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Finding prime factorization of ideals in number rings

asked 8 years ago

nebuckandazzer gravatar image

updated 8 years ago

Let K be a number field and OK its ring of algebraic integers. Let pZ be a rational prime. I want to find the factorization of the ideal pOK. What is the syntax for this ?

For clarity, I request you to demonstrate with an example (say K=Q(2+i) and p=2 and p=3).

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answered 8 years ago

A.Alharbi gravatar image

Define your number field Q(α),α=2+i .

K.<a> = NumberField(definingPolynomial)

In your particular case

Z.<x> = ZZ[] #Makes x lives in Z[x]
K.<a> = NumberField( minpoly(sqrt(2)+i, x))
I = K.ideal(2)
factor(I)
#(Fractional ideal (1/12*a^3 - 1/4*a^2 - 5/12*a + 5/4))^4
#Even fancier
latex(factor(I))
#(\left(\frac{1}{12} a^{3} - \frac{1}{4} a^{2} - \frac{5}{12} a + \frac{5}{4}\right))^{4}
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What is a ?

nebuckandazzer gravatar imagenebuckandazzer ( 8 years ago )

a is the root of definingPolynomial i.e. f(a)=0

A.Alharbi gravatar imageA.Alharbi ( 8 years ago )

The discriminant of K=Q(2+i) is 256. As 3, the ideal \langle 3\rangle should remain inert in O_K. So how come it splits in O_K ?

nebuckandazzer gravatar imagenebuckandazzer ( 8 years ago )

The theorem says a prime p ramifies iff p|\Delta. However, this does NOT imply that a prime p which does not ramify must inert, because there is a third case in which the prime splits.

Look at problem for section 3.4, or for detailed treatment (Kummer-Dedekind theorem) see Stevenhagen's lecture notes Number Rings

A.Alharbi gravatar imageA.Alharbi ( 8 years ago )

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Asked: 8 years ago

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Last updated: Oct 21 '16