# Thursday, August 25

## Projective modules 

:::{.remark}
Recall

- For my own edification: $R \in \Dom$ is a domain iff $R$ is reduced (no nonzero nilpotents) and irreducible (only one minimal prime).
- We'll be considering modules equipped with a form, e.g. the trace form -- a very important construction for this course!
- $P\in \rmod$ is projective iff $P \oplus Q$ is free for some $Q$.
  - Equivalently: every $A\to\B\to P$ splits.
  We'll soon find a local characterization.
- If $R$ is commutative, $\rmod(M, N)\in \rmod$ (noting these are $R\dash$equivariant functions).
- Duals: $M\dual \da \rmod(M, R)$. 
  See the theorem and exercise below.
- Over a domain $R$, define the rank of $M\in \rmod^{\fg}$ as $\rank(M) \da \dim_k (M\tensor_R K)\in \ZZ_{\geq 0}$.
  - Geometric interpretation: finitely-generated projective modules correspond to vector bundles, and ranks correspond to fiber dimensions.
  - Rank one projective modules correspond to line bundles/invertible sheaves.
- For $P_i\in \rmod^{\fg, \proj}$ with $R\in\Dom$ then $\rank_R(\bigoplus P_i) = \sum \rank_R(P_i)$ and $\rank_R(\bigotimes P_i) = \prod \rank_R(P_i)$.
- If $P_i$ are rank 1 projective then $\bigotimes P_i$ is again rank 1 projective.
- $\rmod(R, M) \cong M$ and $R\tensor_R M = M\tensor_R R = R$ canonically.
- There are inverses for tensoring rank 1 projectives, i.e. for $P$ there exists some $P'$ with $P\tensor_R P' \cong R$.
  See exercise below, use that projectives are locally free.
- **Very important invariant of a ring**: taking isomorphism classes forms a commutative group under $\tensor_R$, the Picard group $\Pic R$.
  - $\Pic(R) = \cl(R)$ for $R \in \Dedekind$, the ideal class group.
  - Over domains, projective \(R\dash\)modules correspond to fractional ideals, and $\Pic R$ can be formulated in these terms.

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:::{.slogan}
Projectives are locally free.
:::

:::{.theorem title="\autocite[7.32]{CA}"}
For $A\in \rmod$, TFAE:

- $A \in \rmod^{\fg, \proj}$
- For all $B\in \rmod$, there is an isomorphism
\[
\Phi: A\dual \tensor_R B &\to \rmod(A, B) \\
(f, b) &\mapsto (a\mapsto f(a) b)
.\]

> See reflexive and torsionless modules.

:::

:::{.exercise title="?"}
Show that if $M\cong R^n$ is free then $M\dual \cong R^n$, and $\rmod^{\fg, \proj}$ is closed under duals.
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:::{.exercise title="?"}
If $R\in\Dom$ and $I\in \Id(R)$ is nonzero,

- Show $I$ is principal iff $I \in \rmod^{\free}$ and this $I\isoas_{\rmod} R$.
- Show that if $I$ is projective then $\rank_R(I) = 1$.

:::

:::{.exercise title="?"}
Let $P\in\rmod{\fg, \proj}$, then show
\[
P\tensor_R P\dual \iso R
,\]
so $P\dual$ is a inverse with respect to $\tensor_R$.
:::

:::{.solution}
Check $P\dual \tensor_R P = \rmod(P, P) = \endo_R(P) \cong R$.
This is more clearly true for $P$ free of rank 1, since $\Hom_R(P, R) = R$.
:::

## Localization

:::{.remark}
We'll lean heavily on localization -- this is not a particularly fancy tool!

- Local Dedekind domains are DVRs.
- Step 1, the prototypical example: $R\in \Dom$ admits a fraction field $K = \ff(R)$.
    Define $K = (R\times R\smz)/\sim$ where $\sim$ denotes equivalence under cross-multiplication.
    Obtain a map $\iota: R\injects K$ where $a\maps [(a, 1)]$.
- Step 2: this inverts everything, and localization inverts a subset.
    For $B\subseteq R\smz$, can take $R \subseteq R\adjoin{\ts{1/b \st b\in B}} \subseteq \ff(R)$.
    - This yields many more denominators than just $2,3,6$, so take the submonoid of $R\smz$ generated by $B = \ts{2,3}$, denote this $S_B$.
  - Multiplicative subsets of rings are submonoids of $(R, \cdot)$, i.e. subsets $S$ with $1\in S$ and $SS \subseteq S$.
    Then $R\adjoin{{1\over b}_{b\in B}} = R\adjoin{{1\over s}_{s\in S_B}}$.
  - Define $S\inv R \da R\adjoin{ {1\over s} \st s\in S } = \ts{{a\over s} \st a\in R, s\in S}$.
- Step 3: for $R\in \Dom, S \subseteq R\smz$, define $S\inv R = (R\times S)/\sim$ with $(a_1, s_1)\sim(a_2, s_2) \iff s_2a_1 = s_1a_2$, so $S\inv R = R\adjoin{{1\over s}_{s\in S}} \subseteq \ff(R)$.
- Step 4: for $R\in \CRing$ arbitrary, $S \subseteq R\smz$, the same thing works if $S$ contains no zero divisors and we get an injective ring morphism $\iota(R) = [(r, 1)]$.
  - If $S$ does contain zero divisors, fix the equivalence relation: $\exists s$ such that $ss_1a_2 = ss_2 a_1$.
  - $s\in R$ is a zero divisor iff $\cdot s$ is not injective.
- This universally makes $s\in S$ a unit, but can kill other elements.
- Localization at $I\normal R$ -- this is not quite a multiplicative set, but $I' = I\union\ts{1}$ is.
  One could take the *saturation* of $I$ by adding in all divisors, but here $I' = 0$, so $(I')\inv R = 0$.
  - Note that $R\sm I$ is multiplicative iff $I\in \spec R$, so denote $R_p \da (R\sm p)\inv R$.
  - If $p=\gens{0}$ then $R_p = \ff(R)$.
- $R_p$ is local with maximal ideal corresponding to $p$.
- Semilocalization: consider localizing at finitely many $p_i\in \spec R$, non containing any other.
  Note that $S \da \Intersect R\sm p_i$ is a multiplicative set and define $R_{p_1, \cdots, p_r} = S\inv R$, then $\mspec R_{p_1, \cdots, p_r} = \ts{p_1, \cdots, p_r}$.
  - This will be useful to us for Dedekind domains.
- Push/pull: for $f\in \CRing(A, B)$ and $I\in \Id(A)$, define $f_*(I) = \gens{f(I)}_B\in \Id(B)$ and $f^*(J) \da f\inv(J) \in \Id(A)$.
  See exercise.
  - Consider $a\normal R$ and $q: R\to R/a$, then $q_*(I) = (I+a)/a, q^*J) = J$, and $J$ corresponds to $a \subseteq J \normal R$.
    Then $q^* q_* I = I+a$ and $q_* q^* J = J$, so $q^*$ identifies $\Id(R/a) = \ts{J\in \Id(R) \st J\contains a}$.
- Localization is roughly "dual" to taking quotients in some sense.

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:::{.example title="of inverting elements"}
$\ZZ\adjoin{{1\over 2}, {1\over 3}} =\ts{ {a\over 2^{b_1}3^{b_2} } \st a\in \ZZ, b_i \geq 0}$.
:::

:::{.exercise title="?"}
Show that if $S$ contains zero divisors, then $s_2a_1 = s_1a_2$ need not be an equivalence relation.
Check transitivity.
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:::{.exercise title="?"}
Show that if $0\in S$, then $S\inv R = 0$.
Show that this also holds if $S$ contains a nilpotent element.
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:::{.exercise title="?"}
Show that if $s\in S$ is a zero divisor and $s\in \Ann(a)$, so $sa = 0$, then $\iota(a) = 0$.
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:::{.exercise title="?"}
Show that $\ker(R\to S\inv R)$ are elements $x\in R$ such that $\Ann_R(x)$ intersects $S$.
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:::{.exercise title="?"}
Show that $\iota(S) \subseteq (S\inv R)\units$ universally, i.e. if $R \mapsvia{\phi} T$ with $\phi(S) \subseteq T\units$ then $\phi$ factors through $\iota$.
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:::{.exercise title="?"}
Show that $f^* f_* I \contains I$ and $f_* f^* I \subseteq I$.
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:::{.proposition title="\autocite[7.3]{CA}"}
For all $J\normal S\inv R$, one has $i_* i^* J = J$.
Moreover $i^*$ identifies $\spec S\inv R$ with $\ts{p\in \spec R \st p \text{ is disjoint from } S}$.
:::