# Monday March 30th

We'll cover localization and Noetherian rings.

We'll need some local to global results.

Corollary
: If $R$ is a domain with fraction field $K$, then
    $$
    \intersect_{\mfm \in\maxspec R} R_\mfm = R
    $$

It follows from this that the analogous statement for prime ideals holds.

Proposition (7.11: Expressing Colon Ideal as Annihilator)
:   Let $M_1, M_2 <_R M$ be sub $R\dash$modules of $M$, then $$(M_1 :_M M_2) = \theset{x\in R \suchthat xM_2 \subseteq M_1} = \ann\qty{M_1 + M_2 \over M_1}.$$
    So the colon is an ideal.
    Suppose that $S \subset R$ be a multiplicatively closed subset satisfying

    a. If $M$ is finitely generated, $S\inv \ann M = \ann S\inv M$.
    b. If $M_2$ is finitely generated, then $S\inv(M_1 : M_2) = (S\inv M_1: S\inv M_2)$.

Proof (of Proposition)
: Omitted, see notes.

Proof (of Corollary)
:   We want to show that $R \injects \intersect R_\mfm$.
    If $x\in K \setminus R$ and $I \definedas (R : Rx) \definedas \theset{r\in R \suchthat rx \in R}$, note that $1\not\in I$.
    Thus $I$ is a proper ideal, so let $\mfm \in \maxspec R$ with $\mfm \supset I$.
    Then $(R\mfm: R\mfm x) = I_\mfm = ? \subset \mfm R_\mfm$ which is a proper ideal in $R\mfm$.
    So $1$ is not in the colon ideal.

These colon ideals aren't the obvious thing to look at, but come up in applications to algebraic geometry and number theory.

Remark
:   For $I, J \normal R$, we have $(I:_R J) = \theset{x\in R \suchthat xJ \subset I}$.
    Thus this construction formalizes the idea of a "quotient $I/J$".
    This works for ideals in a domain, but also for *fractional ideals*.

Definition (Fractional Ideal)
: A *fractional $R\dash$ideal* is a nonzero $R\dash$submodule $I$ of $K$ such that there exists an $x\in R^\bullet$ such that $xI \subset R$.

Any ideal is a fractional ideal by taking $x=1$.
Note that some books define fractional ideals as finitely generated $R\dash$submodules, but this isn't a great definition.


Exercise (Easy)
:   If $I \subset K$ is finitely generated, then $I$ is a fractional ideal.

    > Idea: scale all generators.

Note that $I$ is a fractional $R\dash$ideal iff $(R:_K I) = \theset{x\in K \suchthat xI \subset R}$ is nonzero.


Next up: local-global theory for lattices.

Theorem (Local-Global Principle for Lattices)
:   Let $R$ be a domain with fraction field $K$ and $V$ a finite dimensional $K\dash$vector space.
    Let $\Lambda \subset V$ be a finitely generated $R\dash$submodule.
    Then $\intersect_{\mfm \in \maxspec R} \Lambda_\mfm = \Lambda$.

Note that if $V=K$ and $\Lambda = R$, this recovers the previous theorem.
Thus a global lattice over an integral domain can be recovered in terms of its localizations.

Next up: rounding out some theorems about projective and free modules.

Theorem (Kaplanksy, Very Important: Projective Over Local Implies Free)
: A projective module over a local ring  is free.


We proved this for finitely generated modules, which is the most important case.
Note that projective modules are *locally free*, i.e. if $M$ is a projective $R\dash$module then for all $\mfp \in \spec R$, $M_\mfp$ is a free $R_\mfp\dash$module.

We'll now define a notion of "the least number of generators" locally.

Definition (Rank Function)
:   Suppose $M$ is a finitely generated $R\dash$module.
    For $\mfp \in \spec R$, denote $k(\mfp) = ff(R/\mfp)$ the *residue field* at $\mfp$.
    We have $R \surjects R/\mfp \injects k(\mfp)$, so we define the rank function as
    \begin{align*}
    \rk_M: \spec R &\to \NN \\
    \mfp &\mapsto \dim_{k(\mfp)} M \tensor_R k(\mfp)
    .\end{align*}
    where the RHS is base-changing to $k(\mfp)$ to get a finite dimensional vector space over $k(\mfp)$.

Exercise
:   Show the following properties of the rank function for $M, N$ finitely generated $R\dash$modules:

    a. $\rk_{M\oplus N} = \rk_M + \rk_N$ and $\rk_{M\tensor_R N} = \rk_M \cdot \rk_N$.
    b. Compute the rank function on $\ZZ/n\ZZ$ for $R = \ZZ$.
    c. For $R$ a PID, compute $\rk_M$.

- Taking $\mfp = p\ZZ$ in $\ZZ$ yields a delta function at $p$.
- If $M$ is finitely generated and free, then
\begin{align*}
\rk_M(\mfp) = \dim_{k(\mfp)} R^n \tensor_R k(\mfp) = \dim_{k(\mfp)} \bigoplus_{i=1}^n R \tensor_R k(\mfp) = \dim_{k(\mfp)} k(\mfp)^n = n
.\end{align*}
- If $M$ is locally free, then for all $\mfp \in \spec R$, we have $M_\mfp \cong R_\mfp^{\rk_M(\mfp)}$.
\begin{center}
\begin{tikzcd}
R \ar[rd] \ar[r] & R_\mfp\ar[d] \\
& k(\mfp)
\end{tikzcd}
\end{center}
    - Thus the rank can be thought of as the fiberwise dimension for bundles.

- If $M$ is *stably free*, i.e. there exists an $m, n\in \NN$ such that $M \oplus R^m \cong R^n$, then $\rk_M = n-m$.

Note that projective implies locally free.
In order for a finitely generated projective module to be free, it must have constant rank function.
The geometric analog here would be that the fibers having constant dimension is necessary for a bundle to be trivial.

Proposition (Determining if a Projective is Free)
:   Suppose $M$ is finitely generated projective of constant rank $n$.
    Then $M$ is free iff $M$ can be generated by $n$ elements.