Hecke's notation for fractional ideals

Question

I'm confused about some of the notation appearing in Hecke's Lectures on the Theory of Algebraic Numbers, in Section 54 of Chapter 7.

Let $\mathfrak{d}$ be the different of $K$ (the inverse ideal to the dual lattice of $\mathcal{O}_{K}$); then for each $\omega \in K$ we may write $\mathfrak{d}\omega=\frac{\mathfrak{b}}{\mathfrak{a}}$ where $\mathfrak{a}$ and $\mathfrak{b}$ are integral ideals and $(\mathfrak{a},\mathfrak{b})=1$. Suppose $\mathfrak{a}=\mathfrak{a}_{1}\mathfrak{a}_{2}$, where $(\mathfrak{a}_1,\mathfrak{a}_{2})=1$. Then Hecke says that we may find auxiliary ideals $\mathfrak{c}_1$ and $\mathfrak{c_2}$ such that $\mathfrak{a}_{1}\mathfrak{c}_{1}=\alpha_{1}$ and $\mathfrak{a}_{2}\mathfrak{c}_{2}=\alpha_{2}$ are integers and $(\mathfrak{a},\mathfrak{c}_1 \mathfrak{c}_2)=1$.

$\textbf{Question 1}$: The text states that $\alpha_1$ and $\alpha_2$ are elements of $\mathcal{O}_K$, so the equations involving $\alpha_1$ and $\alpha_2$ are shorthand for equalities of integral ideals: $\mathfrak{a}_{1}\mathfrak{c}_{1}=(\alpha_{1})$. How do we know that such a $\mathfrak{c}_1$ exists?

$\textbf{Question 2}$: Shortly afterwards, Hecke says that $\beta:=\frac{\mathfrak{b}\mathfrak{c}_1 \mathfrak{c}_2}{\mathfrak{d}}$ is an element of $k$. Why is this true?

Thanks.

Answer 1

For your question 1

• Given a non-zero ideal $I$, for non-zero $b\in I$ then $(b)=IJ$, and there is such a $b\in I$ such that $(I,J)=(1)$

  Proof : one of the properties of Dedekind domains is that $I$ becomes principal $=(b\bmod I^2)$ in $R/I^2$, then $(b,I^2)=I$ and $(b)=IJ$ means $I=(IJ,I^2)=I(I,J)$ thus $(I,J)=(1)$.

• With $b_1\in \mathfrak{a}_1 (N(\mathfrak{a}))$

  then $\mathfrak{c}_1=(b_1) (\mathfrak{a}_1 (N(\mathfrak{a})))^{-1}$ is coprime with $\mathfrak{a}_1 (N(\mathfrak{a}))$ thus with $\mathfrak{a}$,

• With $b_2\in \mathfrak{a}_2 (N(\mathfrak{a}))$

  then $\mathfrak{c}_2=(b_2) (\mathfrak{a}_2 (N(\mathfrak{a})))^{-1}$ is coprime with $\mathfrak{a}_2 (N(\mathfrak{a}))$ thus with $\mathfrak{a}$,

and hence $$( \mathfrak{a},\mathfrak{c}_1\mathfrak{c}_2)=(1),\qquad \mathfrak{a}_j\mathfrak{c}_j= (b_j/N(\mathfrak{a}))$$

For your question 2 it is because $\mathfrak{b}\mathfrak{c}_1\mathfrak{c}_2$ has the same ideal class as $\mathfrak{b}\mathfrak{a}^{-1}$

Answer 2

Let $\newcommand\aa{\mathfrak{a}}\newcommand\bb{\mathfrak{b}}\newcommand\dd{\mathfrak{d}}\newcommand\cc{\mathfrak{c}} \newcommand\pp{\mathfrak{p}}\aa=\aa_1\aa_2$, with $(\aa_1,\aa_2)=1$, all integral.

Actually, I realized that I think the easiest construction of $\cc_1$, $\cc_2$ is by Chebotarev's density theorem (no idea if this is at all reasonable to cite at this point in the book). The argument here is that if $H_K$ is the Hilbert class field for $K$, and we identify $\newcommand\Gal{\operatorname{Gal}}\Gal(H_K/K)\simeq C_K$, with $C_K$ the ideal class group of $K$ by the Artin map (which sends a prime ideal in the group of fractional ideals to its Frobenius element). Then by Chebotarev's density theorem, there are infinitely many prime ideals $\newcommand\qq{\mathfrak{q}}\qq$ in the ideal class of the inverse of $\aa_1$. Choose a prime $\qq_1$ that doesn't divide $\aa$. Do the same for $\aa_2$. Then by definition of the ideal class group, $\qq_i\aa_i$ is principal (and integral, since both $\qq_i$ and $\aa_i$ are), and by choice of $\qq_i$, $(\aa,\qq_1\qq_2)=1$.

Side note

Constructing $\cc_1$ and $\cc_2$ with the strategy I was advocating in the comments (take the inverse of $\aa_i$ then multiply by a large enough $\alpha_i$ to kill denominators) is hard to do and ensure that $(\aa,\cc_1\cc_2)=1$. However, I think Chebotarev's density theorem is definitely overkill here. Unfortunately it's gotten late, so I can't come up with a better argument right now.

Answer to question 2

Then regardless of how you construct $\cc_1,\cc_2$, 2 follows from their definition. We have $$\frac{\bb\cc_1\cc_2}{\dd} = \frac{\bb\cc_1\cc_2 \omega}{\bb/\aa} = \aa\cc_1\cc_2\omega = \alpha_1\alpha_2 \omega \in K$$