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Problem:

Let $K$ be a quadratic field, $\mathcal{O}_K$ its ring of integers. Now I want to prove the equality

$$D(\mathfrak{a}) = \mathcal{N}(\mathfrak{a})^2 \cdot \Delta_K $$

where $\mathfrak{a} \subseteq K$ is a fractional ideal, $D(\mathfrak{a})$ is the Idealdiscriminant, $\mathcal{N}(\mathfrak{a})$ is the Idealnorm and $\Delta_K$ is the discriminant of $K$.

Attempt:

Let $(1, \omega)$ be an integral basis of $\mathcal{O}_K$, $n \in \mathbb{N}$ such that $n \mathfrak{a} \subseteq \mathcal{O}_K$, $(x, y)$ an integral basis of $n\mathfrak{a}$. (Disclaimer: I'm not required to show that such a basis, $n$ exist).

$$\implies \hspace{15pt}(\frac{x}{n}, \frac{x}{n}) \text{ is an integral basis of } \mathfrak{a}$$ There are also $a_1, a_2, b_1, b_2 \in \mathbb{Z}$ satisfying: $$x = a_1 + a_2 \omega \hspace{30pt} y = b_1 + b_2\omega$$ Using the definition of $D(\mathfrak{a})$:

$$ D(\mathfrak{a}) = \text{det}\left[ {\begin{array}{cc} \frac{x}{n} & \frac{y}{n} \\ \frac{\bar x}{n} & \frac{\bar y}{n} \\ \end{array} } \right]^2 = \frac{1}{n^4} \cdot \text{det}\left[ {\begin{array}{cc} x & y \\ \bar x & \bar y \\ \end{array} } \right]^2 = \frac{(x\bar y - \bar x y)^2}{n^4} \\ = \frac{((a_1 + a_2\omega)(b_1 + b_2\bar\omega) - (a_1 + a_2\bar\omega)(b_1 + b_2\omega))^2}{n^4} \\ = \frac{(a_1b_2(\bar\omega - \omega) + a_2b_1(\omega - \bar\omega))^2}{n^4} \\ = \frac{(\bar\omega - \omega)^2(a_1b_2 - a_2b_1)^2}{n^4} \\ = \frac{\Delta_K(a_1b_2 - a_2b_1)^2}{n^4} $$

At this point it would be really nice if $\mathcal{N}(\mathfrak{a}) = \frac{a_1b_2 - a_2b_1}{n^2}$. This is the case iff $\mathcal{N}(n\mathfrak{a}) = a_1b_2 - a_2b_1$. Sadly I have no clue how to prove this last equality. I would be happy as well if someone provided a simpler proof to the initial equality.

1 Answers1

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It is true that $\mathcal N(n\mathfrak a) = a_1 b_2 - a_2 b_1$.

This follows from the following fact about abelian groups: if $G = \mathbb Z \oplus \mathbb Z$, and if $H$ is a subgroup of $G$ generated by the linearly-independent elements $(a_1, a_2), (b_1, b_2) \in G$, then the order of the quotient group $G/H$ is $$ |G/H| = \left| \det \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix}\right|.$$ (See this post, which derives this result by considering the Smith normal form of the matrix describing the embedding of $H$ into $G$ with respect to your chosen bases.)

Now apply this with $G = \mathcal O_K$ and $H = n\mathfrak a$ (both viewed as abelian groups)…

By the way, you probably don't need to be so explicit in your matrix manipulations. The key idea that makes your proof work is that $$ \begin{bmatrix} x & y \\ \bar x & \bar y \end{bmatrix} = \begin{bmatrix} 1 & \omega \\ 1 & \bar\omega \end{bmatrix}\begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix} . $$ Now observe that

$$ D(n\mathfrak a) = \det\begin{bmatrix} x & y \\ \bar x & \bar y \end{bmatrix}^2, \ \ \ \Delta_K = \det \begin{bmatrix} 1 & \omega \\ 1 & \bar\omega \end{bmatrix}^2, \ \ \ \mathcal N(n\mathfrak a)^2 = \det\begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix}^2 .$$

Kenny Wong
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  • Thank you! The post you linked provides a beautiful general viewpoint. I'm working through Zetafunktionen und quadratische Körper by Zagier (1981). He writes that the equality in question can be proved using elementary linear algebra only. Do you think he intended to use this general viewpoint or can you come up with a simpler proof of $\mathcal{N}(n\mathfrak{a}) = a_1b_2 - a_2b_1$? I'm just curious but really satisfied with your answer! –  Jul 05 '18 at 23:13
  • @Herickson I think he is probably referring to the argument with Smith norm form. Essentially, the linear algebra fact is that, regardless of what the embedding matrix for $H \to G$ looks like, you can always find new bases for $H$ and $G$ such that the embedding matrix becomes diagonal. Moreover, the original matrix and the new matrix have the same determinant. Finally, in the diagonal matrix case, it's obvious that $|G/H|$ is the determinant. – Kenny Wong Jul 05 '18 at 23:18
  • @Herickson Here's another explanation... Think of $\mathbb Z \oplus \mathbb Z$ as a lattice embedded in $\mathbb R \oplus \mathbb R$. The unit cell has area one. Now consider the sublattice generated by the vectors $(a_1, a_2)$ and $(b_1, b_2)$. The area of the unit cell of this subattice is the norm of vector cross product of $(a_1, a_2)$ and $(b_1, b_2)$, which is $a_1 b_2 - b_1 a_2$.. Since the unit cell of the sublattice is larger than the unit cell of the original lattice by a factor of $a_1 b_2 - b_1 a_2$, it is obvious that the quotient group has order $a_1 b_2 - b_1 a_2$. – Kenny Wong Jul 06 '18 at 06:44
  • Thanks again. For me, this is a more intuitive way to think about the order of the quotient group. –  Jul 06 '18 at 11:09