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Let $L$ be an extension field of $\mathbb Q$ with $[L:\mathbb Q]=n$. Let $A$ be the ring of integers of $L$ i.e. the Integral closure of $\mathbb Z$ in $L$. Then $A$ is a Dedekind domain. If $I$ is a non-zero ideal of $A$, then I can show that $I$ is a free $\mathbb Z$-module of rank $n$ and $aA \subseteq I$ for some $0 \ne a \mathbb Z$ . Moreover $A/I$ is finite.

My question is : If $I=(x)$ is a non-zero principal ideal, then how to show that $|A/I|=| N_{L|\mathbb Q } (x)|$ ? where $N_{L|\mathbb Q}$ denotes the field Norm function .

user
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1 Answers1

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Choose an integral basis for $O_L$ (which will denote the integral closure of $\mathbb{Z}$ in the extension $L$), denoted $\{ \alpha_1 , \dots , \alpha_n \}$. Then, certainly $\{ x \alpha_1 , \dots , x \alpha_n \}$ is an integral basis for the principal ideal $(x)$. Then, by definition of discriminant $$\textrm{disc} ( (x)) = \det ( \sigma_i ( x \alpha_j )_{i,j} )^2$$ Where $\sigma_i$ denote embeddings of our $\alpha_j$ into some algebraic closure of $L$. Note that by properties of determinants, $$\det ( \sigma_i ( x \alpha_j )_{i,j} ) = \prod_{i=1}^n \sigma_i (x) \cdot \det ( \sigma_i (\alpha_j)_{i,j} )$$ But of course the product on the right is precisely the norm of $x$. We then see $$N_{L/\mathbb{Q}} (x)^2 = \frac{\textrm{disc} (O_L) }{\textrm{disc}( (x))}$$ But the quotient on the right is already known to be precisely $|O_L /(x) |^2$, whence we conclude that $N_{L/\mathbb{Q}} (x) = |O_L / (x)|$, as desired.

Rellek
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  • Don't you need to multiply $\alpha_i$ s by a common factor to first make it a basis for $A$ over $\mathbb Z$ ? otherwise $x\alpha_i$ s may not be in $I$ ... – user Feb 17 '18 at 20:33
  • I'm assuming that you are already allowed the fact that integral bases exist over PID's. If you have to prove that fact as well, this would be a long problem since you would then need to employ standard facts about modules of PID's, etc, etc. – Rellek Feb 17 '18 at 20:36
  • Yes I'm allowed to use that ... – user Feb 17 '18 at 20:41
  • I am confused as to how you got the norm expression ... and what is $O_L/(x)$ ... what is $O_L$ ? Is your $O_L$ my $A$ , the ring of integers of $L$ ? – user Feb 17 '18 at 20:42
  • Yes $O_L$ is the standard notation when working over $\mathbb{Z}$. $O_L / (x)$ is precisely $A/I$. – Rellek Feb 17 '18 at 20:45
  • And the definition of norm is the product over all embeddings into the algebraic closure of $L$, correct? Or did you use a different definition. – Rellek Feb 17 '18 at 20:46
  • shouldn't there be a power of $[L:\mathbb Q(x)]$ in the product for the norm expression ? https://en.wikipedia.org/wiki/Field_norm – user Feb 17 '18 at 20:50