# Wednesday February 12th

## Projective Modules and Ideals

**Summary:**
Free $\implies$ projective $\implies$ flat $\implies_{R \text{ a domain }}$ torsion free.
Moreover, projective $\implies$ reflexive.

If $M, N$ are cyclic $R\dash$modules, then $\ann(M\tensor_R N) = \ann M + \ann N$.
Does this hold for every $M, N$? 
The answer is no; we have $$\ann(M\tensor_R N) \supseteq \ann M + \ann N.$$
See MSE post: let $I\normal R$ and $M$ an $R\dash$module, we have $M \tensor_R R/I = M/IM$.
Is there an equality $\ann(M/IM) = \ann(M) + I$?
No, take $R = \CC[x, y]$.

Recall that an $R\dash$module is *reflexive* iff $\iota: M \to (M\dual)\dual$ is an isomorphism, where $M\dual = \hom_R(M, R)$.
This is injective for $R$ a field, and then surjective iff $R$ is finite-dimensional.
As shown in the problem sessions, finitely generated free modules are reflexive.

**Exercise:**
Show that direct summands of reflexive modules are reflexive, and $M_1 \oplus M_2$ is reflexive iff $M_i$ are reflexive.
Conclude that finitely generated projective modules are reflexive. 

Example
: To get a projective module that is not free, take
$\CC^2 = \qty{ \CC \cross 0  } \oplus \qty{ 0\cross \CC  } = \CC^2$, which is free, so the summands are projective, but not free.

Note: this corresponds to taking a vector bundle over a disconnected space, and letting the fibers just be different dimensions.

Letting the summands above be $I, J$, note that $I+J = R$ and $IJ = 0$, which is a comaximality condition.

Lemma (3.17)
:   Let $I, J, K_1, \cdots, K_n \normal R$.
    Then

    a. $(I+J)(I\intersect J) \subseteq IJ$

    b. If $I+J = R$ (so $I,J$ are comaximal), then $I\intersect J = IJ$. 

    c. If $I+ k_i = R$ for all $1\leq i \leq n$ then $I + K_1\cdots K_n = R$.

Proof
: Omitted.

Theorem (First Isomorphism Theorem for Rings)
:   Let $R$ be a domain, $IJ \normal R$ such that $I+J = R$.
    We can form a map:

    \begin{center}
    \begin{tikzcd}
    I \arrow[rdd, bend right] \arrow[rrrdd, bend left] &                            &  &   \\
    &                            &  &   \\
    & I \oplus J \arrow[rr, "q"] &  & R \\
    &                            &  &   \\
    J \arrow[ruu, bend left] \arrow[rrruu, bend right] &                            &  &  
    \end{tikzcd}
    \end{center}

    where

    \begin{align*}
    q: I \oplus J &\to R \\
    (i, j) &\mapsto i+j
    .\end{align*}

    Then

    a. $q$ is surjective

    b. $\ker q = \theset{(x, -x) \suchthat x\in I\intersect J} \cong_R I\intersect J = IJ$.

    c. There is a SES $0 \to IJ \to I\oplus J \mapsvia{q} R \to 0$, and since $R$ is projective, $I \oplus J \cong_R IJ \oplus R$.

    d. If $IJ$ is principal (so $IJ \cong_R R$) then $I$ is projective.

See "monogenic". 
This gives a criterion for determining if ideals are projective.

Exercise
:   Let $R =  \ZZ[\sqrt{-5}]$ with $\mfp_1 = \generators{3, 1 + \sqrt{-5}}$ and $\mfp_2 = \generators{3, 1-\sqrt{-5}}$.

    a. Show that $R/\mfp_1 \cong R/\mfp_2 \cong \ZZ/3\ZZ$.

    b. Show $\mfp_1 + \mfp_2 = R$.

    c. Show $\mfp_1 \mfp_2 = \generators{3}$.

    d. Show neither $\mfp_1, \mfp_2$ are not principal.

    e. Conclude $\mfp_1 \cong_R \mfp_2$ is a finitely generated projective but *not* free $R\dash$module.

## The Picard Group

Definition (Picard Group)
:   Let $R$ be a domain with fraction field $K$, we'll define *picard group* $\pic (R)$ in the following way:
  
    For $I\normal R$ with $I\neq 0$. we say $I$ is invertible iff there exists a $J\normal R$ such that $IJ$ is principal.
    
    Then $\pic (R)$ is the set of invertible ideals modulo the equivalence $I\sim J$ iff there exist $a, b\in R^\bullet$ such that $\generators{a}I = \generators{b}J$.

    This is a group under $[I] + [J] = [IJ]$ (check that this is well-defined).

Note that if $I$ is principal, then $[I] = 1$ is the identity, and if $IJ = \generators{x}$, then $[I] [J] = [\generators{x}] = 1$.

Fact
: If $I$ is invertible, then $I$ is a projective $R\dash$module.

Fact (Stronger)
:   If $I\normal R$ in a domain, then 

    - $I$ is invertible iff $I$ is a projective $R\dash$module.
  
    - $[I] = 1$ in $\pic R$ iff $I$ is principal iff $I$ is a free $R\dash$module.

Proof
: Later!

Every nontrivial element gives a projective but not free $R\dash$module!
Note that $\pic R = 0$ for $R$ a PID.

Definition (Dedekind Domain and Class Group)
: $R$ is a *Dedekind domain* iff every nonzero $I\normal R$ is invertible, and $\pic R$ is referred to as the *class group* of $R$.
  
In this case, $\pic R = 0$ iff every ideal is principal iff $R$ is a PID.

So the class group measures how far $R$ is from a PID.
Any Dedekind domain that is not a PID yields projectives that aren't free.

Rings of integers over number fields are Dedekind domains.

> Embarrassingly open problem: are there are infinitely many number fields $K$ such that the ring of integers $\ZZ_K$ is a PID, or equivalently $\pic \ZZ_K = 0$?

Example (Important)
:   Let $k$ be a field and $n\in \ZZ^+$, and define $R\definedas k[t_1, \cdots, t_n]$.
    Since $k$ is a PID, $R$ is a PID, and every finitely generated module over a PID is free.

Theorem (Bass, 1962)
: Let $R$ be connected (recall: rules out silly case!) and noetherian.
  Then every infinitely generated (i.e. *not* finitely generated) projective module is free.

So we can restrict our attention to the finitely generated case.

> Analogy: is every topological vector bundle trivial? E.g. for $\CC^n$, yes. 
> Are all holomorphic bundles trivial? In general, no, see Stein manifolds.

*Question (Serre, 1950s):*
Is every projective $R\dash$module free?

*Answer:* 
Yes, showed by Quillen, Suslin 1976.
See book about this by T.Y. Lam.

Upcoming:
Algebraic $K\dash$theory, built from f.g. projective $R\dash$modules.
Trivial in $K_0$ doesn't quite imply free, usually weaker.
Tries to analyze distinction between projective and free.
Also some results about flat modules.