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Let $K=\mathbb{Q}(\alpha)$ be a number field of degree $n$ and R an order in it.

Let $I$ be a $R$-fractional ideal, that is, a $R$-submodule of $K$ that span the whole $K$ when tensored with $\mathbb{Q}$. Equivalently, $I$ is a fractional $R$-ideal if and only if $I$ is an $R$-module that has rank equal to n as an abelian group. This means that we can write $I = x_1\mathbb{Z}\oplus...\oplus x_n\mathbb{Z}$, for some $x_i \in K$.

Let $\gamma \in K$ and let $m_\gamma \in \mathcal{M}_n(\mathbb{Q})$ be the matrix that represents the multiplication by $\gamma$ in $K$ as a $\mathbb{Q}$-linear map and with respect to the basis $\{ x_i \}$.

  1. Do we have the following equivalences?

(i) $m_\gamma \in \mathcal{M}_n(\mathbb{Z})$

(ii) $\gamma I \subset I$

(iii) $I$ is a $R[\gamma]$-fractional ideal

  1. For any $x,y\in K$, do we have $m_{xy}=m_x m_y$?
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    Is this homework? Can you tell us what work you have done so far in trying to solve these problems? – 2017-01-05
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    No, it's not homework. In fact, I think all assertions are true, but I want to be sure. This is a question related to abelian varieties defined over finite fields, and closer to this question: http://math.stackexchange.com/questions/2077461/ideal-classes-and-matrix-conjugation-in-a-cm-field ... If the assertions are true, I(Z[a,f(0)/a])=I(Z[a]), when I(R) is the monoid of fractionals ideals of R and f(t) is the minimal polynomial of a, and considering K=Q(a) – 2017-01-05

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