Assume that $q$ is a power of a prime number. Consider the field extension $F_q \subset F_{q^n}$ . Both fields can be thought as subfields of the algebraic closure of $F_p$, defined with
K=GF(p).algebraic_closure()
The question is: Is it possible to compute a basis of $F_{q^n}$ over $F_{q}$ ? Or at least to have something in the spirit of
K.<x1,...,xn>=GF(q)[]
which can be done when $q$ is prime?
You can always do find these elements
sage: a = K.multiplicative_generator()
sage: basis = [a**i for i in range(dim)]
where K is your finite field and dim is the dimension of your vector space (or the codimension of the ground field sitting inside the big one).
Vincent
Asked: 2015-06-07 22:34:22 +0100
Seen: 1,470 times
Last updated: Jun 09 '15
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What do you mean by a basis? Suppose you have
sage: K = GF(q,'a'). Then you can typesage: (L,f) = K.extension(2,'b',map=True)to get inLthe finite field with q^2 elements, defined as an extensionK. Andfgives you the embedding ofKinsideL.I am interested in finding $n$ elements of $F_{q^n}$ linear-independent over $F_q$, even when $q$ is not prime.