# Monday January 13

## Logistics

Some topics for final projects

- The cardinal Krull dimension of $\mathrm{Hol}(X)$.
- Galois connections
- Ordinal filtrations
- Lam-Reyes prime ideal principal
- $C(X)$
- $\mathrm{Hol}(X)$
- Semigroup rings
- Swan's Theorem 
  - Vector bundles on a compact space
- Boolean rings and Stone duality
- More Nullstellensatz 
  - Beyond Hilbert's usual one
- Hochster's Theorem 
  - Characterizes $\spec R$ as a topological space, i.e. when is a topological space homeomorphic to the spectrum of some commutative ring.
- Invariant theory (quotients of rings under finite group actions, i.e. $R^G$ for $\abs G < \infty$)
  - For $R=k$ a field, this is Galois theory
  - Easy case of geometric invariant theory, when $G$ is infinite
- UFDs
  - What conditions does a ring need to have to ensure unique factorization?
- Euclidean rings
- Claborn (Leedham-Green-Clark): Every commutative group is (up to isomorphism) the class group of some Dedekind domain.
  - A type of inverse problem, class group measures deviation from being a UFD
  - Uses ordinal filtrations, transfinite induction
  - See proof in elliptic curves course


## Rings of Functions

Let $k$ be a field, $X$ a set of cardinality $\abs{X} \geq 2$, and define $k^X \definedas \mathrm{Maps}(X. k) = \theset{f: X\to K}$ is a ring under pointwise addition and multiplication.
As a ring, this is a (big!) cartesian product.

*Some facts:*

- $k^X$ is not a domain (**exercise**), and there are nontrivial idempotents ($e^2 = e$)
  
  > Note: it could be worse and have nilpotents.

- $k^X$ is *reduced*, i.e. it has no nonzero nilpotents, where $z\in R$ is nilpotent iff $\exists n\geq 1$ such that $z^n = 0$.
  - Note: domains are reduced, cartesian products of reduced rings are reduced.

- Every subring of $k^X$ is reduced. 
  
  > Moral: should be viewing every ring as functions on some space, but this can't literally be true because of the above restrictions. 
  > Nilpotent elements are "hard to view as functions".

- For $X$ a topological space, $C(X)$ the ring of continuous functionals to $\RR$, then $C(X) \subset \RR^X$.

Exercise
: When is $C(X)$ a domain? (Note that we can have products of nonzero functions being identically zero.)

*Example:* 
Let $R$ be the ring of holomorphic functions $\CC\selfmap$, i.e. $\mathrm{Hol}(\CC, \CC) \definedas \theset{f:\CC \to \CC \suchthat f \text{ is holomorphic }}$.

The set of zeros of such an $f$ must be discrete, the example of bump functions doesn't go through holomorphically.

This is a domain, not Noetherian, not a PID, but every f.g. ideal is principal (thus this is a Bezout domain, a non-Noetherian analog of a PID).

It has infinite Krull dimension: recall that ideals are prime iff $xy\in \pr \implies x\in\pr$ or $y\in\pr$ iff $R/\pr$ is a domain, and the Krull dimension is the supremum $S$ of lengths of chains of prime ideals (only when $S$ is finite).

If $C \subset (X, \leq)$ is a finite-length chain in a totally ordered set, then the length $\ell(C) = \abs{C} - 1$ (1 less than the number of elements appearing).
The *cardinal Krull dimension* of a ring $R$ is the actual supremum.

> Note: in Noetherian rings, there can still be finite but unbounded length chains.

Letting $X$ be a complex manifold (i.e. covered by subsets of $\CC^n$ with holomorphic transition functions) and let $\mathrm{Hol}(X)$ be the holomorphic functionals $f: X\to \CC$.
Then $\mathrm{Hol}(X)$ is a domain iff $X$ is connected.

> Note that if $X$ is disconnected, we can take a function that is constant on one component and zero on another, then switch, then multiply to get a zero function.

If $X$ is a compact connected projective variety, then $\mathrm{Hol}(X)$ is just constant functions by the open mapping functions.
So $\mathrm{Hol}(X) = \CC$, and $\mathrm{card dim} \CC = 0$ because for any field there are only two ideals, and here $(0)$ is prime.
Moreover, $\mathrm{card dim}\mathrm{Hol}(\CC) \geq \alpha_0$.

> Note that for complex manifolds, $X$ is either compact or supports many holomorphic functions.

Theorem (Function Spaces Can Have Large Unbounded Chains)
: If $X$ is a connected complex manifold which has a nontrivial holomorphic function, i.e. $\mathrm{Hol}(X) \supset \CC$, then there exists a chain of prime ideals in $\mathrm{Hol}(X)$ of length $\abs{\RR} > \aleph_0$, i.e. it has at least the cardinality of the continuum.

Note: the cardinality could be even bigger!

> Maximals are prime: equivalent to fields are integral domains.

## Rings

Take all rings to be unital, i.e. containing $1$.
A ring without identity is referred to as an *rng*.
In this course, all rings are commutative.

*Example:* 
This is a fairly special restriction.
Take $(A, +)$ a commutative group and define $\mathrm{End}(A) = \theset{f: A\to A}$ the ring of group homomorphisms under pointwise addition and composition.
This is generally not commutative, i.e. $\mathrm{End}(\ZZ/(2) \oplus \ZZ/(2)) = M_2(\ZZ/(2))$ the ring of matrices with $\ZZ/(2)$ entries, which is not commutative.

Exercise
: Given $(A, +)$, show that $\mathrm{End}(\bigoplus^n A) = M_n(\mathrm{End}(A))$.

Generally, if $R$ is a ring and $M$ is as $R\dash$module, then $\mathrm{End}_R(M) = \theset{f: M \to M}$ of $R\dash$module homomorphisms is always a ring under pointwise addition and composition, and is (probably) non-commutative.