Thanks for the reply! Just one more question. In general, if k is a finite field, what is k(i), where i is an integer, suppose to be ?
The following example suggests k(i) will be the elements in the base field as i varies. sage: k.<a> = GF(9) sage: for i in [0..8]: print k(i) ....: 0 1 2 0 1 2 0 1 2 Shing On Sep 22, 4:13 pm, Simon King <[email protected]> wrote: > Hi Shing! > > Two possibilities: > > sage: k.<a> = GF(9) > sage: t=a^2+1 > > Now, you can learn about all possible methods for elements of k by > doing > sage: t.<TAB> > (you type t, dot, and hit the tab key). > This will show you a list of possible methods. > > One of the methods is called int_repr: > sage: t.int_repr() > '5' > So, it gives you a string, that you can easily transform into an > integer: ZZ(t.int_repr()) for a Sage integer, int(t.int_repr()) for a > Python integer. > > There is also log_to_int: > sage: t.log_to_int() > 5 > Probably this last method is easier. > > The inverse way is > sage: k.fetch_int(5) > a + 2 > which is indeed the same as t: > sage: t > a + 2 > > Cheers, > Simon --~--~---------~--~----~------------~-------~--~----~ To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/sage-support URLs: http://www.sagemath.org -~----------~----~----~----~------~----~------~--~---
