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Let $w=\sqrt{-5}$; we work in $\mathbb Z[w]$; which is not a PID or UFD.

Now in Sage (3) factors in $(3,1+w)(3,2+w)$ and (7) in $(7,3+w)(7,4+w)$.

Please explain clearly how factorization of ideals in a quadratic non-PID works.

Is it because this is a non-PID quadratic numberfield that the ideal factors in two non-principal ideals?

The numbers 11 and 13 are irreducible as 3 and 7 are, yet (11) and (13) do not factor as ideals. How do I determine the difference in this context?

EDIT: implementation of solution provided by Dietrich Burde (second cell in Sage Worksheet)

worksheet

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  • $\begingroup$ You might want to read this paper. $\endgroup$ – rogerl Nov 29 '14 at 13:53
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Factorisation of ideals in the ring of integers $\mathcal{O}_K$ of a quadratic number field $K=\mathbb{Q}(\sqrt{d})$ works as follows. Let $D$ be the discriminant, i.e., $D=d$ or $D=4d$.

Theorem: Let $p>2$ be a prime and $K$ a quadratic number field as above.

1.) If the Kronecker symbol $(D/p)=0$ then $(p)=(p,\sqrt{d})^2$ ramifies.
2.) If the Kronecker symbol $(D/p)=1$ then $(p)=P_1P_2$ splits, with two prime ideals $P_1,P_2$.
3.) If the Kronecker symbol $(D/p)=-1$ then $(p)$ is inert, i.e., remains prime.

In the second case, $P_1=(p, x+\sqrt{d})$ and $P_2=(p, x-\sqrt{d})$ with an integer $x$ such that $D\equiv x^2 (p)$. For $p=2$ there is a similar result.

This gives a formula how to factorize ideals, regardless whether or not $\mathcal{O}_K$ is a PID or UFD.

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  • $\begingroup$ Looks awesome, very clear indeed! Will go through my books and see why I missed this. $\endgroup$ – nilo de roock Nov 29 '14 at 17:07
  • $\begingroup$ Your books probably say "Legendre symbol" rather than "Kronecker symbol." And your CAS might expect "Jacobi symbol." $\endgroup$ – Robert Soupe Nov 29 '14 at 18:27
  • $\begingroup$ I mentioned Sage in the question, in Sage Legendre symbol is referred to as kronecker $\endgroup$ – nilo de roock Nov 29 '14 at 18:41
  • $\begingroup$ I added a follow-up question math.stackexchange.com/questions/1044017/… $\endgroup$ – nilo de roock Nov 29 '14 at 19:21
  • $\begingroup$ For what it's worth, Mathematica has always had JacobiSymbol, whereas KroneckerSymbol was introduced in Version 6.0. $\endgroup$ – Robert Soupe Nov 29 '14 at 20:07

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