Factor Rings over Finite Fields
$begingroup$
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
edited 3 hours ago
Servaes
27.8k34098
asked 4 hours ago
Solarflare0Solarflare0
773
$endgroup$
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
edited 3 hours ago
Servaes
27.8k34098
asked 4 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
4 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
edited 3 hours ago
Servaes
27.8k34098
asked 4 hours ago
Solarflare0Solarflare0
773
$endgroup$
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
abstract-algebra ring-theory field-theory finite-fields quotient-spaces
edited 3 hours ago
Servaes
27.8k34098
asked 4 hours ago
Solarflare0Solarflare0
773
edited 3 hours ago
Servaes
27.8k34098
asked 4 hours ago
Solarflare0Solarflare0
773
edited 3 hours ago
Servaes
27.8k34098
edited 3 hours ago
Servaes
27.8k34098
edited 3 hours ago
Servaes
27.8k34098
27.8k34098
asked 4 hours ago
Solarflare0Solarflare0
773
asked 4 hours ago
Solarflare0Solarflare0
773
asked 4 hours ago
Solarflare0Solarflare0
773
773
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
4 hours ago
add a comment |
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
4 hours ago
1
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
4 hours ago
$begingroup$
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
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $r-r'=0$.
answered 4 hours ago
ServaesServaes
27.8k34098
$endgroup$
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?
answered 4 hours ago
Santana AftonSantana Afton
2,9132629
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
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 $r-r'=0$.
answered 4 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 $r-r'=0$.
answered 4 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 $r-r'=0$.
answered 4 hours ago
ServaesServaes
27.8k34098
$endgroup$
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 $r-r'=0$.
answered 4 hours ago
ServaesServaes
27.8k34098
edited 4 hours ago
answered 4 hours ago
ServaesServaes
27.8k34098
answered 4 hours ago
ServaesServaes
27.8k34098
answered 4 hours ago
ServaesServaes
27.8k34098
27.8k34098
add a comment |
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?
answered 4 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?
answered 4 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?
answered 4 hours ago
Santana AftonSantana Afton
2,9132629
$endgroup$
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?
answered 4 hours ago
Santana AftonSantana Afton
2,9132629
answered 4 hours ago
Santana AftonSantana Afton
2,9132629
answered 4 hours ago
Santana AftonSantana Afton
2,9132629
answered 4 hours ago
Santana AftonSantana Afton
2,9132629
2,9132629
add a comment |
add a comment |
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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
4 hours ago