Let $K$ be a number field, $\mathcal{O}_K$ be its ring of integers. Let $\mathfrak{q}$ be a nonzero ideal in $\mathcal{O}_K$. The $\mathfrak{q}$-ideal class group consists of equivalence classes of fractional ideals with equivalence relation: $$ \mathfrak{a} \sim \mathfrak{b}$$ if and only if there are $\alpha, \beta \in \mathcal{O}_K$ such that $\alpha \mathfrak{a} = \beta \mathfrak{b}$, with $\alpha\equiv \beta$ ( mod $\mathfrak{q}$), and $\alpha, \beta$ totally positive.

Let $x\geq 1$, and $[\mathfrak{a}_0]_x$ be the set of integral ideals $\mathfrak{a}$ such that $N\mathfrak{a}\leq x$ and $\mathfrak{a}\sim \mathfrak{a}_0$. Let $h(\mathfrak{q})$ be the order of $\mathfrak{q}$-ideal class group.

I wonder if there is a good upper bound of size of $[\mathfrak{a}_0]_x$:

Suppose that $N\mathfrak{q}>\frac{x}{2}$.

In particular, Is it true that: $$ |[\mathfrak{a}_0]_x|\leq B$$ for some constant $B$ depending only on $K$?

My thoughts on this problem is that

1. This is true when $K=\mathbb{Q}$.

2. Using inverse class of $\mathfrak{a}_0$, we get a constant bound but it depends on $\mathfrak{a}_0$.

3. In Hinz and Lodemann's paper 'On Siegel Zeros of Hecke Landau Zeta Functions', it was mentioned that $$ |[\mathfrak{a}_0]_x|\ll \frac{x+N\mathfrak{q}}{h(\mathfrak{q})}$$ But, this is not quite enough for boundedness.

Let $K$ be a number field, $\mathcal{O}_K$ be its ring of integers. Let $\mathfrak{q}$ be a nonzero ideal in $\mathcal{O}_K$. The $\mathfrak{q}$-ideal class group consists of equivalence classes of fractional ideals with equivalence relation: $$ \mathfrak{a} \sim \mathfrak{b}$$ if and only if there are $\alpha, \beta \in \mathcal{O}_K$ such that $\alpha \mathfrak{a} = \beta \mathfrak{b}$, with $\alpha\equiv \beta$ ( mod $\mathfrak{q}$), and $\alpha, \beta$ totally positive.

Let $x\geq 1$, and $[\mathfrak{a}_0]_x$ be the set of integral ideals $\mathfrak{a}$ such that $N\mathfrak{a}\leq x$ and $\mathfrak{a}\sim \mathfrak{a}_0$. Let $h(\mathfrak{q})$ be the order of $\mathfrak{q}$-ideal class group.

I wonder if there is a good upper bound of size of $[\mathfrak{a}_0]_x$:

Suppose that $N\mathfrak{q}>\frac{x}{2}$.

In particular, Is it true that: $$ |[\mathfrak{a}_0]_x|\leq B$$ for some constant $B$ depending only on $K$?

My thoughts on this problem is that

1. This is true when $K=\mathbb{Q}$.

2. Using inverse class of $\mathfrak{a}_0$, we get a constant bound but it depends on $\mathfrak{a}_0$.

-- This is even not clear to me right now. I found a gap in my argument.

3. In Hinz and Lodemann's paper 'On Siegel Zeros of Hecke Landau Zeta Functions', it was mentioned that $$ |[\mathfrak{a}_0]_x|\ll \frac{x+N\mathfrak{q}}{h(\mathfrak{q})}$$
   But, this is not quite enough for boundedness.

Edit: Hinz&Lodemann's paper cites Tatuzawa's 'On the number of integral ideals whose norms belonging to some norm residue class mod q', Sci. Pap. Coll. Gen. Educ., Univ. Tokyo 27, 1-8 (1977)

I want to find that paper but I could not find it online. There are too many abbreviations too. Does anyone have the link to that paper or does anyone know what the journal actually is?

Let $K$ be a number field, $\mathcal{O}_K$ be its ring of integers. Let $\mathfrak{q}$ be a nonzero ideal in $\mathcal{O}_K$. The $\mathfrak{q}$-ideal class group consists of fractional ideals with equivalence relation: $$ \mathfrak{a} \sim \mathfrak{b}$$ if and only if there are $\alpha, \beta \in \mathcal{O}_K$ such that $\alpha \mathfrak{a} = \beta \mathfrak{b}$, with $\alpha$, and $\beta$ totally positive.

Let $x\geq 1$, and $[\mathfrak{a}_0]_x$ be the set of integral ideals $\mathfrak{a}$ such that $N\mathfrak{a}\leq x$ and $\mathfrak{a}\sim \mathfrak{a}_0$. Let $h(\mathfrak{q})$ be the order of $\mathfrak{q}$-ideal class group.

I wonder if there is a good upper bound of size of $[\mathfrak{a}_0]_x$:

Suppose that $N\mathfrak{q}>\frac{x}{2}$.

In particular, Is it true that: $$ |[\mathfrak{a}_0]_x|\leq B$$ for some constant $B$ depending only on $K$?

My thoughts on this problem is that

1. This is true when $K=\mathbb{Q}$.

2. Using inverse class of $\mathfrak{a}_0$, we get a constant bound but it depends on $\mathfrak{a}_0$.

3. In Hinz and Lodemann's paper 'On Siegel Zeros of Hecke Landau Zeta Functions', it was mentioned that $$ |[\mathfrak{a}_0]_x|\ll \frac{x+N\mathfrak{q}}{h(\mathfrak{q})}$$ But, this is not quite enough for boundedness.

Let $K$ be a number field, $\mathcal{O}_K$ be its ring of integers. Let $\mathfrak{q}$ be a nonzero ideal in $\mathcal{O}_K$. The $\mathfrak{q}$-ideal class group consists of equivalence classes of fractional ideals with equivalence relation: $$ \mathfrak{a} \sim \mathfrak{b}$$ if and only if there are $\alpha, \beta \in \mathcal{O}_K$ such that $\alpha \mathfrak{a} = \beta \mathfrak{b}$, with $\alpha\equiv \beta$ ( mod $\mathfrak{q}$), and $\alpha, \beta$ totally positive.

Let $x\geq 1$, and $[\mathfrak{a}_0]_x$ be the set of integral ideals $\mathfrak{a}$ such that $N\mathfrak{a}\leq x$ and $\mathfrak{a}\sim \mathfrak{a}_0$. Let $h(\mathfrak{q})$ be the order of $\mathfrak{q}$-ideal class group.

I wonder if there is a good upper bound of size of $[\mathfrak{a}_0]_x$:

Suppose that $N\mathfrak{q}>\frac{x}{2}$.

In particular, Is it true that: $$ |[\mathfrak{a}_0]_x|\leq B$$ for some constant $B$ depending only on $K$?

My thoughts on this problem is that

1. This is true when $K=\mathbb{Q}$.

2. Using inverse class of $\mathfrak{a}_0$, we get a constant bound but it depends on $\mathfrak{a}_0$.

3. In Hinz and Lodemann's paper 'On Siegel Zeros of Hecke Landau Zeta Functions', it was mentioned that $$ |[\mathfrak{a}_0]_x|\ll \frac{x+N\mathfrak{q}}{h(\mathfrak{q})}$$ But, this is not quite enough for boundedness.