| Version current |
Version 6 |
|
Let $A$ be an algebra (not necessarily commutative), finitely generated over
|
Let $A$ be an algebra, finitely generated over
|
| $\mathbb{Q}$. An {\it order} $R$ of $A$ is a subring |
$\mathbb{Q}$. An {\it order} $R$ of $A$ is a subring |
| of $A$ which is finitely generated as a |
of $A$ which is finitely generated as a |
| $\mathbb{Z}$-module and which satisfies $R \otimes \mathbb{Q}= A$. |
$\mathbb{Z}$-module and which satisfies $R \otimes \mathbb{Q}= A$. |
|
|
|
Remark: The algebra $A$ is not necessarily commutative.\\ |
|
|
| {\bf Examples:} |
{\bf Examples:} |
| \begin{enumerate} |
\begin{enumerate} |
| \item The ring of integers in a number field is an order, known as |
\item The ring of integers in a number field is an order, known as |
| the {\it maximal order}. |
the {\it maximal order}. |
|
|
|
\item Let $K$ be a quadratic imaginary field and $\mathcal{O}_K$ its
|
\item Let $K$ be a quadratic imaginary field and $O$ its
|
|
ring of integers. For each integer $n\geq 1$ the ring
|
ring of integers. Then for each integer $n$, the ring
|
|
$\mathcal{O}={\mathbb{Z}}+n\mathcal{O}_K$ is an order of $K$ (in fact it can be
|
${\mathbb{Z}}+nO$ is an order of $K$ (in fact it can be
|
|
proved that every order of $K$ is of this form). The number $n$ is called the {\it \PMlinkescapetext{conductor}} of the order $\mathcal{O}$.
|
proved that every order of $K$ is of this form)..
|
| \end{enumerate} |
\end{enumerate} |
|
|
| {\it Reference}: Joseph H. Silverman, {\it The arithmetic of |
{\it Reference}: Joseph H. Silverman, {\it The arithmetic of |
| elliptic curves}, Springer-Verlag, New York, 1986. |
elliptic curves}, Springer-Verlag, New York, 1986. |
