What does it mean that field $mathbb{F}_{p^n}$ “contains” the prime field $mathbb{Z}_p$?
$begingroup$
I have read in few books (example Computational Number Theory, page 77) that any extension field $mathbb{F}_{p^n}$ "contains" as a subfield the prime field $mathbb{Z}_p$?
What exactly does "contains" mean? Since the representations of the two fields are not same (we have to use polynomials or some other structure to represent $mathbb{F}_{p^n}$, while we use modular arithmetic to represent $mathbb{Z}_p$), how can we say that one contains the other? Does it actually mean that it contains a subfield that is isomorphic to $mathbb{Z}_p$?
For instance, we can represent $mathbb{F}_{3^2}$ using arbitrary symbols ${0, 1, a, b, c, d, e, f, g}$ or by pairs ${(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)}$. How does this "contain" $mathbb{Z}_3$?
This is more related to the computational perspective, so for example if I am working in $mathbb{F}_{p^n}$, how do I represent the subfield and work directly on integers?
Additionally if the subfield is itself an extension field (i.e. $mathbb{F}_{p^m}$ for some $m$ that divides $n$), how can I use a smaller data structure to represent it while at the same time keeping it also part of $mathbb{F}_{p^n}$?
number-theory field-theory finite-fields extension-field computational-algebra
asked Jan 6 at 7:30
Jus12Jus12
1808
$endgroup$
add a comment |
$begingroup$
I have read in few books (example Computational Number Theory, page 77) that any extension field $mathbb{F}_{p^n}$ "contains" as a subfield the prime field $mathbb{Z}_p$?
What exactly does "contains" mean? Since the representations of the two fields are not same (we have to use polynomials or some other structure to represent $mathbb{F}_{p^n}$, while we use modular arithmetic to represent $mathbb{Z}_p$), how can we say that one contains the other? Does it actually mean that it contains a subfield that is isomorphic to $mathbb{Z}_p$?
For instance, we can represent $mathbb{F}_{3^2}$ using arbitrary symbols ${0, 1, a, b, c, d, e, f, g}$ or by pairs ${(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)}$. How does this "contain" $mathbb{Z}_3$?
This is more related to the computational perspective, so for example if I am working in $mathbb{F}_{p^n}$, how do I represent the subfield and work directly on integers?
Additionally if the subfield is itself an extension field (i.e. $mathbb{F}_{p^m}$ for some $m$ that divides $n$), how can I use a smaller data structure to represent it while at the same time keeping it also part of $mathbb{F}_{p^n}$?
number-theory field-theory finite-fields extension-field computational-algebra
asked Jan 6 at 7:30
Jus12Jus12
1808
$endgroup$
1
$begingroup$
Not an answer, just two comments. (i) $mathbb{F}_p$ cannot be a (proper) extension field; (ii) equally important, though you don't ask about it, is that $mathbb{F}_{p^n}$ contains exactly one $mathbb{F}_p$, and there is only one way of setting up an isomorphism $mathbb{F}_ptomathbb{F}_{p^n}$.
$endgroup$
– ancientmathematician
Jan 6 at 7:51
$begingroup$
@ancientmathematician for (i), I meant some other extension field $mathbb{F}_{p^m}$, where $m$ divides $n$. Also, good point (ii).
$endgroup$
– Jus12
Jan 6 at 8:09
add a comment |
$begingroup$
I have read in few books (example Computational Number Theory, page 77) that any extension field $mathbb{F}_{p^n}$ "contains" as a subfield the prime field $mathbb{Z}_p$?
What exactly does "contains" mean? Since the representations of the two fields are not same (we have to use polynomials or some other structure to represent $mathbb{F}_{p^n}$, while we use modular arithmetic to represent $mathbb{Z}_p$), how can we say that one contains the other? Does it actually mean that it contains a subfield that is isomorphic to $mathbb{Z}_p$?
For instance, we can represent $mathbb{F}_{3^2}$ using arbitrary symbols ${0, 1, a, b, c, d, e, f, g}$ or by pairs ${(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)}$. How does this "contain" $mathbb{Z}_3$?
This is more related to the computational perspective, so for example if I am working in $mathbb{F}_{p^n}$, how do I represent the subfield and work directly on integers?
Additionally if the subfield is itself an extension field (i.e. $mathbb{F}_{p^m}$ for some $m$ that divides $n$), how can I use a smaller data structure to represent it while at the same time keeping it also part of $mathbb{F}_{p^n}$?
number-theory field-theory finite-fields extension-field computational-algebra
asked Jan 6 at 7:30
Jus12Jus12
1808
$endgroup$
I have read in few books (example Computational Number Theory, page 77) that any extension field $mathbb{F}_{p^n}$ "contains" as a subfield the prime field $mathbb{Z}_p$?
What exactly does "contains" mean? Since the representations of the two fields are not same (we have to use polynomials or some other structure to represent $mathbb{F}_{p^n}$, while we use modular arithmetic to represent $mathbb{Z}_p$), how can we say that one contains the other? Does it actually mean that it contains a subfield that is isomorphic to $mathbb{Z}_p$?
For instance, we can represent $mathbb{F}_{3^2}$ using arbitrary symbols ${0, 1, a, b, c, d, e, f, g}$ or by pairs ${(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)}$. How does this "contain" $mathbb{Z}_3$?
This is more related to the computational perspective, so for example if I am working in $mathbb{F}_{p^n}$, how do I represent the subfield and work directly on integers?
Additionally if the subfield is itself an extension field (i.e. $mathbb{F}_{p^m}$ for some $m$ that divides $n$), how can I use a smaller data structure to represent it while at the same time keeping it also part of $mathbb{F}_{p^n}$?
number-theory field-theory finite-fields extension-field computational-algebra
number-theory field-theory finite-fields extension-field computational-algebra
asked Jan 6 at 7:30
Jus12Jus12
1808
asked Jan 6 at 7:30
Jus12Jus12
1808
edited Jan 6 at 8:08
Jus12
asked Jan 6 at 7:30
Jus12Jus12
1808
asked Jan 6 at 7:30
Jus12Jus12
1808
asked Jan 6 at 7:30
Jus12Jus12
1808
1808
1
$begingroup$
Not an answer, just two comments. (i) $mathbb{F}_p$ cannot be a (proper) extension field; (ii) equally important, though you don't ask about it, is that $mathbb{F}_{p^n}$ contains exactly one $mathbb{F}_p$, and there is only one way of setting up an isomorphism $mathbb{F}_ptomathbb{F}_{p^n}$.
$endgroup$
– ancientmathematician
Jan 6 at 7:51
$begingroup$
@ancientmathematician for (i), I meant some other extension field $mathbb{F}_{p^m}$, where $m$ divides $n$. Also, good point (ii).
$endgroup$
– Jus12
Jan 6 at 8:09
add a comment |
1
$begingroup$
Not an answer, just two comments. (i) $mathbb{F}_p$ cannot be a (proper) extension field; (ii) equally important, though you don't ask about it, is that $mathbb{F}_{p^n}$ contains exactly one $mathbb{F}_p$, and there is only one way of setting up an isomorphism $mathbb{F}_ptomathbb{F}_{p^n}$.
$endgroup$
– ancientmathematician
Jan 6 at 7:51
$begingroup$
@ancientmathematician for (i), I meant some other extension field $mathbb{F}_{p^m}$, where $m$ divides $n$. Also, good point (ii).
$endgroup$
– Jus12
Jan 6 at 8:09
1
1
$begingroup$
Not an answer, just two comments. (i) $mathbb{F}_p$ cannot be a (proper) extension field; (ii) equally important, though you don't ask about it, is that $mathbb{F}_{p^n}$ contains exactly one $mathbb{F}_p$, and there is only one way of setting up an isomorphism $mathbb{F}_ptomathbb{F}_{p^n}$.
$endgroup$
– ancientmathematician
Jan 6 at 7:51
$begingroup$
Not an answer, just two comments. (i) $mathbb{F}_p$ cannot be a (proper) extension field; (ii) equally important, though you don't ask about it, is that $mathbb{F}_{p^n}$ contains exactly one $mathbb{F}_p$, and there is only one way of setting up an isomorphism $mathbb{F}_ptomathbb{F}_{p^n}$.
$endgroup$
– ancientmathematician
Jan 6 at 7:51
$begingroup$
@ancientmathematician for (i), I meant some other extension field $mathbb{F}_{p^m}$, where $m$ divides $n$. Also, good point (ii).
$endgroup$
– Jus12
Jan 6 at 8:09
$begingroup$
@ancientmathematician for (i), I meant some other extension field $mathbb{F}_{p^m}$, where $m$ divides $n$. Also, good point (ii).
$endgroup$
– Jus12
Jan 6 at 8:09
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Technically it means that an isomorphic copy of $Bbb{Z}_p$ is in $Bbb{F}_{p^n}$. For example, in $Bbb{F}_9$, the constant polynomials will be an isomorphic copy of $Bbb{Z}_3$.
begin{align*}
Bbb{F}_9 & =Bbb{Z}_3[x]/langle x^2+1rangle\
& ={ax+b , | , a,b in Bbb{Z}_3, x^2+1 equiv 0}\
&={0,1,2,x,x+1,x+2,2x,2x+1,2x+2}
end{align*}
Then the set of constant polynomials ${0,1,2}$ is isomorphic to $Bbb{Z}_3$.
With your edited version:
$Bbb{F}_{p^n}$ can be thought of as a vector space over the field of scalars $Bbb{Z}_p$. Thus each object in $Bbb{F}_{p^n}$ is a $n-$tuple vector (this is very similar to the polynomial representation I have used above).
So consider the set of the vectors
$$S={(0,0,ldots ,c) , | , c in Bbb{Z}_p}.$$
This set in $Bbb{F}_{p^n}$ is isomorphic to $Bbb{Z}_p$.
answered Jan 6 at 7:42
Anurag AAnurag A
26k12250
$endgroup$
add a comment |
$begingroup$
The mapping ${Bbb Z}_prightarrow {Bbb F}_{p^n}$ given by $amapsto acdot 1$, where $acdot 1 = 1+ldots+1$ ($a$-times) is the $a$-multiple of $1$ and $1$ is the unit element of ${Bbb F}_{p^n}$, is a (ring) monomorphism. In this way, ${Bbb F}_{p^n}$ contains an isomorphic copy of ${Bbb Z}_p$.
answered Jan 6 at 8:10
WuestenfuxWuestenfux
4,4981413
$endgroup$
$begingroup$
I wanted a representation-independent way to define this subset. Hence, this useful. We can use "double and repeat" if $a$ is large.
$endgroup$
– Jus12
Jan 6 at 8:28
$begingroup$
Indeed, ''double and repeat'' can be used.
$endgroup$
– Wuestenfux
Jan 6 at 8:28
add a comment |
Your Answer
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2 Answers
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active
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2 Answers
2
active
oldest
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active
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$begingroup$
Technically it means that an isomorphic copy of $Bbb{Z}_p$ is in $Bbb{F}_{p^n}$. For example, in $Bbb{F}_9$, the constant polynomials will be an isomorphic copy of $Bbb{Z}_3$.
begin{align*}
Bbb{F}_9 & =Bbb{Z}_3[x]/langle x^2+1rangle\
& ={ax+b , | , a,b in Bbb{Z}_3, x^2+1 equiv 0}\
&={0,1,2,x,x+1,x+2,2x,2x+1,2x+2}
end{align*}
Then the set of constant polynomials ${0,1,2}$ is isomorphic to $Bbb{Z}_3$.
With your edited version:
$Bbb{F}_{p^n}$ can be thought of as a vector space over the field of scalars $Bbb{Z}_p$. Thus each object in $Bbb{F}_{p^n}$ is a $n-$tuple vector (this is very similar to the polynomial representation I have used above).
So consider the set of the vectors
$$S={(0,0,ldots ,c) , | , c in Bbb{Z}_p}.$$
This set in $Bbb{F}_{p^n}$ is isomorphic to $Bbb{Z}_p$.
answered Jan 6 at 7:42
Anurag AAnurag A
26k12250
$endgroup$
add a comment |
$begingroup$
Technically it means that an isomorphic copy of $Bbb{Z}_p$ is in $Bbb{F}_{p^n}$. For example, in $Bbb{F}_9$, the constant polynomials will be an isomorphic copy of $Bbb{Z}_3$.
begin{align*}
Bbb{F}_9 & =Bbb{Z}_3[x]/langle x^2+1rangle\
& ={ax+b , | , a,b in Bbb{Z}_3, x^2+1 equiv 0}\
&={0,1,2,x,x+1,x+2,2x,2x+1,2x+2}
end{align*}
Then the set of constant polynomials ${0,1,2}$ is isomorphic to $Bbb{Z}_3$.
With your edited version:
$Bbb{F}_{p^n}$ can be thought of as a vector space over the field of scalars $Bbb{Z}_p$. Thus each object in $Bbb{F}_{p^n}$ is a $n-$tuple vector (this is very similar to the polynomial representation I have used above).
So consider the set of the vectors
$$S={(0,0,ldots ,c) , | , c in Bbb{Z}_p}.$$
This set in $Bbb{F}_{p^n}$ is isomorphic to $Bbb{Z}_p$.
answered Jan 6 at 7:42
Anurag AAnurag A
26k12250
$endgroup$
add a comment |
$begingroup$
Technically it means that an isomorphic copy of $Bbb{Z}_p$ is in $Bbb{F}_{p^n}$. For example, in $Bbb{F}_9$, the constant polynomials will be an isomorphic copy of $Bbb{Z}_3$.
begin{align*}
Bbb{F}_9 & =Bbb{Z}_3[x]/langle x^2+1rangle\
& ={ax+b , | , a,b in Bbb{Z}_3, x^2+1 equiv 0}\
&={0,1,2,x,x+1,x+2,2x,2x+1,2x+2}
end{align*}
Then the set of constant polynomials ${0,1,2}$ is isomorphic to $Bbb{Z}_3$.
With your edited version:
$Bbb{F}_{p^n}$ can be thought of as a vector space over the field of scalars $Bbb{Z}_p$. Thus each object in $Bbb{F}_{p^n}$ is a $n-$tuple vector (this is very similar to the polynomial representation I have used above).
So consider the set of the vectors
$$S={(0,0,ldots ,c) , | , c in Bbb{Z}_p}.$$
This set in $Bbb{F}_{p^n}$ is isomorphic to $Bbb{Z}_p$.
answered Jan 6 at 7:42
Anurag AAnurag A
26k12250
$endgroup$
Technically it means that an isomorphic copy of $Bbb{Z}_p$ is in $Bbb{F}_{p^n}$. For example, in $Bbb{F}_9$, the constant polynomials will be an isomorphic copy of $Bbb{Z}_3$.
begin{align*}
Bbb{F}_9 & =Bbb{Z}_3[x]/langle x^2+1rangle\
& ={ax+b , | , a,b in Bbb{Z}_3, x^2+1 equiv 0}\
&={0,1,2,x,x+1,x+2,2x,2x+1,2x+2}
end{align*}
Then the set of constant polynomials ${0,1,2}$ is isomorphic to $Bbb{Z}_3$.
With your edited version:
$Bbb{F}_{p^n}$ can be thought of as a vector space over the field of scalars $Bbb{Z}_p$. Thus each object in $Bbb{F}_{p^n}$ is a $n-$tuple vector (this is very similar to the polynomial representation I have used above).
So consider the set of the vectors
$$S={(0,0,ldots ,c) , | , c in Bbb{Z}_p}.$$
This set in $Bbb{F}_{p^n}$ is isomorphic to $Bbb{Z}_p$.
answered Jan 6 at 7:42
Anurag AAnurag A
26k12250
edited Jan 6 at 7:50
answered Jan 6 at 7:42
Anurag AAnurag A
26k12250
answered Jan 6 at 7:42
Anurag AAnurag A
26k12250
answered Jan 6 at 7:42
Anurag AAnurag A
26k12250
26k12250
add a comment |
add a comment |
$begingroup$
The mapping ${Bbb Z}_prightarrow {Bbb F}_{p^n}$ given by $amapsto acdot 1$, where $acdot 1 = 1+ldots+1$ ($a$-times) is the $a$-multiple of $1$ and $1$ is the unit element of ${Bbb F}_{p^n}$, is a (ring) monomorphism. In this way, ${Bbb F}_{p^n}$ contains an isomorphic copy of ${Bbb Z}_p$.
answered Jan 6 at 8:10
WuestenfuxWuestenfux
4,4981413
$endgroup$
$begingroup$
I wanted a representation-independent way to define this subset. Hence, this useful. We can use "double and repeat" if $a$ is large.
$endgroup$
– Jus12
Jan 6 at 8:28
$begingroup$
Indeed, ''double and repeat'' can be used.
$endgroup$
– Wuestenfux
Jan 6 at 8:28
add a comment |
$begingroup$
The mapping ${Bbb Z}_prightarrow {Bbb F}_{p^n}$ given by $amapsto acdot 1$, where $acdot 1 = 1+ldots+1$ ($a$-times) is the $a$-multiple of $1$ and $1$ is the unit element of ${Bbb F}_{p^n}$, is a (ring) monomorphism. In this way, ${Bbb F}_{p^n}$ contains an isomorphic copy of ${Bbb Z}_p$.
answered Jan 6 at 8:10
WuestenfuxWuestenfux
4,4981413
$endgroup$
$begingroup$
I wanted a representation-independent way to define this subset. Hence, this useful. We can use "double and repeat" if $a$ is large.
$endgroup$
– Jus12
Jan 6 at 8:28
$begingroup$
Indeed, ''double and repeat'' can be used.
$endgroup$
– Wuestenfux
Jan 6 at 8:28
add a comment |
$begingroup$
The mapping ${Bbb Z}_prightarrow {Bbb F}_{p^n}$ given by $amapsto acdot 1$, where $acdot 1 = 1+ldots+1$ ($a$-times) is the $a$-multiple of $1$ and $1$ is the unit element of ${Bbb F}_{p^n}$, is a (ring) monomorphism. In this way, ${Bbb F}_{p^n}$ contains an isomorphic copy of ${Bbb Z}_p$.
answered Jan 6 at 8:10
WuestenfuxWuestenfux
4,4981413
$endgroup$
The mapping ${Bbb Z}_prightarrow {Bbb F}_{p^n}$ given by $amapsto acdot 1$, where $acdot 1 = 1+ldots+1$ ($a$-times) is the $a$-multiple of $1$ and $1$ is the unit element of ${Bbb F}_{p^n}$, is a (ring) monomorphism. In this way, ${Bbb F}_{p^n}$ contains an isomorphic copy of ${Bbb Z}_p$.
answered Jan 6 at 8:10
WuestenfuxWuestenfux
4,4981413
answered Jan 6 at 8:10
WuestenfuxWuestenfux
4,4981413
answered Jan 6 at 8:10
WuestenfuxWuestenfux
4,4981413
answered Jan 6 at 8:10
WuestenfuxWuestenfux
4,4981413
4,4981413
$begingroup$
I wanted a representation-independent way to define this subset. Hence, this useful. We can use "double and repeat" if $a$ is large.
$endgroup$
– Jus12
Jan 6 at 8:28
$begingroup$
Indeed, ''double and repeat'' can be used.
$endgroup$
– Wuestenfux
Jan 6 at 8:28
add a comment |
$begingroup$
I wanted a representation-independent way to define this subset. Hence, this useful. We can use "double and repeat" if $a$ is large.
$endgroup$
– Jus12
Jan 6 at 8:28
$begingroup$
Indeed, ''double and repeat'' can be used.
$endgroup$
– Wuestenfux
Jan 6 at 8:28
$begingroup$
I wanted a representation-independent way to define this subset. Hence, this useful. We can use "double and repeat" if $a$ is large.
$endgroup$
– Jus12
Jan 6 at 8:28
$begingroup$
I wanted a representation-independent way to define this subset. Hence, this useful. We can use "double and repeat" if $a$ is large.
$endgroup$
– Jus12
Jan 6 at 8:28
$begingroup$
Indeed, ''double and repeat'' can be used.
$endgroup$
– Wuestenfux
Jan 6 at 8:28
$begingroup$
Indeed, ''double and repeat'' can be used.
$endgroup$
– Wuestenfux
Jan 6 at 8:28
add a comment |
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Not an answer, just two comments. (i) $mathbb{F}_p$ cannot be a (proper) extension field; (ii) equally important, though you don't ask about it, is that $mathbb{F}_{p^n}$ contains exactly one $mathbb{F}_p$, and there is only one way of setting up an isomorphism $mathbb{F}_ptomathbb{F}_{p^n}$.
$endgroup$
– ancientmathematician
Jan 6 at 7:51
$begingroup$
@ancientmathematician for (i), I meant some other extension field $mathbb{F}_{p^m}$, where $m$ divides $n$. Also, good point (ii).
$endgroup$
– Jus12
Jan 6 at 8:09