Definition of the d-th power residue symbol
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I started reading about the reciprocity law for polynomials over a finite field, and I came across the definition of the d-th power residue symbol:
Let $Bbb F$ be a finite field with $q$ elements, $q=p^f$ ($p$ is the characteristic of $Bbb F$), and let $A = Bbb F[T]$. Let $Pin A$ be an irreducible polynomial, $d$ a divisor of $q-1$ and $a in A$.
Definition: If $P$ does not divide $a$, let $(a/P)_d$ be the unique element of $Bbb F^*$ such that $$a^{frac{|P|-1}{d}}=left(frac{a}{P} right)_d pmod P$$
If $P|a$ define $(a/P)_d=0$. The symbol $(a/P)_d$ is called the d-th power residue symbol.
My question is: How do we know that $(a/P)_d$ is in $Bbb F^*$ ? Why can't it be just any element of $(A/PA)^*$?
polynomials algebraic-number-theory finite-fields
asked Nov 14 at 19:28
user401516
84728
add a comment |
up vote
1
down vote
favorite
I started reading about the reciprocity law for polynomials over a finite field, and I came across the definition of the d-th power residue symbol:
Let $Bbb F$ be a finite field with $q$ elements, $q=p^f$ ($p$ is the characteristic of $Bbb F$), and let $A = Bbb F[T]$. Let $Pin A$ be an irreducible polynomial, $d$ a divisor of $q-1$ and $a in A$.
Definition: If $P$ does not divide $a$, let $(a/P)_d$ be the unique element of $Bbb F^*$ such that $$a^{frac{|P|-1}{d}}=left(frac{a}{P} right)_d pmod P$$
If $P|a$ define $(a/P)_d=0$. The symbol $(a/P)_d$ is called the d-th power residue symbol.
My question is: How do we know that $(a/P)_d$ is in $Bbb F^*$ ? Why can't it be just any element of $(A/PA)^*$?
polynomials algebraic-number-theory finite-fields
asked Nov 14 at 19:28
user401516
84728
Since $P in mathbf{F}_q[T]$ is irreducible $(P)$ is a maximal ideal and $mathbf{F}_q[T]/(P) simeq mathbf{F}_{q^m}$ where $m = deg(P)$. For $a in mathbf{F}_{q^m}^*$, $a^{q^m-1 }=1$ so the order of $a^{(q^m-1)/d}$ divides $d$ and for $d | q-1$ then $a^{(q^m-1)/d} in mathbf{F}_q^*$.
– reuns
Nov 14 at 19:44
How did you conclude that $a^{(q^m-1})/d} in F_q^*$? I understand everything else you said
– user401516
Nov 15 at 13:47
$mathbf{F}_q^*$ is a group with $q-1$ elements, so all those elements satisfy $x^{q-1}=1$. The polynomial $x^{q-1}-1$ has at most $q-1$ distinct roots in the field $mathbf{F}_{q^m}$. Thus $c in mathbf{F}_{q^m}^*$ is in $mathbf{F}_{q}^*$ iff it is a root of $x^{q-1}-1$ that is its order divides $q-1$ (this is also the way we prove those groups are cyclic)
– reuns
Nov 15 at 18:36
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I started reading about the reciprocity law for polynomials over a finite field, and I came across the definition of the d-th power residue symbol:
Let $Bbb F$ be a finite field with $q$ elements, $q=p^f$ ($p$ is the characteristic of $Bbb F$), and let $A = Bbb F[T]$. Let $Pin A$ be an irreducible polynomial, $d$ a divisor of $q-1$ and $a in A$.
Definition: If $P$ does not divide $a$, let $(a/P)_d$ be the unique element of $Bbb F^*$ such that $$a^{frac{|P|-1}{d}}=left(frac{a}{P} right)_d pmod P$$
If $P|a$ define $(a/P)_d=0$. The symbol $(a/P)_d$ is called the d-th power residue symbol.
My question is: How do we know that $(a/P)_d$ is in $Bbb F^*$ ? Why can't it be just any element of $(A/PA)^*$?
polynomials algebraic-number-theory finite-fields
asked Nov 14 at 19:28
user401516
84728
I started reading about the reciprocity law for polynomials over a finite field, and I came across the definition of the d-th power residue symbol:
Let $Bbb F$ be a finite field with $q$ elements, $q=p^f$ ($p$ is the characteristic of $Bbb F$), and let $A = Bbb F[T]$. Let $Pin A$ be an irreducible polynomial, $d$ a divisor of $q-1$ and $a in A$.
Definition: If $P$ does not divide $a$, let $(a/P)_d$ be the unique element of $Bbb F^*$ such that $$a^{frac{|P|-1}{d}}=left(frac{a}{P} right)_d pmod P$$
If $P|a$ define $(a/P)_d=0$. The symbol $(a/P)_d$ is called the d-th power residue symbol.
My question is: How do we know that $(a/P)_d$ is in $Bbb F^*$ ? Why can't it be just any element of $(A/PA)^*$?
polynomials algebraic-number-theory finite-fields
polynomials algebraic-number-theory finite-fields
asked Nov 14 at 19:28
user401516
84728
asked Nov 14 at 19:28
user401516
84728
asked Nov 14 at 19:28
user401516
84728
asked Nov 14 at 19:28
user401516
84728
asked Nov 14 at 19:28
user401516
84728
84728
Since $P in mathbf{F}_q[T]$ is irreducible $(P)$ is a maximal ideal and $mathbf{F}_q[T]/(P) simeq mathbf{F}_{q^m}$ where $m = deg(P)$. For $a in mathbf{F}_{q^m}^*$, $a^{q^m-1 }=1$ so the order of $a^{(q^m-1)/d}$ divides $d$ and for $d | q-1$ then $a^{(q^m-1)/d} in mathbf{F}_q^*$.
– reuns
Nov 14 at 19:44
How did you conclude that $a^{(q^m-1})/d} in F_q^*$? I understand everything else you said
– user401516
Nov 15 at 13:47
$mathbf{F}_q^*$ is a group with $q-1$ elements, so all those elements satisfy $x^{q-1}=1$. The polynomial $x^{q-1}-1$ has at most $q-1$ distinct roots in the field $mathbf{F}_{q^m}$. Thus $c in mathbf{F}_{q^m}^*$ is in $mathbf{F}_{q}^*$ iff it is a root of $x^{q-1}-1$ that is its order divides $q-1$ (this is also the way we prove those groups are cyclic)
– reuns
Nov 15 at 18:36
add a comment |
Since $P in mathbf{F}_q[T]$ is irreducible $(P)$ is a maximal ideal and $mathbf{F}_q[T]/(P) simeq mathbf{F}_{q^m}$ where $m = deg(P)$. For $a in mathbf{F}_{q^m}^*$, $a^{q^m-1 }=1$ so the order of $a^{(q^m-1)/d}$ divides $d$ and for $d | q-1$ then $a^{(q^m-1)/d} in mathbf{F}_q^*$.
– reuns
Nov 14 at 19:44
How did you conclude that $a^{(q^m-1})/d} in F_q^*$? I understand everything else you said
– user401516
Nov 15 at 13:47
$mathbf{F}_q^*$ is a group with $q-1$ elements, so all those elements satisfy $x^{q-1}=1$. The polynomial $x^{q-1}-1$ has at most $q-1$ distinct roots in the field $mathbf{F}_{q^m}$. Thus $c in mathbf{F}_{q^m}^*$ is in $mathbf{F}_{q}^*$ iff it is a root of $x^{q-1}-1$ that is its order divides $q-1$ (this is also the way we prove those groups are cyclic)
– reuns
Nov 15 at 18:36
Since $P in mathbf{F}_q[T]$ is irreducible $(P)$ is a maximal ideal and $mathbf{F}_q[T]/(P) simeq mathbf{F}_{q^m}$ where $m = deg(P)$. For $a in mathbf{F}_{q^m}^*$, $a^{q^m-1 }=1$ so the order of $a^{(q^m-1)/d}$ divides $d$ and for $d | q-1$ then $a^{(q^m-1)/d} in mathbf{F}_q^*$.
– reuns
Nov 14 at 19:44
Since $P in mathbf{F}_q[T]$ is irreducible $(P)$ is a maximal ideal and $mathbf{F}_q[T]/(P) simeq mathbf{F}_{q^m}$ where $m = deg(P)$. For $a in mathbf{F}_{q^m}^*$, $a^{q^m-1 }=1$ so the order of $a^{(q^m-1)/d}$ divides $d$ and for $d | q-1$ then $a^{(q^m-1)/d} in mathbf{F}_q^*$.
– reuns
Nov 14 at 19:44
How did you conclude that $a^{(q^m-1})/d} in F_q^*$? I understand everything else you said
– user401516
Nov 15 at 13:47
How did you conclude that $a^{(q^m-1})/d} in F_q^*$? I understand everything else you said
– user401516
Nov 15 at 13:47
$mathbf{F}_q^*$ is a group with $q-1$ elements, so all those elements satisfy $x^{q-1}=1$. The polynomial $x^{q-1}-1$ has at most $q-1$ distinct roots in the field $mathbf{F}_{q^m}$. Thus $c in mathbf{F}_{q^m}^*$ is in $mathbf{F}_{q}^*$ iff it is a root of $x^{q-1}-1$ that is its order divides $q-1$ (this is also the way we prove those groups are cyclic)
– reuns
Nov 15 at 18:36
$mathbf{F}_q^*$ is a group with $q-1$ elements, so all those elements satisfy $x^{q-1}=1$. The polynomial $x^{q-1}-1$ has at most $q-1$ distinct roots in the field $mathbf{F}_{q^m}$. Thus $c in mathbf{F}_{q^m}^*$ is in $mathbf{F}_{q}^*$ iff it is a root of $x^{q-1}-1$ that is its order divides $q-1$ (this is also the way we prove those groups are cyclic)
– reuns
Nov 15 at 18:36
add a comment |
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Since $P in mathbf{F}_q[T]$ is irreducible $(P)$ is a maximal ideal and $mathbf{F}_q[T]/(P) simeq mathbf{F}_{q^m}$ where $m = deg(P)$. For $a in mathbf{F}_{q^m}^*$, $a^{q^m-1 }=1$ so the order of $a^{(q^m-1)/d}$ divides $d$ and for $d | q-1$ then $a^{(q^m-1)/d} in mathbf{F}_q^*$.
– reuns
Nov 14 at 19:44
How did you conclude that $a^{(q^m-1})/d} in F_q^*$? I understand everything else you said
– user401516
Nov 15 at 13:47
$mathbf{F}_q^*$ is a group with $q-1$ elements, so all those elements satisfy $x^{q-1}=1$. The polynomial $x^{q-1}-1$ has at most $q-1$ distinct roots in the field $mathbf{F}_{q^m}$. Thus $c in mathbf{F}_{q^m}^*$ is in $mathbf{F}_{q}^*$ iff it is a root of $x^{q-1}-1$ that is its order divides $q-1$ (this is also the way we prove those groups are cyclic)
– reuns
Nov 15 at 18:36