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To show that, in $\mathbb Z$, we have the factorization of ideals $$(6) = (2, 1+\sqrt {-5})^2 (3, 1+\sqrt{-5}) (3, 1-\sqrt{-5})$$

is there any better way than write down the generic product with indeterminates? (that should be 8 indeterminate coefficients that I should end up showing simplifying and find a multiple of 6)?

TheStudent
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    Have you tried showing that $(2,1+\sqrt{-5})^2=(2)$ and the other pair produces $(3)$? Then use associativity of the product of ideals :-) – Jyrki Lahtonen Nov 21 '20 at 18:10
  • @JyrkiLahtonen Thank you very much for the tip! I succeded finding the result for $(2)$, however for the product of the two other ideals, I don't find $3$ in it. Indeed I get $9ac+6bd+3cb+3ad + \sqrt{-5}(3cb-3ad)$ and I cannot make the second factor zero and the first $3$... Am I mistaken somewhere? – TheStudent Nov 22 '20 at 02:46
  • The product rule for ideals given in terms of generators is that if $I=(a_1,a_2,...,a_n)$ and $J=(b_1,b_2,...,b_m)$ then $IJ=(a_1b_1,a_1b_2,...,a_ib_j,...,a_nb_m)$. That gives $$(3,1+\sqrt{-5})(3,1-\sqrt{-5})=(9,3+3\sqrt{-5},3-3\sqrt{-5},6).$$ Then apply the rule that you can replace any one of the generators $x_i$ with $x_i-rx_j$ where $r$ can be any element of the ring, and $x_j$ is some other generator (so $j\neq i$). – Jyrki Lahtonen Nov 22 '20 at 05:08
  • (cont'd) Can you write $3$ as a $\Bbb{Z}[\sqrt{-5}]$-linear combinatioin of those generators? And the observe that the others are all multiples of $3$, and can then be disposed of. – Jyrki Lahtonen Nov 22 '20 at 05:10
  • Here are more complicated example calculations of products of ideals. That time of $\Bbb{Z}[\sqrt{223}]$. – Jyrki Lahtonen Nov 22 '20 at 05:12
  • @JyrkiLahtonen Thanks a lot for these details, I understand a lot better now! Indeed, I was trying to find a single product equal to $3$ rather than a linear combination of them. Thank you very much – TheStudent Nov 22 '20 at 10:06
  • You are welcome. Feel free to post all your calculations as an answer. That way you get a final check and/or more feedback. Also, then you can mark this question as handled (by accepting your answer). – Jyrki Lahtonen Nov 22 '20 at 11:18

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