Ggthjy

Translation of 答えを知っている人はいませんでした

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Factor Rings over Finite Fields

Quotient Rings of Polynomials Over Finite FieldsPolynomial rings — Inherited properties from coefficient ringAddition in finite fieldsFactor Ring fundamental questionIs division allowed in rings and fields?Extensions Fields and Zeros of PolynomialsAre these rings fields?Finitely generated modules over non-commutative principal ideal ringsIrreducible polynomials of degree greater than 4 over finite fieldsSplitting Fields over arbitrary fields

2

$begingroup$

Given a polynomial ring over a field $F[x]$, I can factor, for example, the ideal generated by an irreducible polynomial $ax^2 + bx + c$: $F[x]/left<ax^2 + bx + cright>$, and guarantee that this factor ring is also a field.

My question concerns the structure of this factor ring. For example, if I consider the factor ring $Z_p[x] / left<ax^2 + bx + cright>$ for some irreducible polynomial $ax^2 + bx + c$, I can guarantee, for example, that this field has $p^2$ elements.

I am unsure why this is the case. My understanding is that the coset representitives of this factor ring are possible remainders by division by $ax^2 + bx + c$. Is this the right idea, and how would I know that two different remainders aren't in the same coset? Thanks.

abstract-algebra ring-theory field-theory finite-fields quotient-spaces

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edited 3 hours ago

Servaes

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$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Roughly speaking, elements in your said ring are linear polynomials $Ax+B$. There are $p$ many choices for each of the coefficients $A$ and $B$.
  $endgroup$
  – thedilated
  3 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Given a polynomial ring over a field $F[x]$, I can factor, for example, the ideal generated by an irreducible polynomial $ax^2 + bx + c$: $F[x]/left<ax^2 + bx + cright>$, and guarantee that this factor ring is also a field.

My question concerns the structure of this factor ring. For example, if I consider the factor ring $Z_p[x] / left<ax^2 + bx + cright>$ for some irreducible polynomial $ax^2 + bx + c$, I can guarantee, for example, that this field has $p^2$ elements.

I am unsure why this is the case. My understanding is that the coset representitives of this factor ring are possible remainders by division by $ax^2 + bx + c$. Is this the right idea, and how would I know that two different remainders aren't in the same coset? Thanks.

abstract-algebra ring-theory field-theory finite-fields quotient-spaces

share|cite|improve this question

edited 3 hours ago

Servaes

27.8k34098

asked 3 hours ago

Solarflare0Solarflare0

773

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Roughly speaking, elements in your said ring are linear polynomials $Ax+B$. There are $p$ many choices for each of the coefficients $A$ and $B$.
  $endgroup$
  – thedilated
  3 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Given a polynomial ring over a field $F[x]$, I can factor, for example, the ideal generated by an irreducible polynomial $ax^2 + bx + c$: $F[x]/left<ax^2 + bx + cright>$, and guarantee that this factor ring is also a field.

My question concerns the structure of this factor ring. For example, if I consider the factor ring $Z_p[x] / left<ax^2 + bx + cright>$ for some irreducible polynomial $ax^2 + bx + c$, I can guarantee, for example, that this field has $p^2$ elements.

I am unsure why this is the case. My understanding is that the coset representitives of this factor ring are possible remainders by division by $ax^2 + bx + c$. Is this the right idea, and how would I know that two different remainders aren't in the same coset? Thanks.

abstract-algebra ring-theory field-theory finite-fields quotient-spaces

share|cite|improve this question

edited 3 hours ago

Servaes

27.8k34098

asked 3 hours ago

Solarflare0Solarflare0

773

$endgroup$

share|cite|improve this question

edited 3 hours ago

Servaes

27.8k34098

asked 3 hours ago

Solarflare0Solarflare0

773

share|cite|improve this question

edited 3 hours ago

Servaes

27.8k34098

asked 3 hours ago

Solarflare0Solarflare0

773

edited 3 hours ago

Servaes

27.8k34098

edited 3 hours ago

Servaes

27.8k34098

asked 3 hours ago

Solarflare0Solarflare0

773

asked 3 hours ago

Solarflare0Solarflare0

773

2 Answers
2

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oldest

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2

$begingroup$

Indeed the elements of the factor ring $Bbb{F}_p[x]/langle ax^2+bx^2+crangle$ can be represented by the remainders by division by $ax^2+bx+c$. This is true because we can divide polynomials in $Bbb{F}_p[x]$ by $ax^2+bx+c$ with remainder. What this means is that

For every polynomial $finBbb{F}_p[x]$ there exist unique $q,rinBbb{F}_p[x]$ with $deg r<2$ such that
$$f=q(ax^2+bx+c)+r.tag{1}$$

This equality shows that $f$ and $r$ are in the same coset of $langle ax^2+bx+crangle$, and hence they are mapped to the same element of the factor ring $Bbb{F}_p[x]/langle ax^2+bx+crangle$. Hence the image of
$f$ in the factor ring is represented by $r$, and so every element of the factor ring is represented by a linear polynomial.

To see that no two linear polynomials represent the same element of $Bbb{F}_p[x]/langle ax^2+bx+crangle$, it suffices to note that the remainder $r$ in $(1)$ is unique for every $finBbb{F}_p[x]$, meaning in particular that every linear polynomial is represented only by itself.

Alternatively, if two linear polynomials $r$ and $r'$ represent the same coset of $langle ax^2+bx+crangle$ in the factor ring, then $r-r'$ is a multiple of $ax^2+bx+c$. Because $deg r-r'<deg(ax^2+bx+c)$ it follows that $r-r'=0$.

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

ServaesServaes

27.8k34098

$endgroup$

add a comment | 

3

$begingroup$

The quotient you get is a two-dimensional vector space over your field. Can you show that any such vector space must have $p^2$ elements?

share|cite|improve this answer

answered 3 hours ago

Santana AftonSantana Afton

2,9132629

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add a comment | 

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2

$begingroup$

Indeed the elements of the factor ring $Bbb{F}_p[x]/langle ax^2+bx^2+crangle$ can be represented by the remainders by division by $ax^2+bx+c$. This is true because we can divide polynomials in $Bbb{F}_p[x]$ by $ax^2+bx+c$ with remainder. What this means is that

For every polynomial $finBbb{F}_p[x]$ there exist unique $q,rinBbb{F}_p[x]$ with $deg r<2$ such that
$$f=q(ax^2+bx+c)+r.tag{1}$$

This equality shows that $f$ and $r$ are in the same coset of $langle ax^2+bx+crangle$, and hence they are mapped to the same element of the factor ring $Bbb{F}_p[x]/langle ax^2+bx+crangle$. Hence the image of
$f$ in the factor ring is represented by $r$, and so every element of the factor ring is represented by a linear polynomial.

To see that no two linear polynomials represent the same element of $Bbb{F}_p[x]/langle ax^2+bx+crangle$, it suffices to note that the remainder $r$ in $(1)$ is unique for every $finBbb{F}_p[x]$, meaning in particular that every linear polynomial is represented only by itself.

Alternatively, if two linear polynomials $r$ and $r'$ represent the same coset of $langle ax^2+bx+crangle$ in the factor ring, then $r-r'$ is a multiple of $ax^2+bx+c$. Because $deg r-r'<deg(ax^2+bx+c)$ it follows that $r-r'=0$.

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

ServaesServaes

27.8k34098

$endgroup$

add a comment | 

2

$begingroup$

Indeed the elements of the factor ring $Bbb{F}_p[x]/langle ax^2+bx^2+crangle$ can be represented by the remainders by division by $ax^2+bx+c$. This is true because we can divide polynomials in $Bbb{F}_p[x]$ by $ax^2+bx+c$ with remainder. What this means is that

For every polynomial $finBbb{F}_p[x]$ there exist unique $q,rinBbb{F}_p[x]$ with $deg r<2$ such that
$$f=q(ax^2+bx+c)+r.tag{1}$$

This equality shows that $f$ and $r$ are in the same coset of $langle ax^2+bx+crangle$, and hence they are mapped to the same element of the factor ring $Bbb{F}_p[x]/langle ax^2+bx+crangle$. Hence the image of
$f$ in the factor ring is represented by $r$, and so every element of the factor ring is represented by a linear polynomial.

To see that no two linear polynomials represent the same element of $Bbb{F}_p[x]/langle ax^2+bx+crangle$, it suffices to note that the remainder $r$ in $(1)$ is unique for every $finBbb{F}_p[x]$, meaning in particular that every linear polynomial is represented only by itself.

Alternatively, if two linear polynomials $r$ and $r'$ represent the same coset of $langle ax^2+bx+crangle$ in the factor ring, then $r-r'$ is a multiple of $ax^2+bx+c$. Because $deg r-r'<deg(ax^2+bx+c)$ it follows that $r-r'=0$.

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

ServaesServaes

27.8k34098

$endgroup$

add a comment | 

$begingroup$

Indeed the elements of the factor ring $Bbb{F}_p[x]/langle ax^2+bx^2+crangle$ can be represented by the remainders by division by $ax^2+bx+c$. This is true because we can divide polynomials in $Bbb{F}_p[x]$ by $ax^2+bx+c$ with remainder. What this means is that

For every polynomial $finBbb{F}_p[x]$ there exist unique $q,rinBbb{F}_p[x]$ with $deg r<2$ such that
$$f=q(ax^2+bx+c)+r.tag{1}$$

This equality shows that $f$ and $r$ are in the same coset of $langle ax^2+bx+crangle$, and hence they are mapped to the same element of the factor ring $Bbb{F}_p[x]/langle ax^2+bx+crangle$. Hence the image of
$f$ in the factor ring is represented by $r$, and so every element of the factor ring is represented by a linear polynomial.

To see that no two linear polynomials represent the same element of $Bbb{F}_p[x]/langle ax^2+bx+crangle$, it suffices to note that the remainder $r$ in $(1)$ is unique for every $finBbb{F}_p[x]$, meaning in particular that every linear polynomial is represented only by itself.

Alternatively, if two linear polynomials $r$ and $r'$ represent the same coset of $langle ax^2+bx+crangle$ in the factor ring, then $r-r'$ is a multiple of $ax^2+bx+c$. Because $deg r-r'<deg(ax^2+bx+c)$ it follows that $r-r'=0$.

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

ServaesServaes

27.8k34098

$endgroup$

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

ServaesServaes

27.8k34098

answered 3 hours ago

ServaesServaes

27.8k34098

answered 3 hours ago

ServaesServaes

27.8k34098

3

$begingroup$

The quotient you get is a two-dimensional vector space over your field. Can you show that any such vector space must have $p^2$ elements?

share|cite|improve this answer

answered 3 hours ago

Santana AftonSantana Afton

2,9132629

$endgroup$

add a comment | 

3

$begingroup$

The quotient you get is a two-dimensional vector space over your field. Can you show that any such vector space must have $p^2$ elements?

share|cite|improve this answer

answered 3 hours ago

Santana AftonSantana Afton

2,9132629

$endgroup$

add a comment | 

$begingroup$

The quotient you get is a two-dimensional vector space over your field. Can you show that any such vector space must have $p^2$ elements?

share|cite|improve this answer

answered 3 hours ago

Santana AftonSantana Afton

2,9132629

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Santana AftonSantana Afton

2,9132629

answered 3 hours ago

Santana AftonSantana Afton

2,9132629

answered 3 hours ago

Santana AftonSantana Afton

2,9132629