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Finding prime factorization of ideals in number rings

asked 2016-10-21 08:28:30 -0500

nebuckandazzer gravatar image

updated 2016-10-21 09:11:14 -0500

Let $K$ be a number field and $O_K$ its ring of algebraic integers. Let $p\in\mathbb{Z}$ be a rational prime. I want to find the factorization of the ideal $pO_K$. What is the syntax for this ?

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

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answered 2016-10-21 10:00:48 -0500

A.Alharbi gravatar image

Define your number field $\mathbb{Q}(\alpha), \alpha = \sqrt{2} + i$ .

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 ( 2016-10-22 02:32:15 -0500 )edit

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

A.Alharbi gravatar imageA.Alharbi ( 2016-10-26 13:32:50 -0500 )edit

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

nebuckandazzer gravatar imagenebuckandazzer ( 2016-10-31 01:51:08 -0500 )edit

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 ( 2016-11-20 16:14:53 -0500 )edit

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Asked: 2016-10-21 08:28:30 -0500

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Last updated: Oct 21 '16