Let $E/F$ be an extension of algebraic number fields and $\mathcal{O}_E$ and $\mathcal{O}_F$ be the ring of integers. Define $$A^*=\mathcal{O}_EA,$$ where A is an ideal of $\mathcal{O}_F$.
Prove that $A^*$ is and ideal of of $\mathcal{O}_E$ and $(AB)^*=A^*B^*$.
I tried to show the following:
- $A^*$ is non-empty
- $A^*$ is closed under addition
- $A^*$ "absorbs" the multiplication with the elements from the ring $\mathcal{O}_E$
Sure $A^*$ is non-empty because $A$ and $\mathcal{O}_E$ are. And the third property is trivial. But I'm struggling to show that it's closed under addition. I took two elements $x,y \in A^*$ which are of the form $e_ia_i$ $(i=1,2 )$ where $e_i \in \mathcal{O}_E$ and $a_i \in A$. Now $x+y=e_1a_1+e_2a_2$. How to show this belongs to $A^*$?
To show that $(AB)^*=A^*B^*$ I simply use the fact that the ideals of ring of integers are finitely generated and then just multiply. Let $A=[a_1,\, \dots ,\,a_n]$ and $B=[b_1,\, \dots ,\, b_k]$. Then $$ A^*B^*= (\mathcal{O}_EA)(\mathcal{O}_EB)= \mathcal{O}_E[a_1,\, \dots ,\,a_n]\mathcal{O}_E[b_1,\, \dots ,\, b_k]= \mathcal{O}_E(\mathcal{O}_F a_1 + \dots + \mathcal{O}a_n)\mathcal{O}_E(\mathcal{O}_F b_1 + \dots + \mathcal{O}_Fb_k) \quad = (\mathcal{O}_E a_1 + \dots + \mathcal{O}_Ea_n)(\mathcal{O}_E b_1 + \dots + \mathcal{O}_Eb_k) \quad = \mathcal{O}_E[a_1b_1 ,\, a_1b_2,\, \dots,\, a_1b_k,\, a_2b_1,\, \dots,\, a_nb_k]= (AB)^*.$$
Is the work I've managed to do correct and how to complete the first proof? I also wonder if there is name for this operation "$*$".