# How make Kummer extensions

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I want to calculate the relative discriminant of field extensions of this kind: $$\mathbb{Q}(\zeta_5)(\sqrt[5]{a})$$ Where $a \in \mathbb{Q}(\zeta_5)$. So I use SAGE and make this calculations:

K.<b>=CyclotomicField(5);    //my field base
alpha=1+3*b^2;               //an element of my field base
f=(1+3*b^2).minpoly();       //its minimal polynomial
f.is_irreducible()           //is it irreducible?
R.<a>=K.extension(f)         //the field extension of my field base
R.relative_discriminant()    //the calculation of the relative discriminant


But when I execute it, appears this error

defining polynomial (x^4 - x^3 + 6*x^2 + 14*x + 61) must be irreducible


But it is irreducible, what am I doing wrong? Or how can I solve this?

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Note: I have NOT studied cyclotomic fields.

I believe the problem lies in minpoly() function. It has been reported that minpoly sometimes doesn not return an irreducible polynomial.

Also, if you check where f belongs, you will see that it is not in $\mathbb{Q}(\xi)_5[x]$ !.

sage: K.<b>=CyclotomicField(5);
sage: alpha=1+3*b^2;
sage: f=(1+3*b^2).minpoly()
sage: f.parent()
Univariate Polynomial Ring in x over Rational Field
sage: factor(f)
x^4 - x^3 + 6*x^2 + 14*x + 61
sage: G.<x> = K[] #Define a polynomial ring over K
sage: f = G(f) #Coerce f into G
sage: f.parent()
Univariate Polynomial Ring in x over Cyclotomic Field of order 5 and degree 4
sage: factor(f)
(x - 3*b - 1) * (x - 3*b^2 - 1) * (x - 3*b^3 - 1) * (x + 3*b^3 + 3*b^2 + 3*b + 2)
sage: f.is_irreducible()
False


Update: Could you please elaborate which element would you like to append it to $\mathbb{Q}(\xi)_5$? It seems to me that $\alpha$ is already in $\mathbb{Q}(\xi)_5$ but $\sqrt[5]{\alpha}$ is not. If you use fractional expontiation or the function nth_root in minpoly, it will gives you an error. I am afraid that you have to hardcode the equation $x^5 - \alpha$

sage: alpha in K
True
sage: alpha^(1/5) in K
False
sage: L.<a> = K.extension(x^5 - alpha)
sage: L
Number Field in a with defining polynomial x^5 - 3*b^2 - 1 over its base field


Second update

If you use fractional exponentiation or the function nth_root in minpoly, it will gives you an error.

It seems an issue need to be solved.

by the way, what f = G(f) means?

$f$ was living in $\mathbb{Q}$,i.e. arithmetic on $f$ would be carry over $\mathbb{Q}$, writing f = G(f) tells sage $f$ is an element of $G$

Example:

sage: a = 7
sage: a^2
49


But,

sage: R = IntegerModRing(13)
sage: a = R(a)
sage: a
7
sage: a^2
10
sage: a + 6
0


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1

minpoly behaves as it should in this case. minpoly(alpha) is irreducible over Q, and (alpha being a root of it), it does have a root over K.

In your update you suggest the proper approach: adjoin the fifth root of alpha by constructing a polynomial that has the appropriate root.

( 2016-05-28 20:23:40 +0200 )edit

This post is a wiki. Anyone with karma >750 is welcome to improve it.

Ok, what is happening is that minoply returns de minimal polynomial over $\mathbb{Q}$ and not over a specific field. This answered me. by the way, what

f = G(f)

means?

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