# Thursday, September 15

## $r\dash$generation and orders

:::{.proposition title="?"}
Let $I \normal R$ be a proper ideal for $R\in \Dedekind$.
Then

- $R/I$ is a principal ring.
- $R/I$ is Artinian.

:::

:::{.proof title="?"}
Write $I = \prod p_i^{a_i}$, then $R/I$ is a quotient of the semilocalization $R_{p_1,\cdots, p_r}$ which is a PID.
A quotient of a principal ring is principal, and using the ideal correspondence, ideals of $R/I$ biject with ideals $J \subseteq I$ in $R$, which uniquely factors as $J = \prod p_i^{b_i}$ with $0\leq b_i \leq a_i$. 
Since there are finitely many ideals of $R/I$, namely $\prod (a_i + 1)$, it must be Artinian.
:::

:::{.definition title="r-generation"}
A ring $R$ has the $r\dash$generation property if it can be generated by at most $r$ elements.
Note that 1-generation is equivalent to being principal.
Say that $R$ has the $(r+\eps)\dash$generation property iff for every nonzero $I\normal R$ and any $x\in I\smz$ there exist $y_i$ such that $I = \gens{x, y_1,\cdots, y_r}_R$.
I.e. for any nonzero element of $I$, there are $r$ more elements that together generate the ideal.
:::

:::{.remark}
We'll show Dedekind domains have the 2-generation property, and in fact the $1+\eps$ generation property.
:::

:::{.proof title="?"}
$2\implies 1$:
Let $I\normal R$ for $R \in \Dedekind$ and $x\in I\smz$, then there is a SES
\[
0\to \gens{x} \to I \to I/\gens{x} \to 0
.\]
Since $R/\gens{x}$ is principal, there is a generator $\bar y$ of $I/\gens{x}$.
Lift this to $y$ in $I$, then $I = \gens{x, y}_R$.

$1\implies 2$:
See [@CA Theorem 20.12]; roughly $1+\eps$ generation implies Noetherian, so it STS $R_p$ is a DVR for any nonzero $p\in \spec R$.
Apply Nakayama's lemma.
:::

:::{.exercise title="?"}
Prove $R$ is a Dedekind domains iff it has the $1+\eps$ generation property? Fair question for an oral exam.
:::

:::{.remark}
So Dedekind domain are close to principal: every ideal can be generated by 2 elements, and the first can be whatever you want.
:::

:::{.definition title="Orders and the NT1 Square"}
Let $A \in \Dedekind$ with $K = \ff(A)$ and $L/K$ a finite separable field extensions.
Let $B = \intcl_L(A)$, then an $A\dash$order in $L$ is a ring $\OO$ with $A \subseteq \OO \subseteq B$ and $\ff(\OO) = L$.

\begin{tikzcd}
	B && L \\
	\\
	A && K
	\arrow["\subseteq", hook, from=1-1, to=1-3]
	\arrow["\subseteq", hook, from=3-1, to=3-3]
	\arrow["{[L:K] =n}"', from=3-3, to=1-3]
	\arrow[from=3-1, to=1-1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJCIl0sWzIsMCwiTCJdLFswLDIsIkEiXSxbMiwyLCJLIl0sWzAsMSwiXFxzdWJzZXRlcSIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzIsMywiXFxzdWJzZXRlcSIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzMsMSwiW0w6S10gPW4iLDJdLFsyLDBdXQ==)

:::

:::{.remark}
Suppose $A$ is a PID, then $\OO \cong A^n\in \mods{A}$, so $\rank_A(\OO)\leq n$.
Here the rank is the minimal $r$ such that $\OO$ has the $r\dash$generation property.
This is an important invariant of $\OO$: for fixed $A, L$, all values $a \leq \rank \OO \leq n$ can arise.
There is a local to global principal for ranks: if $A$ is Dedekind but not a PID, see the Foster-Swan theorem.
:::

:::{.theorem title="?"}
Let $R \in \Dedekind, M\in \rmod^\fg$.
Then

a. $P \da M/M_\tors \in \rmod^{\fg, \proj}$, say of rank $r$.
b. If $r=0$ then $M = M_\tors$ and if $r\geq 1$ then $\exists I \normal R$ such that $M\cong M_\tors \oplus P = M_\tors \oplus R^{r-1} \oplus I$.
c. The class $[I] \in \Pic(R)$ is an invariant of $R$ called the Steinitz class $\St(M)$.
d. If $M_\tors\neq 0$ then there exists an $N$ and $\ts{n_i}_{i\leq N} \in \ZZ_{\geq 0}$ such that and $\ts{p_i}_{i\leq N} \in \mspec R$ such that $M_\tors \cong \bigoplus _{i\leq N} R/p_i^{n_i}$.

:::

:::{.proof title="Sketch"}
In steps:

1. Ascertain the structure of $M_\tors$ by reducing to PIDs.
2. Show $M$ finitely-generated torsionfree implies projective.
3. Show $P$ rank $r$ projective implies $P \cong \oplus _{i\leq r} I_i$ for $I_i \normal R$
4. For $I, J \in \Frac(R)$, show $I \oplus J \cong IJ \oplus R$.
5. Show that $P$ rank $r$ projective implies $P \cong R^{r-1} \oplus I$ using steps 3 and 4, then show $[I]$ is an isomorphism invariant of $P$.

We'll come back to proving this.

**Step 2**: Letting $M\in \rmod^{\fg}$ be torsionfree, $M$ is projective iff \( \bigoplus _{p\in \mspec R} M\localize{p} \) is a sum of frees in $\mods{R\localize{p}}$.
For any $S$, if $M$ is finitely-generated torsionfree then $M_S \da M \tensor_R S\inv R$ is a finitely-generated torsionfree $S\inv R$ modules, so $M\localize{p}$ is a finitely-generated torsionfree modules over the PID $R_p$, which must be free.

> See a consequence in the proposition below.

**Step 3**:
For $P\in \rmod^{\fg, \proj}$ of rank $r\geq 1$, $V\da P\tensor_R K \in \Vect\slice K^\fd$.
Let \( \lambda: V\to K \) be a surjective $K\dash$linear map, then $Q\da \lambda(P)$ is projective by the proposition below.
Then there is a SES
\[
0\to K \da \ker \ro{ \lambda }{P} \to P\to Q\da \lambda(P) \to 0
,\]
and since images of projectives are projective here, this splits as $P \cong K \oplus Q$.
This splits off a rank 1 direct summand, and we continue inductively.
[^splitting_vbs]

[^splitting_vbs]: 
This is like saying a vector bundle splits into line bundles, which is nontrivial (and not true) in general.
However, here we're essentially splitting a vector space into 1-dimensional subspaces.

**Step 4**:
Skipped! Not so conceptually interesting, but see [@CA].

**Step 5**:
Letting $I, J\in \Frac(R)$; we want to show stable equivalence implies equivalence, so $R^n \oplus I \cong R^n \oplus J$ implies $[I] = [J]$, or equivalently $I \cong_R J$.
Note that the converse is immediate.

A computation:
\[
R^{n+1} \oplus R \cong R^{n+2} \cong (R^n \oplus I) \oplus I\inv \cong (R^n \oplus J) \oplus I\inv \cong R^{n+1} \oplus JI\inv
.\]
Thus $JI\inv$ is stably free, i.e. summing with finitely-generated free module yields a finitely-generated free module. 
By [@CA Prop 7.17] $JI\inv$ is free, so principal, so $[I] = [J]$.
:::

:::{.proposition title="Projective modules over DDs are lattices"}
Let $R\in \Dedekind$ with $\ff(R) = K$. For $M\in \rmod^{\fg}$, TFAE

- $M$ is projective
- There exists a $V\in \Vect\slice k^\fd$ and an embedding $\phi \in \rmod{M \embeds V}$.

:::

:::{.proof title="?"}
$1\implies 2$:
$M$ projective implies torsionfree implies $M\injects V\da M \tensor_R K$ (the kernel is the torsion submodule, which vanishes), and $K$ is the desired vector space.

$2\implies 1$:
$V\in \kmod$ on which $R$ acts invertibly, hence $V\in \rmod$ is torsionfree.
If $M \embeds V$ then $M$ is finitely-generated torsionfree and this projective by step 2 in the theorem.
:::

## Characteristic Ideals

:::{.remark}
Our next major topic: lattices in Dedekind domains.
For $V \in \kmod$, to a pair of $R\dash$lattices \( \Lambda_i \) we will attach an invariant \( \chi( \Lambda_1, \Lambda_2) \).
:::

:::{.definition title="Characteristic ideals"}
Let $R\in \CRing$ and $M\in \rmod$, then $M$ has finite length iff $M$ has a Jordan-Holder series.
The JH theorem shows that any two JH series yield the same multiset of simple modules $R/\mfm_i$ and multiplicities.
The **characteristic ideal** of $M$ is the product
\[
\chi(M) = \prod_{i\leq r} \mfm_i
.\]
:::

:::{.exercise title="?"}
Let $M\in \rmod^{\fl}$.

- Show $\chi(M)$ annihilates $M$.
- Show $M\in\mods{R/\chi(M)}$, but need not be faithful (i.e. one could have a proper containment $\chi(M) \subset \Ann_R(M) )$.

:::

:::{.exercise title="?"}
Let $M_1 \injects M_2 \surjects M_3$ be a SES in $\rmod$.

- Show that $M_1$ has finite length iff $M_1, M_3$ have finite length.
- If $M_2$ has finite length, show $\chi(M_2) = \chi(M_1) \chi(M_3)$.

:::

:::{.exercise title="?"}
For $R$ a domain, $M\in \rmod$,

- Show if $M$ is finite length then $M$ is finitely-generated torsion.
- Find a domain $R$ such that some finitely-generated torsion module does not have finite length.

Thus finite length is strictly stronger than finitely-generated torsion.
:::

:::{.exercise title="?"}
For $R$ a **Dedekind domain**, $M\in \rmod$,

- Show $M$ has finite length iff $M$ is finitely-generated torsion.
- Suppose $M$ is finitely-generated torsion and write $M \cong \bigoplus R/I_i$ using the structure theorem.
  Show $\chi(M) = \prod I_i$, and in particular $\chi(R/I) = I$.

:::

:::{.remark}
How to think about $\chi(M)$?
It is an ideal that annihilates $M$, but if $M\cong R/I_i$ then $\chi(M) = \prod I_i$ while $\Ann_R(M) = \lcm(I_i)$.
The difference is adding exponents vs taking the maximum exponents, and the former can yield a smaller ideal.
Note that we can only recover $M$ from $\chi(M)$ when $\chi(M)$ is squarefree - consider $\chi(M) = p^2$ and $R/p^2$ vs $(R/p)\sumpower{2}$.
:::

:::{.exercise title="?"}
Two parts:

a. Let $R = \ZZ$, so finite length is equivalent to being a finite group.
  Decomposing into cyclic subgroups $M \cong \bigoplus Z/n_i Z$, show that $\chi(M) = \prod n_i = \size M$ is the cardinality, and thus one can interpret $\chi(M)$ as a measure of the size of $M$.

b. Let $L \in\Field, R= k[t]$ and $M\in \rmod$ nonzero.
  Show $M$ is finitely-generated torsion iff $\dim_k M < \infty$, and when this holds one may choose a $k\dash$basis and identify $M\cong k^n$ and the $R\dash$module structure is determined by $t\cdot \in \Endo_k(M) \cong \Mat_{n\times n}(k)$.
  Then $\chi(M)$ has a unique monic generator given by the characteristic polynomial $p(t) \da \det(t\cdot I_n - M)$.

This explains the terminology, and shows that $\chi(M)$ contains more refined information since $\dim_k M = \def p(t)$.
:::

:::{.exercise title="Categorical fun, not difficult"}
Let $R\in \Dedekind$, show that $M\to \chi(M)$ induces an isomorphism $\K_0(R) \to \Frac(R)$ in a universal way.
:::

:::{.exercise title="?"}
Let $R\in \Dedekind$ and $I \subseteq J$ fractional ideals. 
Show that $J/I$ has finite length and $\chi(J/I) = I J\inv$.
:::

:::{.exercise title="More important"}
Show that over a Dedekind domain, $\chi(M)$ can be computed locally:
take $p\in \mspec R$ and localize to $M\localize{p} \da M_\tensor_R R\localize{p}$.

- Show $M\localize{p}$ is finitely-generated torsion in $\mods{R\localize p}$
- Since $R \localize p \in \DVR$, write $\chi(M\localize p) = (p R\localize p)^{a_i}$ for some $a_i \geq 0$.

:::

:::{.remark}
Next time: lattices over Dedekind domains, discriminants for such lattices (when one has a quadratic form).
:::