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order

Idea: orders are like lattices!

Order for general rings

An order ${\mathcal{O}}$ is a Noetherian integral domain of dimension one with nonzero conductor. Equivalently, \begin{align*} \operatorname{cl}^{\mathrm{int}} {\mathcal{O}}\in {}_{{\mathcal{O}}}{\mathsf{Mod}} ^{\mathrm{fg}}.\end{align*} attachments/Pasted%20image%2020220224124457.png

Order for k-algebras

Setup: take $R$ a Noetherian integral domain with $K\coloneqq\operatorname{ff}(R)$, and let $\Lambda\in \mathsf{Alg} _{/ {K}} ^{\mathrm{fd}}$. Then an $R{\hbox{-}}$order in $\Lambda$ is a ring $X$ with the structure of an $R{\hbox{-}}$lattice, so $X\leq \Lambda \in {}_{R}{\mathsf{Mod}} ^{\mathrm{fg}}$ such that $\mathop{\mathrm{span}}_K X =\Lambda$, i.e. $X$ is an $R{\hbox{-}}$submodule of $\Lambda$ which spans it as a $K{\hbox{-}}$vector space.

Equivalently, if $K\in \mathsf{Alg} _{/ {{\mathbf{Q}}}} ^{\mathrm{fd}}$ with $\dim_{\mathbf{Q}}K = n$, an order in $K$ is a subring ${\mathcal{O}}\leq K$ that is a free ${\mathbf{Z}}{\hbox{-}}$module of rank $n$, so $K = {\mathcal{O}}\otimes_{\mathbf{Z}}{\mathbf{Q}}$.

Must be a lattice and a unital ring, so must contain 1.

Examples

Any Dedekind domain which is not a field is an order.
$X\coloneqq{ \operatorname{End} }(E)$ for $E\in \mathrm{Ell} _{/ {{\mathbf{Q}}}} $ an elliptic curve is a ${\mathbf{Z}}{\hbox{-}}$order in some $\Lambda \in \mathsf{Alg} _{/ {{\mathbf{Q}}}} $ with $\dim_{\mathbf{Q}}X \in \left\{{1,2,4}\right\}$.
- In dimensions 1 and 2, $\Lambda = {\mathbf{Q}}$ or ${\mathbf{Q}}(\sqrt d)$ is an imaginary quadratic field. In dimension 4, $L$ is a quaternion algebra.
${\mathbf{Z}}$ is the only order in ${\mathbf{Q}}$.
$2{\mathbf{Z}}\leq {\mathbf{Q}}$ is not an order: it is a lattice but not a ring.
$X\coloneqq\left\{{{a\over 2^n} {~\mathrel{\Big\vert}~}a,n \in {\mathbf{Z}}}\right\} \leq {\mathbf{Q}}$ is not an order: it is a ring but not a lattice since it is infinitely generated as a ${\mathbf{Z}}{\hbox{-}}$module.
For number fields $K\in \mathsf{Field}_{/ {{\mathbf{Q}}}} $:
- ${\mathcal{O}}_K \subseteq K$ is the unique maximal order. Note that generally, maximal orders for arbitrary $K\in \mathsf{Alg} _{/ {{\mathbf{Q}}}} $ need not be unique.
- Every other order ${\mathcal{O}}$ must be a subset of ${\mathcal{O}}_K$ and $[{\mathcal{O}}_K: {\mathcal{O}}]$ must be finite since ${\mathcal{O}}$ contains a ${\mathbf{Q}}{\hbox{-}}$basis of $K$
- For $K$ imaginary quadratic, the orders ${\mathcal{O}}$ are all of the form ${\mathbf{Z}}+ n{\mathcal{O}}_K$ for $n\in {\mathbf{Z}}_{\geq 0}$. The integer $n$ here is the conductor of the order, and equals the index $[{\mathcal{O}}_K: {\mathcal{O}}] < \infty$.
- For $K = {\mathbf{Q}}(a_1,\cdots, a_n)$, the ring ${\mathcal{O}}\coloneqq{\mathbf{Z}}[a_1,\cdots, a_n]$ is an order.
The integral closure of an order is always Dedekind.
Not every ring whose integral closure is Dedekind is an order: Nagata constructs Noetherian domains of dimension one whose conductor is zero.
- In common ANT situation (see AKLB setup), $B_{/ {A}} $ is finitely generated so every intermediate ring is finitely generated, including those whose integral closure is $B$ – so if $A[\alpha], B$ have the same fraction field, so $L = K(\alpha)$, then $A[\alpha]$ is an order in $B$.

Orders in k-algebras

attachments/Pasted%20image%2020220123190527.png

attachments/Pasted%20image%2020220126093016.png

Orders for elliptic curves

The ring of endomorphisms of an elliptic curve can be of one of three forms: - ${\mathbf{Z}}$ - ${\mathfrak{o}}\subseteq K$ an order in an imaginary quadratic field - ${\mathfrak{o}}\subseteq A \in \mathsf{Alg} _{/ {{\mathbf{Q}}}} $ a quaternion algebra

Conductors

attachments/Pasted%20image%2020220730134641.png Relation to discriminants:

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number fields in Unsorted/order

quadratic field in Unsorted/order
ring of integers

For a number field $K_{/ {{\mathbf{Q}}}} $, define the ring of integers as \begin{align*}{\mathcal{O}}_K \coloneqq \operatorname{cl}^{\mathrm{int}} _K({\mathbf{Z}}).\end{align*} More generally, for $R$ a ring, \begin{align*} {\mathcal{O}}_R \coloneqq \operatorname{cl}^{\mathrm{int}} _{\operatorname{ff}(R)}(R) \end{align*} or equivalently a maximal order.
class group

For ${\mathcal{O}}\in{ \mathrm{Ord} }(K)$, let ${ \ddot{\operatorname{Id}} }^{\times}({\mathcal{O}})$ be the invertible fractional ideals and $\mathop{\mathrm{Prin}}{ \ddot{\operatorname{Id}} }^{\times}({\mathcal{O}}) \subseteq { \ddot{\operatorname{Id}} }^{\times}({\mathcal{O}})$ be the principal fractional ideals of ${\mathcal{O}}$. The quotient $\operatorname{Cl} ({\mathcal{O}}) \coloneqq{ \ddot{\operatorname{Id}} }^{\times}({\mathcal{O}}) / \mathop{\mathrm{Prin}}{ \ddot{\operatorname{Id}} }^{\times}({\mathcal{O}})$ is the ideal class group of the order ${\mathcal{O}}$.