f(x) in mathbbZ[x]$ s.t. $f(alpha)=0$.

Let $R:=mathbbZ[alpha]$. For some positive integer $m$, prove that $R/mR$ is finite and determine its order.

I'll appreciate any help/hint.

Thanks.

abstract-algebra free-modules

share|cite|improve this question

edited Mar 21 at 17:03

Andrews

1,2812422

asked Mar 21 at 15:09

Charles SeifeCharles Seife

837

$endgroup$

share|cite|improve this question

edited Mar 21 at 17:03

Andrews

1,2812422

asked Mar 21 at 15:09

Charles SeifeCharles Seife

837

share|cite|improve this question

edited Mar 21 at 17:03

Andrews

1,2812422

asked Mar 21 at 15:09

Charles SeifeCharles Seife

837

edited Mar 21 at 17:03

Andrews

1,2812422

edited Mar 21 at 17:03

Andrews

1,2812422

asked Mar 21 at 15:09

Charles SeifeCharles Seife

837

asked Mar 21 at 15:09

Charles SeifeCharles Seife

837

1 Answer
1

active

oldest

votes

0

$begingroup$

Sorry for my late reply.

Following your idea: $F[alpha]=a_0+a_1alpha+cdots+a_n-1alpha^n-1mid a_j in mathbbZ$. It looks like $alpha^0,alpha, alpha^2,...,alpha^n-1$ is a basis for $F[alpha]$. Moreover, $F[alpha]cong mathbbZ^n$ under this basis. Thus, given a positive integer $m$,
$$
R/mR cong mathbbZ^n/mmathbbZ^ncong fracmathbbZ oplus mathbbZoplus cdots oplus mathbbZm mathbbZoplus mmathbbZopluscdots oplus m mathbbZ quad textrm By invariant factor form.
$$
Therefore,
$$
left|R/mRright|=m^n.
$$
I would like to be more precisely in each step, it's like the sketch of a possible proof.
Thanks so much for any extra information that you see might be needed.

share|cite|improve this answer

answered Mar 24 at 2:03

Charles SeifeCharles Seife

837

$endgroup$

add a comment | 

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0

$begingroup$

Sorry for my late reply.

Following your idea: $F[alpha]=a_0+a_1alpha+cdots+a_n-1alpha^n-1mid a_j in mathbbZ$. It looks like $alpha^0,alpha, alpha^2,...,alpha^n-1$ is a basis for $F[alpha]$. Moreover, $F[alpha]cong mathbbZ^n$ under this basis. Thus, given a positive integer $m$,
$$
R/mR cong mathbbZ^n/mmathbbZ^ncong fracmathbbZ oplus mathbbZoplus cdots oplus mathbbZm mathbbZoplus mmathbbZopluscdots oplus m mathbbZ quad textrm By invariant factor form.
$$
Therefore,
$$
left|R/mRright|=m^n.
$$
I would like to be more precisely in each step, it's like the sketch of a possible proof.
Thanks so much for any extra information that you see might be needed.

share|cite|improve this answer

answered Mar 24 at 2:03

Charles SeifeCharles Seife

837

$endgroup$

add a comment | 

0

$begingroup$

Sorry for my late reply.

Following your idea: $F[alpha]=a_0+a_1alpha+cdots+a_n-1alpha^n-1mid a_j in mathbbZ$. It looks like $alpha^0,alpha, alpha^2,...,alpha^n-1$ is a basis for $F[alpha]$. Moreover, $F[alpha]cong mathbbZ^n$ under this basis. Thus, given a positive integer $m$,
$$
R/mR cong mathbbZ^n/mmathbbZ^ncong fracmathbbZ oplus mathbbZoplus cdots oplus mathbbZm mathbbZoplus mmathbbZopluscdots oplus m mathbbZ quad textrm By invariant factor form.
$$
Therefore,
$$
left|R/mRright|=m^n.
$$
I would like to be more precisely in each step, it's like the sketch of a possible proof.
Thanks so much for any extra information that you see might be needed.

share|cite|improve this answer

answered Mar 24 at 2:03

Charles SeifeCharles Seife

837

$endgroup$

add a comment | 

$begingroup$

Sorry for my late reply.

Following your idea: $F[alpha]=a_0+a_1alpha+cdots+a_n-1alpha^n-1mid a_j in mathbbZ$. It looks like $alpha^0,alpha, alpha^2,...,alpha^n-1$ is a basis for $F[alpha]$. Moreover, $F[alpha]cong mathbbZ^n$ under this basis. Thus, given a positive integer $m$,
$$
R/mR cong mathbbZ^n/mmathbbZ^ncong fracmathbbZ oplus mathbbZoplus cdots oplus mathbbZm mathbbZoplus mmathbbZopluscdots oplus m mathbbZ quad textrm By invariant factor form.
$$
Therefore,
$$
left|R/mRright|=m^n.
$$
I would like to be more precisely in each step, it's like the sketch of a possible proof.
Thanks so much for any extra information that you see might be needed.

share|cite|improve this answer

answered Mar 24 at 2:03

Charles SeifeCharles Seife

837

$endgroup$

share|cite|improve this answer

answered Mar 24 at 2:03

Charles SeifeCharles Seife

837

answered Mar 24 at 2:03

Charles SeifeCharles Seife

837

answered Mar 24 at 2:03

Charles SeifeCharles Seife

837