Galois field elements as Integers in $mathbb{Z}$ [closed]
$begingroup$
In the case of $GF(3)$ its elements are
gap> Elements(GF(3));
of which I get
[ 0*Z(3), Z(3)^0, Z(3)]. Further, as an example, to get the integer value of Z(3), I use
gap> Int(Z(3)); of which I get 2.
Now, how do I find the equivalent values of the elements of $GF(3^2)$ as integers of $mathbb{Z}$ through GAP? Int(Z(3^2)) gives me an error. And in general, the elements of $GF(p^n)$ as integers? Thank you so much!
gap> Elements(GF(3^2));
[ 0*Z(3), Z(3)^0, Z(3), Z(3^2), Z(3^2)^2, Z(3^2)^3, Z(3^2)^5, Z(3^2)^6, Z(3^2)^7 ]
abstract-algebra gap
asked Jan 19 at 7:25
primerprimer
92
$endgroup$
closed as unclear what you're asking by Eric Wofsey, A. Pongrácz, Cesareo, José Carlos Santos, metamorphy Jan 19 at 15:21
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 1 more comment
$begingroup$
In the case of $GF(3)$ its elements are
gap> Elements(GF(3));
of which I get
[ 0*Z(3), Z(3)^0, Z(3)]. Further, as an example, to get the integer value of Z(3), I use
gap> Int(Z(3)); of which I get 2.
Now, how do I find the equivalent values of the elements of $GF(3^2)$ as integers of $mathbb{Z}$ through GAP? Int(Z(3^2)) gives me an error. And in general, the elements of $GF(p^n)$ as integers? Thank you so much!
gap> Elements(GF(3^2));
[ 0*Z(3), Z(3)^0, Z(3), Z(3^2), Z(3^2)^2, Z(3^2)^3, Z(3^2)^5, Z(3^2)^6, Z(3^2)^7 ]
abstract-algebra gap
asked Jan 19 at 7:25
primerprimer
92
$endgroup$
closed as unclear what you're asking by Eric Wofsey, A. Pongrácz, Cesareo, José Carlos Santos, metamorphy Jan 19 at 15:21
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What do you mean by "as integers"?
$endgroup$
– Eric Wofsey
Jan 19 at 7:28
$begingroup$
And what do you mean by "equivalent"? Every element $x$ of $GF(p^n)$ has the property that $pcdot x = 0$ in that field; you won't find a set of integers with a similar property.
$endgroup$
– Greg Martin
Jan 19 at 7:53
$begingroup$
@EricWofsey For instance, in $GF(3),$ gap> Elements(GF(3)); [0*Z(3), Z(3)^0, Z(3)]. Using GAP gap> Int(Z(3)); 2. the number 2 is what i'm referring to 'as integers' in $mathbb{Z}.$
$endgroup$
– primer
Jan 19 at 11:18
$begingroup$
If $n>1$ the elements of $GF(p^n)$ outside $GF(p)$ cannot be represented by integers, but are algebraic elements over the prime field. Thus what you ask cannot be done (which is the reason for the error message you get).
$endgroup$
– ahulpke
Jan 19 at 17:38
$begingroup$
@ahulpke thanks. now, as an element of GF(3^2), I get Z(3^2)^3. how do I get to understand this? thanks
$endgroup$
– primer
Jan 19 at 18:10
|
show 1 more comment
$begingroup$
In the case of $GF(3)$ its elements are
gap> Elements(GF(3));
of which I get
[ 0*Z(3), Z(3)^0, Z(3)]. Further, as an example, to get the integer value of Z(3), I use
gap> Int(Z(3)); of which I get 2.
Now, how do I find the equivalent values of the elements of $GF(3^2)$ as integers of $mathbb{Z}$ through GAP? Int(Z(3^2)) gives me an error. And in general, the elements of $GF(p^n)$ as integers? Thank you so much!
gap> Elements(GF(3^2));
[ 0*Z(3), Z(3)^0, Z(3), Z(3^2), Z(3^2)^2, Z(3^2)^3, Z(3^2)^5, Z(3^2)^6, Z(3^2)^7 ]
abstract-algebra gap
asked Jan 19 at 7:25
primerprimer
92
$endgroup$
In the case of $GF(3)$ its elements are
gap> Elements(GF(3));
of which I get
[ 0*Z(3), Z(3)^0, Z(3)]. Further, as an example, to get the integer value of Z(3), I use
gap> Int(Z(3)); of which I get 2.
Now, how do I find the equivalent values of the elements of $GF(3^2)$ as integers of $mathbb{Z}$ through GAP? Int(Z(3^2)) gives me an error. And in general, the elements of $GF(p^n)$ as integers? Thank you so much!
gap> Elements(GF(3^2));
[ 0*Z(3), Z(3)^0, Z(3), Z(3^2), Z(3^2)^2, Z(3^2)^3, Z(3^2)^5, Z(3^2)^6, Z(3^2)^7 ]
abstract-algebra gap
abstract-algebra gap
asked Jan 19 at 7:25
primerprimer
92
asked Jan 19 at 7:25
primerprimer
92
edited Jan 19 at 16:19
primer
asked Jan 19 at 7:25
primerprimer
92
asked Jan 19 at 7:25
primerprimer
92
asked Jan 19 at 7:25
primerprimer
92
92
closed as unclear what you're asking by Eric Wofsey, A. Pongrácz, Cesareo, José Carlos Santos, metamorphy Jan 19 at 15:21
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Eric Wofsey, A. Pongrácz, Cesareo, José Carlos Santos, metamorphy Jan 19 at 15:21
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What do you mean by "as integers"?
$endgroup$
– Eric Wofsey
Jan 19 at 7:28
$begingroup$
And what do you mean by "equivalent"? Every element $x$ of $GF(p^n)$ has the property that $pcdot x = 0$ in that field; you won't find a set of integers with a similar property.
$endgroup$
– Greg Martin
Jan 19 at 7:53
$begingroup$
@EricWofsey For instance, in $GF(3),$ gap> Elements(GF(3)); [0*Z(3), Z(3)^0, Z(3)]. Using GAP gap> Int(Z(3)); 2. the number 2 is what i'm referring to 'as integers' in $mathbb{Z}.$
$endgroup$
– primer
Jan 19 at 11:18
$begingroup$
If $n>1$ the elements of $GF(p^n)$ outside $GF(p)$ cannot be represented by integers, but are algebraic elements over the prime field. Thus what you ask cannot be done (which is the reason for the error message you get).
$endgroup$
– ahulpke
Jan 19 at 17:38
$begingroup$
@ahulpke thanks. now, as an element of GF(3^2), I get Z(3^2)^3. how do I get to understand this? thanks
$endgroup$
– primer
Jan 19 at 18:10
|
show 1 more comment
$begingroup$
What do you mean by "as integers"?
$endgroup$
– Eric Wofsey
Jan 19 at 7:28
$begingroup$
And what do you mean by "equivalent"? Every element $x$ of $GF(p^n)$ has the property that $pcdot x = 0$ in that field; you won't find a set of integers with a similar property.
$endgroup$
– Greg Martin
Jan 19 at 7:53
$begingroup$
@EricWofsey For instance, in $GF(3),$ gap> Elements(GF(3)); [0*Z(3), Z(3)^0, Z(3)]. Using GAP gap> Int(Z(3)); 2. the number 2 is what i'm referring to 'as integers' in $mathbb{Z}.$
$endgroup$
– primer
Jan 19 at 11:18
$begingroup$
If $n>1$ the elements of $GF(p^n)$ outside $GF(p)$ cannot be represented by integers, but are algebraic elements over the prime field. Thus what you ask cannot be done (which is the reason for the error message you get).
$endgroup$
– ahulpke
Jan 19 at 17:38
$begingroup$
@ahulpke thanks. now, as an element of GF(3^2), I get Z(3^2)^3. how do I get to understand this? thanks
$endgroup$
– primer
Jan 19 at 18:10
$begingroup$
What do you mean by "as integers"?
$endgroup$
– Eric Wofsey
Jan 19 at 7:28
$begingroup$
What do you mean by "as integers"?
$endgroup$
– Eric Wofsey
Jan 19 at 7:28
$begingroup$
And what do you mean by "equivalent"? Every element $x$ of $GF(p^n)$ has the property that $pcdot x = 0$ in that field; you won't find a set of integers with a similar property.
$endgroup$
– Greg Martin
Jan 19 at 7:53
$begingroup$
And what do you mean by "equivalent"? Every element $x$ of $GF(p^n)$ has the property that $pcdot x = 0$ in that field; you won't find a set of integers with a similar property.
$endgroup$
– Greg Martin
Jan 19 at 7:53
$begingroup$
@EricWofsey For instance, in $GF(3),$ gap> Elements(GF(3)); [0*Z(3), Z(3)^0, Z(3)]. Using GAP gap> Int(Z(3)); 2. the number 2 is what i'm referring to 'as integers' in $mathbb{Z}.$
$endgroup$
– primer
Jan 19 at 11:18
$begingroup$
@EricWofsey For instance, in $GF(3),$ gap> Elements(GF(3)); [0*Z(3), Z(3)^0, Z(3)]. Using GAP gap> Int(Z(3)); 2. the number 2 is what i'm referring to 'as integers' in $mathbb{Z}.$
$endgroup$
– primer
Jan 19 at 11:18
$begingroup$
If $n>1$ the elements of $GF(p^n)$ outside $GF(p)$ cannot be represented by integers, but are algebraic elements over the prime field. Thus what you ask cannot be done (which is the reason for the error message you get).
$endgroup$
– ahulpke
Jan 19 at 17:38
$begingroup$
If $n>1$ the elements of $GF(p^n)$ outside $GF(p)$ cannot be represented by integers, but are algebraic elements over the prime field. Thus what you ask cannot be done (which is the reason for the error message you get).
$endgroup$
– ahulpke
Jan 19 at 17:38
$begingroup$
@ahulpke thanks. now, as an element of GF(3^2), I get Z(3^2)^3. how do I get to understand this? thanks
$endgroup$
– primer
Jan 19 at 18:10
$begingroup$
@ahulpke thanks. now, as an element of GF(3^2), I get Z(3^2)^3. how do I get to understand this? thanks
$endgroup$
– primer
Jan 19 at 18:10
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
If $n=1$, then the field GF(p) is a quotient of the integeres, but if $n>1$ it is not. In general $GF(p^n)$ can be expressed as $GF(p)[x]/f(x)$ for some polynomial. You can ask GAP for this polynomial $f(x)$ by writting
ConwayPolynomial( p, n )
Any element of $GF(p^n)$ can be express as
$$a_0+a_1z+cdots+a_{n-1}z^{n-1}$$
for some integers $0le a_i <p$, where $z$ is a fixed root of $f(x)$.
There is no direct function in GAP that gives you that.
answered Jan 19 at 8:44
xarlesxarles
1,50079
$endgroup$
add a comment |
$begingroup$
$[0*Z(3), Z(3)^0, Z(3), Z(3^2), Z(3^2)^2, Z(3^2)^3, Z(3^2)^5, Z(3^2)^6, Z(3^2)^7 ]$
What its meant is that
$$[0,alpha^0,alpha,alpha^2,(alpha^2)^2=alpha^4,alpha^3,alpha^5,alpha^6,alpha^7=1].
$$
The ordering comes from the fact that $alpha,alpha^2,alpha^4$ and $alpha^3,alpha^5,alpha^6$ are the zeros of the two conjugate (primitive) polynomials $x^3+x+1$ and $x^3+x^2+1$, resp.
answered Jan 19 at 15:16
WuestenfuxWuestenfux
4,6591413
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $n=1$, then the field GF(p) is a quotient of the integeres, but if $n>1$ it is not. In general $GF(p^n)$ can be expressed as $GF(p)[x]/f(x)$ for some polynomial. You can ask GAP for this polynomial $f(x)$ by writting
ConwayPolynomial( p, n )
Any element of $GF(p^n)$ can be express as
$$a_0+a_1z+cdots+a_{n-1}z^{n-1}$$
for some integers $0le a_i <p$, where $z$ is a fixed root of $f(x)$.
There is no direct function in GAP that gives you that.
answered Jan 19 at 8:44
xarlesxarles
1,50079
$endgroup$
add a comment |
$begingroup$
If $n=1$, then the field GF(p) is a quotient of the integeres, but if $n>1$ it is not. In general $GF(p^n)$ can be expressed as $GF(p)[x]/f(x)$ for some polynomial. You can ask GAP for this polynomial $f(x)$ by writting
ConwayPolynomial( p, n )
Any element of $GF(p^n)$ can be express as
$$a_0+a_1z+cdots+a_{n-1}z^{n-1}$$
for some integers $0le a_i <p$, where $z$ is a fixed root of $f(x)$.
There is no direct function in GAP that gives you that.
answered Jan 19 at 8:44
xarlesxarles
1,50079
$endgroup$
add a comment |
$begingroup$
If $n=1$, then the field GF(p) is a quotient of the integeres, but if $n>1$ it is not. In general $GF(p^n)$ can be expressed as $GF(p)[x]/f(x)$ for some polynomial. You can ask GAP for this polynomial $f(x)$ by writting
ConwayPolynomial( p, n )
Any element of $GF(p^n)$ can be express as
$$a_0+a_1z+cdots+a_{n-1}z^{n-1}$$
for some integers $0le a_i <p$, where $z$ is a fixed root of $f(x)$.
There is no direct function in GAP that gives you that.
answered Jan 19 at 8:44
xarlesxarles
1,50079
$endgroup$
If $n=1$, then the field GF(p) is a quotient of the integeres, but if $n>1$ it is not. In general $GF(p^n)$ can be expressed as $GF(p)[x]/f(x)$ for some polynomial. You can ask GAP for this polynomial $f(x)$ by writting
ConwayPolynomial( p, n )
Any element of $GF(p^n)$ can be express as
$$a_0+a_1z+cdots+a_{n-1}z^{n-1}$$
for some integers $0le a_i <p$, where $z$ is a fixed root of $f(x)$.
There is no direct function in GAP that gives you that.
answered Jan 19 at 8:44
xarlesxarles
1,50079
answered Jan 19 at 8:44
xarlesxarles
1,50079
answered Jan 19 at 8:44
xarlesxarles
1,50079
answered Jan 19 at 8:44
xarlesxarles
1,50079
1,50079
add a comment |
add a comment |
$begingroup$
$[0*Z(3), Z(3)^0, Z(3), Z(3^2), Z(3^2)^2, Z(3^2)^3, Z(3^2)^5, Z(3^2)^6, Z(3^2)^7 ]$
What its meant is that
$$[0,alpha^0,alpha,alpha^2,(alpha^2)^2=alpha^4,alpha^3,alpha^5,alpha^6,alpha^7=1].
$$
The ordering comes from the fact that $alpha,alpha^2,alpha^4$ and $alpha^3,alpha^5,alpha^6$ are the zeros of the two conjugate (primitive) polynomials $x^3+x+1$ and $x^3+x^2+1$, resp.
answered Jan 19 at 15:16
WuestenfuxWuestenfux
4,6591413
$endgroup$
add a comment |
$begingroup$
$[0*Z(3), Z(3)^0, Z(3), Z(3^2), Z(3^2)^2, Z(3^2)^3, Z(3^2)^5, Z(3^2)^6, Z(3^2)^7 ]$
What its meant is that
$$[0,alpha^0,alpha,alpha^2,(alpha^2)^2=alpha^4,alpha^3,alpha^5,alpha^6,alpha^7=1].
$$
The ordering comes from the fact that $alpha,alpha^2,alpha^4$ and $alpha^3,alpha^5,alpha^6$ are the zeros of the two conjugate (primitive) polynomials $x^3+x+1$ and $x^3+x^2+1$, resp.
answered Jan 19 at 15:16
WuestenfuxWuestenfux
4,6591413
$endgroup$
add a comment |
$begingroup$
$[0*Z(3), Z(3)^0, Z(3), Z(3^2), Z(3^2)^2, Z(3^2)^3, Z(3^2)^5, Z(3^2)^6, Z(3^2)^7 ]$
What its meant is that
$$[0,alpha^0,alpha,alpha^2,(alpha^2)^2=alpha^4,alpha^3,alpha^5,alpha^6,alpha^7=1].
$$
The ordering comes from the fact that $alpha,alpha^2,alpha^4$ and $alpha^3,alpha^5,alpha^6$ are the zeros of the two conjugate (primitive) polynomials $x^3+x+1$ and $x^3+x^2+1$, resp.
answered Jan 19 at 15:16
WuestenfuxWuestenfux
4,6591413
$endgroup$
$[0*Z(3), Z(3)^0, Z(3), Z(3^2), Z(3^2)^2, Z(3^2)^3, Z(3^2)^5, Z(3^2)^6, Z(3^2)^7 ]$
What its meant is that
$$[0,alpha^0,alpha,alpha^2,(alpha^2)^2=alpha^4,alpha^3,alpha^5,alpha^6,alpha^7=1].
$$
The ordering comes from the fact that $alpha,alpha^2,alpha^4$ and $alpha^3,alpha^5,alpha^6$ are the zeros of the two conjugate (primitive) polynomials $x^3+x+1$ and $x^3+x^2+1$, resp.
answered Jan 19 at 15:16
WuestenfuxWuestenfux
4,6591413
answered Jan 19 at 15:16
WuestenfuxWuestenfux
4,6591413
answered Jan 19 at 15:16
WuestenfuxWuestenfux
4,6591413
answered Jan 19 at 15:16
WuestenfuxWuestenfux
4,6591413
4,6591413
add a comment |
add a comment |
$begingroup$
What do you mean by "as integers"?
$endgroup$
– Eric Wofsey
Jan 19 at 7:28
$begingroup$
And what do you mean by "equivalent"? Every element $x$ of $GF(p^n)$ has the property that $pcdot x = 0$ in that field; you won't find a set of integers with a similar property.
$endgroup$
– Greg Martin
Jan 19 at 7:53
$begingroup$
@EricWofsey For instance, in $GF(3),$ gap> Elements(GF(3)); [0*Z(3), Z(3)^0, Z(3)]. Using GAP gap> Int(Z(3)); 2. the number 2 is what i'm referring to 'as integers' in $mathbb{Z}.$
$endgroup$
– primer
Jan 19 at 11:18
$begingroup$
If $n>1$ the elements of $GF(p^n)$ outside $GF(p)$ cannot be represented by integers, but are algebraic elements over the prime field. Thus what you ask cannot be done (which is the reason for the error message you get).
$endgroup$
– ahulpke
Jan 19 at 17:38
$begingroup$
@ahulpke thanks. now, as an element of GF(3^2), I get Z(3^2)^3. how do I get to understand this? thanks
$endgroup$
– primer
Jan 19 at 18:10