Yes I wanted the splitting field. I didn't know exactly the terminology of
all this things as I'm still learning most of it.
One more question. Would you happen to know any source that I can look into
it to see how or at least the vague idea of how the splitting field and the
factors are calculated given the polynomial? Not necessarily the algorithm
into its full details but at least the general idea. Mainly I'm more
interested into the complexity of it working if it's bounded to some
approximation of the factors or it's determined in some way. Thank you for
the answer anyways. It was very helpful and I'm sorry for the late reply as
after I've found it and your answer I went to look more into these things
and forgot to answer.
Em quinta-feira, 2 de janeiro de 2020 13:47:44 UTC-3, Brent W. Baccala
escreveu:
>
> On Tuesday, November 19, 2019 at 1:35:24 AM UTC-5, Pedro Vinícius Ferreira
> Baptista wrote:
>>
>> Given an irreducible polynomial P from Q[x] I want to find the primitive
>> element A of the extension field defined by it.
>> Moreover I want to know how to write the roots of P in terms of A(per
>> example, the coefficients of the polynomial of powers of A).
>>
>
> The extension field defined by the polynomial might not be able to factor
> the polynomial completely, so you might not be able to write the roots of P
> in terms of A.
>
> Probably what you want is a splitting field for the polynomial.
>
>
>> Finally I want to find the dimension of the vectors space of the
>> extension field of the roots of P.
>>
>
> Sounds like a splitting field. The dimension of the extension field by P
> would just be the degree of P. You want the dimension of the splitting
> field, right?
>
> Here's how I'd do it:
>
> sage: R.<x> = QQ[]
> sage: P = irreducible polynomial in x
> sage: A = P.splitting_field('a')
> sage: RA = PolynomialRing(A, 'x')
> sage: RA(P).factor()
>
>
> The degree of A's defining polynomial is the dimension of the splitting
> field as a vector space over QQ.
>
> The last step (factor) gives the roots of the polynomial in terms of the
> generator 'a'.
>
> P.S. Sorry for the late response. I don't read this group very often.
>
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