Prove that a module is finitely generated.

Question

Let $K \subseteq L$ be number fields. Let $I$ be a fractional ideal of $\mathcal{O}_L$. Define $$ I^{*}=\{x \in L \mid Tr^L_K(xI) \subseteq \mathcal{O}_K\}.$$ I want to show that $I^*$ is also a fractional ideal of $\mathcal{O}_L$, but I can't show that it is finitely generated. Please help me.

Answer 1

As you suspecteded, this is purely a question of commutative algebra. Let $R$ be a Dedekind domain, $K$ its quotient field, $V$ a finite dimensional $K$-vector space. Let $B(u,v)$ be a non-degenerate, symmetric, $K$-bilinear form on $V$. If $(v_i)$ is a basis of $V$, its dual basis $(w_j)$ is defined by $B(v_i,w_j)=\delta_{i,j}$ (Kronecker symbol). For any $R$-submodule $M$ of $V$, its dual module $D(M)$ consists of the $v \in V$ s.t. $B(v,M)\subset R$. The following lemmas are immediate: (1) If $M$ is the free $R$-module on $(v_i)$, then $D(M)$ is the free $R$-module on the dual basis $(w_j)$, and $D(D(M))=M$ ; (2) Given two finitely generated $R$- submodules $M,N$ of $V$, there is a non null $a\in K$ s.t. $aM\subset N$ (given a basis $(v_i)$ of $V$ contained in $N$, take $a$ as a common denominator of the coefficients of the $v_i$ w.r.t. the basis $(v_i)$).

Let us show that $D(M)$ is a finitely generated $R$-module which spans $V$: $M$ contains a free $R$-module $N$, and lemma 2 shows that $M$ is contained in a free $R$-module $M'=bN$, both $M'$ and $N$ spanning $V$. Hence $D(N)\subset D(M)\subset D(M')$ and, by lemma 1, $D(M')$ and $D(N)$ are free and span $V$. So $D(M)$ is $R$-finitely generated and spans $V$. Apply this to a finite extension $L/K$ of number fields with respective rings of integers $R$ and $S$, and the bilinear form $B$ defined by the trace. The dual $D_R(S)$ of $S$ is clearly an $S$-module, and we have seen that it is finitely generated over $R$, hence over $S$. So $D_R(S)$ is a fractional ideal of $S$. Note that, as $S \subset D_R(S)$ , its inverse ${D_R(S)}^{-1}$ is an integral ideal of $S$ (called the different of $L/K$). It remains only to check, using the definition, that $D_R(I)=I^{-1}D_R(S)$ for any fractional ideal $I$ of $L$ . More directly, redo the proof of this subsection, just replacing $D_R(S)$ by $D_R(I)$ .