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Question: What is the norm of an ideal $I=(2,1+\sqrt{-5})$ in $\mathbb Z[\sqrt{-5}]$?

Basically, I use the usual norm. $N(a)=a.\overline a=a^2+5db^2$

I know that if ideal is generated by one element then its norm is exactly the same as norm of a generator. For example, if $I_1=(2)$ then Norm$(I_1)=4$ and Norm$(I_2)=6$ where $I_2=(1+\sqrt{-5})$.

What is the norm of non-principle ideal? How can I calculate norm of $I$?

Elise9
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  • Are you asking for the general definition or rather for how to compute an ideal norm in this specific ring? Where did you encounter it without definition? – Bill Dubuque May 15 '24 at 20:47
  • I want to learn that how can I compute norm of non-principle ideal like $I$. – Elise9 May 15 '24 at 20:49
  • Generally we can write the ideal as a module in Hermite normal form - which makes the norm obvious, e.g. see this answer and the posts Linked there, and also here (yours follows the same way). – Bill Dubuque May 15 '24 at 21:02
  • For Galois extensions, the norm to $\mathbb{Z}$ of $I$ is $\mathbb{Z}\cap\prod_{\sigma\in G}\sigma(I)$, where $G$ is the Galois group of $\mathbb{Q}[\sqrt{-5}]$ over $\mathbb{Q}$. This is not hard to do here. – Arturo Magidin May 16 '24 at 03:08

1 Answers1

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You can use Hermite normal form, like Bill Dubuque suggests.

However, you have two alternative ways to proceed.

(1) You can show that $(2)=I\bar{I}$, then use multiplicativity of the norm and the fact that $\bar{I}$ and $I$ have same norm to get $N(I)=2$.

(2) You can simply use the definition, but you have to identify the quotient ring. This is easy here: if $R=\mathbb{Z}[\sqrt{-5}]$, the ring morphism $m\in\mathbb{Z}\mapsto m+I\in R/I$ is surjective (because $a+b\sqrt{-5}\equiv a-b \mod I$), with kernel $2\mathbb{Z}$ (exercise).

Hence $R/I\simeq \mathbb{Z}/2\mathbb{Z}$ and $N(I)=2$ again.

GreginGre
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