# Wednesday January 22nd

Let $R$ be a ring and let $\mathcal{I} (R)$ be the set of ideals $I\normal R$. 
This algebraic structure is

- Partially ordered under inclusion
- Forms a complete lattice with $\sup$ the ideal generated by a family and $\inf$ the intersection.
- Forms a commutative monoid under $I+J$
- Forms a commutative monoid under $IJ$

For any commutative monoid $(M, +)$, there exists a group completion $G(M)$ such that

- $G(M)$ is a commutative group
- $g: M \to G(M)$ is a monoid homomorphism
- For any map $\phi: (M, +) \to (G, +)$  into a commutative group, we have the following diagram

\begin{center}
\begin{tikzcd}
M \arrow[rr, "\forall \phi"] \arrow[rdd, "g"] &                                   & G \\
                                              &                                   &   \\
                                              & M(G) \arrow[ruu, "\exists! \Phi"] &  
\end{tikzcd}
\end{center}

> So $\phi$ factors through $M(G)$.

If this exists, it is unique up to unique isomorphism (as are all objects defined by universal properties).
It remains to construct it.

Exercise
:   For $(M, +)$ a commutative monoid, show that TFAE

    1. There exists an injective $\iota: M \injects G$ monoid homomorphism for $G$ some commutative group.
    2. The map $g: M \to G(M)$ is an injection.
    3. $M$ is cancellative, i.e. $\forall x,y,z\in M$ we have $x+z = y+z \implies x = y$, i.e. the map $p_z(x) = x + z$ is injective.

The content here is in $3 \implies 1$.

Definition (Reduced Monoids)
: A commutative monoid is *reduced* iff $M\units = (0)$, i.e. if "$\forall m\in M\exists n$ such that $m+n = 0$" $\implies$ $m=0$

Example
: $(\NN, +)$ and $(\ZZ^+, \cdot)$ are cancellative and reduced.

Definition (Zero Elements in Monoids)
: $z\in M$ is a **zero element** iff $z+x = z$ for all $x\in M$.

Remark
: If $M$ has a zero element, then $G(M) = \theset{0}$.

$(0)$ is a zero element of $(\mathcal I(R), \cdot)$, so this is not cancellative.
If we take $\mathcal{I}^\bullet$ the set of nonzero ideals with multiplication, then this is a submonoid of $\mathcal{I}(R)$ iff $R$ is a domain.

For $R$ a domain, let $\mathcal{I}_1(R)$ be the set of nonzero principal ideals of $R$, then $\mathcal I_1(R) = R^\bullet/R\units$, so this is reduced and cancellative.

What is the group completion? 
In this case, it will consist of fractional ideals.

If $R$ is a PID, then $\mathcal I_1^\bullet(R) = \mathcal I^\bullet (R)$ is reduced and cancellative.

Example
: $\mathcal I^\bullet \cong (\ZZ^+, \cdot)$.

Warning
: If $R$ is not a PID, then $\mathcal I^\bullet(R)$ need not be cancellative.

Exercise
: Take $R = \ZZ[\sqrt{-3}]$ and $p_2 \definedas \generators{1+\sqrt{-3}, 1-\sqrt{-3}}$.
  Show that $\abs{R/p_2} = 2$, $\abs{R/(2)} = 4$, and $p_2^2 = p_2(2)$ and $\abs{R/p_2^2} = 8$.
  Conclude that $\mathcal I^\bullet(R)$ is not cancellative.

What went wrong here?
Take $K = \QQ[\sqrt{-3}]$, then $\ZZ_k[\frac{1 + \sqrt{-3}}{2}]$ is the integral closure of $\ZZ$ in $K$.
$\ZZ_k$ is a Dedekind domain, and there are inclusions

\begin{align*}
\ZZ \subset \ZZ[\sqrt{-3}] \subsetneq \ZZ[\frac{1 + \sqrt{-3}}{2}] \subseteq K
.\end{align*}

Here the problem is that $\ZZ[\sqrt{-3}]$ is not a Dedekind domain.
If $R$ is a Dedekind domain, then $\mathcal I^\bullet(R)$ is cancellative.

Exercise
: Does the converse hold?

Things that are too small to be the full rings of integers, and things tend to go wrong (??).

## Pushing / Pulling

Let $f: R\to S$ be a ring homomorphism.

We can define a pushforward on the set of ideals $\mathcal{I}(R)$:

\begin{align*}
f_*: \mathcal{I}_R &\to \mathcal{I}(S) \\
I &\mapsto \generators{f(I)}
.\end{align*}

and a pullback

\begin{align*}
f^*: \mathcal{I}(S) &\to \mathcal{I}(R) \\
J &\mapsto f\inv(J)
.\end{align*}

**Exercise:**
Show that $f\inv(J) \normal R$.

For $I \normal R$ and $J\normal S$, then 

\begin{align*}
f^* f_* (I) \supseteq I \\
f_* f^* (J) \subseteq J
.\end{align*}

**Exercise:**
These are not equal in general, and give examples where equality does and does not hold.

If $f$ is surjective, $f_* f^* J = J$.

> Will also hold for localization, which is dual to taking a quotient.

Define $\bar I \definedas f^* f_* (I)$ and $J^\circ \definedas f_* f^* (J)$, the closure and interior respectively.
Show that these operations are idempotent.

Definition (Prime Ideals)
: An ideal $\mfp$ is *prime* iff $ab\in \mfp \implies a\in \mfp$ or $b\in \mfp$.

**Exercise:**
$I$ is prime iff $R/I$ is a domain.

Definition (Prime Spectrum)
$\spec(R) = \theset{\mfp \normal R}$ the collection of prime ideals is the spectrum.

Exercise
: Show that for $I\normal R$, if we define 
  $$
  V(I) \definedas \theset{\mfp \in \spec(R) \suchthat p \supseteq I} \subseteq \spec(R)
  ,$$
  then $\theset{V(I) \suchthat I \in \mathcal{I}(R)}$ are the closed sets for a topology on $\spec(R)$ (the Zariski topology).

Exercise
: If $f: R\to S$ and $J \in \spec(S)$ then $f^*(J) \in \spec(R)$.
  Show that $f^*: \spec(S) \to \spec(R)$ is a continuous map.
  Conclude that $\spec(\wait)$ is a functor.

Definition (Maximal Ideals)
: $I\normal R$ is **maximal** iff $I$ is proper and is not contained in any other proper ideal.

Exercise
: $I$ is maximal iff $R/I$ is a field.

Exercise
: Show that maximal ideals are prime.

Definition (Max Spectrum)
: Let $\spec_{\text{max}}(R)$ be the set of maximal ideals and define $V(I) = \theset{\mfm \in \spec_{\text{max}}(R) \suchthat \mfm \supseteq I}$.

Exercise
: Show that these form the closed sets for a topology, and that this is the subspace topology for the Zariski topology.

Exercise
: Show that if $f: R\to S$ and $\mfm \in \spec_{\text{max}}(S)$ that $f^*(\mfm)$ is prime but need not be maximal.

Exercise
: Show that if $f$ is an integral extension, then maximal ideals pull back to maximal ideals.