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How to calculate the product of the below multiplication of prime ideals in Z[√(-5)]?

(2, 1-√(-5))(3, 1+√(-5))

I know it can be firstly expressed as a non-principal ideal generated by three numbers in Z[√(-5)],

(6, 2+2√(-5), 3-3√(-5))

but I don't know if it can be further simplified (and how, if it can).

Thanks for your help!

Jyrki Lahtonen
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Yuan Liu
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    Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Apr 09 '23 at 07:52
  • Please check out the abridged guide to new askers. It is difficult to give a helpful answer to your question, because your background is in the dark. You do hint at knowing this ring is NOT a principal ideal domain (making your choice of tag very strange). But how familiar are you with manipulating ideals of rings of integers of number fields? In this ring the ideals are necessarily free abelian groups of rank two, so how could you possibly need more than two generators. Also, do search for other questions about this ring on the site. – Jyrki Lahtonen Apr 09 '23 at 08:01
  • The background of my inquiry is, in my understanding, cross-multiplication of prime ideals shall correspond to prime factorization. In this case, (2,1+√(-5))(2,1-√(-5))=(2), (3,1+√(-5))(3,1-√(-5))=(3); (2,1+√(-5))(3,1+√(-5))=(1+√(-5)), (2,1-√(-5))(3,1-√(-5))=(1-√(-5)); these two pairs correspond to the two ways of prime factorization of 6 in Z[√(-5)]. However, I cannot figure out how to interpret the remaining possible cross-multiplication, namely (2,1-√(-5))(3,1+√(-5)) and (2,1+√(-5))(3,1-√(-5)). – Yuan Liu Apr 09 '23 at 08:11
  • This thread has a lot of example calculations on how to manipulate ideals of this ring. – Jyrki Lahtonen Apr 09 '23 at 08:12
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    Looking at this ideal a subgroup of $\Bbb{Z}[\sqrt{-5}]$, using the integral basis ${1,\sqrt{-5}}$, we see that the generators have coordinate vectors $(6,0),(2,2),(3,-3)$ so the vector $(6,0)-(2,2)-(3,-3)=(1,1)$ is in there. The ideal has norm six, so sure looks like it must be the principal ideal $(1+\sqrt{-5})$, doesn't it? Anyway, because the class group has size two, it follows that the product of two non-principal ideals is principal. – Jyrki Lahtonen Apr 09 '23 at 08:20
  • The backgroun belongs to the question body, not into a comment. – Jyrki Lahtonen Apr 09 '23 at 08:26
  • More – Jyrki Lahtonen Apr 09 '23 at 08:32
  • As here in the dupe, put $,a,b,c = 2,3,1+w,\ w=\sqrt{-5},$ in

    $$\color{#c00}{(a,b,c)=(1)},\Rightarrow, (a,c)(b,c) = (ab,\color{#c00}{(a,b,c)}c) = (c),\ \ {\rm by}\ \ c\mid ab\qquad\qquad$$ where we used $,(a,c) = (2,1+w) = (2,1-w)\ \ $

     – Bill Dubuque Apr 09 '23 at 09:33
  • There are many possible dupe targets for this but, alas, they are not easy to locate with the existing search tools. – Bill Dubuque Apr 09 '23 at 09:43
  • See esp. this answer for some key ideas on factorization refinement (generalizing 4NT = Euler's Four Number Theorem). – Bill Dubuque Apr 09 '23 at 23:47

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