# Friday January 17th

For a $R \subset T$ a subring of a ring, the set of intermediate rings is a large/interesting class of rings.
Recall: uncountably many rings between $\ZZ$ and $\QQ$!
Taking $R$ a PID and $T$ its fraction field, a similar result will hold.

Define $I\normal R$ as the kernel of a ring morphism.
This implies that $I \subset(R, +)$ with the absorption property $RI \subset I$.
Conversely, any $I$ satisfying these two properties is the kernel of a ring morphism: namely $R \to R/I$.
This makes sense as a group morphism.

Exercise
: Define $xy + I = (x+I)(y+I)$, need to check well-definedness.
	Write out $$(x+i_1)(y+i_2) = \cdots$$
	Need to check that $$i_1y + i_2 x + i_1 i_2 \in I,$$ but the absorption property does precisely this.

Note that if we were in a non-commutative setting, this would define a left ideal.
These don't have to coincide with right ideals -- there are rings where the former satisfy properties that the latter does not.

*Example:*
The subrings of $R = \ZZ$ are of the form $n\ZZ$ for $n\geq 0$, with the usual quotient.

Definition (Proper Ideals)
: An ideal $I \normal R$ is *proper* iff $I \subsetneq R$.

Exercise
: An ideal $I$ is not proper iff $I$ contains a unit.

Exercise
: $R$ is a field iff the only ideals are $0, R$.

Definition (Lattice Structure of Ideals)
: 	Let $\mathcal{I}(R)$ be the set of all ideals in $R$.
		
    What structure does it have?
		It is partially ordered under inclusion.
		It is a complete lattice, i.e. every element has an infimum (GLB) and a supremum (LUB).
		Namely, for a family of ideals $\theset{I_j}$, the **infimum** is the intersection and **supremum** is defined as $\generators{I_j \suchthat j\in J}$, the smallest ideal containing all of the $I_j$, i.e.
		$$
		\generators{y} = \theset{ \sum^n r_i y_i \suchthat n\in \NN_{> 0},~ r_i \in R,~ y_i\in y}
		.$$

Exercise
: For $I_1, I_2 \normal R$, it is the case that $I_1 + I_2 \definedas \theset{i_1 + i_2} = \generators{I_1, I_2}$.

Theorem (Lattice Isomorphism Theorem for Rings)
: Let $I\normal R$ and $\phi: R \to R/I$, and define $\ell(I) = \theset{I \subset J \normal R}$.
	Then we can define maps

	\begin{align*}
	\Phi: \ell(R) &\to \ell(R/I) \\
	J &\mapsto \frac{I+J}{J}
	,\end{align*}

	and

	\begin{align*}
	\Psi: \ell(R/I) &\to \ell(R) \\
	J \normal R/I &\mapsto \phi\inv(J)
	.\end{align*}

	We can check that $\Psi \circ \Phi)(J) = I+J$, and $\Phi \circ \Psi(J) = J$ (= $J/I$?)
	So $\Psi$ has a left inverse and is thus injective.
	Its image is the collection of ideals that contain $J$, and $\Psi: \ell(R/I) \to \ell_I(R)$ **is a bijection** and is in fact a lattice isomorphism with $\ell_I(R) \subset \ell(R)$.

Note that this gives us everything above a given ideal in the ideal lattice (blue); the dual notion will come from localization (red):

\begin{center}
\begin{tikzpicture}[baseline= (a).base, scale=0.8, every node/.style={scale=0.8}]
\node (a) at (0,0){
\begin{tikzcd}
&                                        &    & \spec(R/I)                             &    &                                        &    &  &  &    &                                        &    & \spec(R_\mfp) \arrow[llddd] \arrow[ddd] \arrow[rrddd] &    &                              &    \\
&                                        &    &                                        &    &                                        &    &  &  &    &                                        &    &                                                       &    &                              &    \\
{} & {} \arrow[rruu]                        & {} & {} \arrow[uu]                          & {} & {} \arrow[lluu]                        & {} &  &  &    &                                        &    &                                                       &    &                              &    \\
& \color{blue}{\mfp_1} \arrow[lu] \arrow[u] \arrow[ru] &    & \color{blue}{\mfp_2} \arrow[lu] \arrow[u] \arrow[ru] &    & \color{blue}{\mfp_3} \arrow[lu] \arrow[u] \arrow[ru] &    &  &  &    & \mfp_1 \arrow[ld] \arrow[d] \arrow[rd] &    & \color{red}{\mfp} \arrow[dd] \arrow[ld] \arrow[rd]                 &    & \mfp_2 \arrow[ld] \arrow[rd] &    \\
&                                        &    &                                        &    &                                        &    &  &  & {} & {}                                     & {} &                                                       & {} &                              & {} \\
&                                        &    & \color{blue}I \arrow[lluu] \arrow[uu] \arrow[rruu] &    &                                        &    &  &  &    &                                        &    & \color{red}{\mfq}\ar[dd]                                                  &    &                              &    \\
&                                        &    &                                        &    &                                        &    &  &  &    &                                        &    &                                                       &    &                              &    \\
&                                        &    &                                        &    &                                        &    &  &  &    &                                        &    & \color{red}{(0)}                                                     &    &                              &
\end{tikzcd}
};
\end{tikzpicture}
\end{center}

> Here we take $\mfp$ to be a maximal ideal.

Remarks
: 	The ideal lattice $\ell(R)$ is

	- A complete lattice under subset inclusion,
	- A commutative monoid under addition
	- A commutative monoid under *multiplication*, which we'll define.

Definition (Product of Ideals)
: For $I, J \normal R$, we define $$IJ = \generators{ij \suchthat i\in I,~j\in J}.$$

Note that we have to take the ideal generated by products here.

For $\generators{x} = (x)$ a principal ideal and $\generators{y}$ principal, we do have $(x)(y) = (xy)$.
Note that $IJ \subset I \intersect J$, whereas the sum was larger than $I, J$.

Exercise
: Note that $( \ell(R), \cdot)$ has an absorbing element, namely $(0) I  = (0)$.
	For $(M, +)$ a commutative monoid and $M \injects G$ a group, then multiplication by $x$ is injective and so for all $y\in M$, $xz = yz \implies x=y$, so $M$ is cancellative.

Question: what if we consider $\mathcal{I}^\bullet(R)$ the set of nonzero ideals of $R$.
Does this help?
We will see next time.