Thank you very much, Dima and John.
As they say in Spanish: "No te acostarás sin saber una cosa más."
Guillermo
On Wed, 27 Nov 2024 at 12:26, John Cremona wrote:
> Yes, b.list() works instead of list(b).
>
> There is also b.polynomial() which gives it as a polynomial over the prime
> field, s
Yes, b.list() works instead of list(b).
There is also b.polynomial() which gives it as a polynomial over the prime
field, so I could have done either Zy(b.list()) or Zy(b.polynomial()) as
you can check.
I agree that b.coeffs() would make sense to have. Also, the parent field
F125 = GF(5^3) ha
I would have never guessed that list(b) works. Should there be b.list() or
b.coefs() ?
On 27 November 2024 02:48:05 GMT-06:00, John Cremona
wrote:
>Use list() to get the coefficients:
>
>sage: P=x^3-2*x^2-x-2
>sage: F125.=GF(5^3,name='a',modulus=P)
>sage: b=a^37;b
>4*a^2 + 3*a + 1
>sage: Zy. =
Use list() to get the coefficients:
sage: P=x^3-2*x^2-x-2
sage: F125.=GF(5^3,name='a',modulus=P)
sage: b=a^37;b
4*a^2 + 3*a + 1
sage: Zy. = ZZ[]
sage: Zy(list(b))
4*y^2 + 3*y + 1
On Wednesday, 27 November 2024 at 06:51:57 UTC Dima Pasechnik wrote:
> A natural way would be to construct the quotie
A natural way would be to construct the quotient ring of GF(5)[x] modulo (P),
then b will be a polynomial in x, and you will have direct access to its
coefficients.
On 26 November 2024 16:46:50 GMT-06:00, "G. M.-S." wrote:
>Already asked on
>https://ask.sagemath.org/question/80389/conversion-
