number fields and irreducible polynomials

Semin
37 ●7 ●8 ●14

Hello gentle people. Let's see an example.

 f=x^2+4
 g=x^4+16
 F=NumberField(f,'a')

The extension field of F by g cannot obtained by F.extension(g(s),'s') although we input s=var('s') before. Because g is reducible in F. ( g=(x^2-4I)(x^2+4I) )

Can I check irreducibility of a polynomial in a number field?

Is there any method which makes a number field by given polynomial?

Thanks in advance.

answered 8 years ago

tmonteil
27483 ●32 ●205 ●519 http://wiki.sagemath.o...

updated 8 years ago

You should define g as an element of the univariate polynomial ring with coefficients in F:

sage: R.<y> = F[]
sage: R
Univariate Polynomial Ring in y over Number Field in a with defining polynomial x^2 + 4
sage: g = y^4+16
sage: g.parent()
Univariate Polynomial Ring in y over Number Field in a with defining polynomial x^2 + 4
sage: g.is_irreducible()
False
sage: g.factor()
(y^2 - 2*a) * (y^2 + 2*a)

EDIT If you want the splitting field of g, together with a map to embed F into it, you can do:

sage: g.splitting_field('b')
Number Field in b with defining polynomial y^4 + 1
sage: g.splitting_field('b', map=True)
(Number Field in b with defining polynomial y^4 + 1, Ring morphism:
   From: Number Field in a with defining polynomial x^2 + 4
   To:   Number Field in b with defining polynomial y^4 + 1
   Defn: a |--> 2*b^2)

link

Comments

Hmm.. That's good information but when I have some more polynomials, it looks hard to control variables. When I want to get the splitting field of f1,f2,...,fn which are irreducible in the field of rationals, then I have to make R1.<y1>, R2.<y2>,...,Rn.<yn>, right?

Semin ( 8 years ago )

Well, you can use the same x all the time, but then you should be careful that along your computation, it soed not belong to the same parent, i chosed another indeterminate to avoid confusions.

tmonteil ( 8 years ago )

Okay, I see. Thank you very much. :-)

Semin ( 8 years ago )

^_^ .

tmonteil ( 8 years ago )

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