There exist an integral ideal prime to a given nonzero integral ideal

Question

Let $\mathfrak{m}$ be a nonzero integral ideal of the dedekind domain $\mathfrak{O}$. Show that in every ideal class of $Cl_K$, there exist an integral ideal prime to $\mathfrak{m}$.

My effort : Actually this is a problem given in Algebraic Number Theory by Neukrich(1.3.8). Let $\mathfrak{a}P_K\in Cl_K=J_K/P_K$. where $\mathfrak{a}$ is an fractional ideal of $K$. Hence there exist $c\in \mathfrak{O}$ such that $c\mathfrak{a}\subset \mathfrak{O}$. But after this point how to find $\mathfrak{b} \subset \mathfrak{O}$ such that $\mathfrak{m}+\mathfrak{b}=\mathfrak{O}$. Any help/hint in this regards would be highly appreciated. Thanks in advance!

Answer 1

Let's introduce a lemma first:

For any finite set of prime ideals $\mathfrak{p}_1,...,\mathfrak{p}_m$ and any non-negative integers $a_1,...,a_m$ (in particular they can be zero), there exists an element $x\in \mathfrak{O}$ s.t. $v_{\mathfrak{p}_i}(x)=a_i$.

Proof: For a general prime ideal $\mathfrak{p}$, pick $t\in \mathfrak{p}\backslash \mathfrak{p}^2$. For $n\geq 1$ it's easy to see that the ideals in $\mathfrak{O}/\mathfrak{p}^n$ are exactly $\{\mathfrak{p}^0/\mathfrak{p}^n,\cdots,\mathfrak{p}^n/\mathfrak{p}^n\}$, which is represented by $\{t^0+\mathfrak{p}^n,\cdots,t^n+\mathfrak{p}^n\}$ respectively. For $0\leq m\leq n$, we have $v_\mathfrak{p}(x)=m\Longleftrightarrow x\equiv t^m \mod \mathfrak{p}^n$.

Fix $t_i\in \mathfrak{p}_i\backslash \mathfrak{p}_i^2$ for all $i$, it suffices to find an element $x$ s.t. $\forall i,x\equiv t_i^{a_i} \mod \mathfrak{p}_i^{a_i+1}$. And it always exists by Chinese remainder theorem (consider the factorization of $\mathfrak{O}/\mathfrak{p}_1^{a_1+1}\cdots \mathfrak{p}_m^{a_m+1}$).$\square$

Say $\mathfrak{m}=\mathfrak{p}_1^{a_1}\cdots \mathfrak{p}_m^{a_m}$. And pick an equivalence class $\mathfrak{a}$ in $Cl_K$, where $\mathfrak{a}$ is a fractional ideal. Pick any $c\in \mathfrak{a}^{-1}$, then $c\mathfrak{a}\subset \mathfrak{O}$. So we may assume $\mathfrak{a}$ is integral and $\mathfrak{a}=\mathfrak{p}_1^{b_1}\cdots \mathfrak{p}_m^{b_m}\mathfrak{q}_1^{c_1}\cdots \mathfrak{q}_n^{c_n}$ where $b_i\geq0$ and $c_i>0$.

Now we would like to neutralize all those $b_i$ by multiplying with suitable element of $K$. Use our lemma on $\{\mathfrak{p}_1,...,\mathfrak{p}_m,\mathfrak{q}_1,...,\mathfrak{q}_n\}$ and $\{b_1,...,b_m,c_1,...,c_n\}$ to get a such $x$, now the valuation of $x^{-1}\mathfrak{a}$ is zero on all $\mathfrak{p}_i$ and non-positive on everything else. Say $x^{-1}\mathfrak{a}=\mathfrak{l}_1^{-d_1}\cdots \mathfrak{l}_w^{-d_w}$.

Now we apply the lemma again on $\{\mathfrak{p}_1,...,\mathfrak{p}_m,\mathfrak{l}_1,...,\mathfrak{l}_w\}$ and $\{0,...,0,d_1,...,d_w\}$ to get an element $y$, then $yx^{-1}\mathfrak{a}$ is integral (by checking the valuation is non-negative on every prime ideal) and it has no common factor of prime ideal with $\mathfrak{m}$, so coprime with $\mathfrak{m}$.