Ask Your Question

# Revision history [back]

Regarding the first question, P is threated as a polynomial over GF(27):

sage: P.parent()
Univariate Polynomial Ring in x over Finite Field in c of size 3^3


Now, when you write P(i) for i in range(0,p**q), you should understand that i is an integer, so it has to be transformed into an element of K to be accepted by P. This morphism maps 1 to 1, so every integer remains stuck int the prime field GF(3).

If you want to compute P on each value of K, you should directly iterate oeve relement of K:

sage: e = [P(i) for i in K]


and you will see that the result is very different.

If you prefer a more "cyclic" view, you can use the multiplicative generator of K, which you named c:

sage: c = K.gen()
sage: e = [P(c**i) for i in range(0,p**q)]


Regarding the first question, P is threated as a polynomial over GF(27):

sage: P.parent()
Univariate Polynomial Ring in x over Finite Field in c of size 3^3


Now, when you write P(i) for i in range(0,p**q), you should understand that i is an integer, so it has to be transformed into an element of K to be accepted by P. This morphism maps 1 to 1, so every integer remains stuck int the prime field GF(3).

If you want to compute P on each value of K, you should directly iterate oeve relement of K:

sage: e = [P(i) for i in K]


and you will see that the result is very different.

If you prefer a more "cyclic" view, you can use the multiplicative generator of K, which you named c:

sage: c = K.gen()
sage: e = [P(c**i) for i in range(0,p**q)]


Regarding the first question, P is a polynomial over GF(27):

sage: P.parent()
Univariate Polynomial Ring in x over Finite Field in c of size 3^3


Now, when you write P(i) for i in range(0,p**q), you should understand that i is an integer, so it has to be transformed into an element of K to be accepted by P. This morphism maps 1 to 1, so every integer remains stuck int the prime field GF(3).. The additive struture of GF(p**q) is that of $(\mathbb{Z}/p\mathbb{Z})^q$, not $\mathbb{Z}/p^q\mathbb{Z}$

If you want to compute P on each value of K, you should directly iterate oeve relement of K:

sage: e = [P(i) for i in K]


and you will see that the result is very different.

If you prefer a more "cyclic" view, you can use the multiplicative generator of K, which you named c:

sage: c = K.gen()
sage: e = [P(c**i) for i in range(0,p**q)]


Regarding the first question, P is a polynomial over GF(27):

sage: P.parent()
Univariate Polynomial Ring in x over Finite Field in c of size 3^3


Now, when you write P(i) for i in range(0,p**q), you should understand that i is an integer, so it has to be transformed into an element of K to be accepted by P. This morphism maps 1 to 1, so every integer remains stuck int the prime field GF(3). The additive struture of GF(p**q) is that of $(\mathbb{Z}/p\mathbb{Z})^q$, not $\mathbb{Z}/p^q\mathbb{Z}$$\mathbb{Z}/p^q\mathbb{Z}$.

If you want to compute P on each value of K, you should directly iterate oeve relement of K:

sage: e = [P(i) for i in K]


and you will see that the result is very different.

If you prefer a more "cyclic" view, you can use the multiplicative generator of K, which you named c:

sage: c = K.gen()
sage: e = [P(c**i) for i in range(0,p**q)]