# Friday January 24th
## Ideals and Products
Recall:
Prime and maximal ideals.
***Fact:***
If $I \normal R$ then there exists a maximal ideal $I \subset \mfm \normal R$.
Proof
: Use Zorn's lemma.
Corollary
: $\maxspec R \neq \emptyset \iff R \neq 0$.
Later:
Multiplicative avoidance, if $S \subset R$ is nonempty with $SS\subset S$, let $I \normal R$ with $I\intersect S = \emptyset$, then
a. There exists an ideal $J \supseteq I$ with $J \intersect S =\emptyset$ which is maximal with respect to being disjoint from $S$.
b. Any such ideal $J$ is prime.
Taking $S = \theset{1}$ recovers the previous fact.
Exercise
: Let $f: R\to S$ be a ring homomorphism and $\pr \in \spec(R)$.
Show that $f_*(\pr)$ need not be prime in $S$.
We can consider products of rings, and correspondingly $\mci(R_1 \cross R_2)$.
Exercise
: Show that if $\phi$ is surjective, $\phi(I)$ is an ideal.
Proposition (Ideals of Products are Products of Ideals)
: Let $I \in \mci(R_1 \cross R_2)$.
Take $\pi_i \to R_i$ the projections, and let $I_i$ be the corresponding images of $I$.
Then $I = I_1 \cross I_2$.
Note: a suspiciously strong result! Not every group is the cartesian product of some subgroups.
It's clear that $I \subset I_1 \cross I_2$.
\
Proof
: Showing $I_1 \cross I_2 \normal R_1 \cross R_2$ is an ideal, since it equals $\generators{ I_1 \cross \theset{0}, \theset{0} \cross I_2 }$.
To show $I_1 \cross I_2 \subseteq I$, show that $I_1 \cross 0, 0 \cross I_2 \subseteq I$.
E.g. $I_1 \cross 0 \subseteq I$: take $(x, 0) \in I_1 \cross 0$ such that there exists a $y\in R_2$ with $(x, y) \in I$.
Then $(x, y) \cdot (1, 0) = (x, 0) \in I$, then similarly $0 \cross I_s \subseteq I$.
Exercise
: Use $\mci(R_1 \cross R_2) = \mci(R_1) \cross \mci(R_2)$ to describe $\spec(R_1 \cross R_2)$ in terms of $\spec(R_1)$ and $\spec(R_2)$.
*Question:*
For a ring $R$, when is $R \cong R_1 \cross R_2$ for some nonzero $R_1, R_2$?
Exercise
: Show that comaximal ideals correspond with coprime ideals when $R = \ZZ$.
Theorem (Chinese Remainder)
: If $I_1, I_2$ are comaximal, so $I_1 + I_2 = R$, then the map
\begin{align*}
\Phi: R \to R/I_1 \cross R/I_2 \\
x \mapsto (x+ I_1, x+ I_2)
.\end{align*}
Then $\ker \Phi = I_1 \intersect I_2 \equalsbecause{CRT} I_1 I_2$ and $\Phi$ is surjective, and
\begin{align*}
R/(I_1 \intersect I_2) = R/I_1 I_2 \cong R/I_1 \cross R/I_2
.\end{align*}
Proof
: **Case 1:**
Let $I_1 + I_2 = R$ and $I_1 \intersect I_2 = 0$ (equivalently $I_1 I_2 = (0)$), then $R \cong R/I_1 \cross R/I_2$.
\
Conversely, let $R = R_1 \cross R_2$ with $R_1, R_2$ nonzero.
Let $e_1 = (1, 0)$ and $e_2 = (0, 1)$.
Then $e_1 e_2 = 0$ and $e_2 = (1 - e_1)$, so $0 = e_1(1 - e_1) = e_1 - e_1^2$ and $e_1$ is idempotent.
\
So $e_1, e_2$ are complementary nontrivial idempotents.
Then $I_1 I_2 = e_1 e_2 = (0)$, $I_1 + I_2 = \generators{e_1, e_2} = R$, and thus $R = R/e_2R \cross R/e_1 R$.
Note that $e_2 R = 0 \cross R_2$ and $e_1 R = R_1 \cross 0$, thus
\begin{align*}
R/e_2 R = \frac{R_1 \cross R_2}{0 \cross R_2} = R_1 \\
R/e_1 R = \frac{R_1 \cross R_2}{R_1 \cross 0} = R_2
.\end{align*}
We thus have a correspondence
\begin{align*}
\correspond{\text{Nontrivial product decompositions }R = R_1 \cross R_2}
&\iff
\correspond{I_1, I_2 \normal R \text{ such that } I_1 I_2 = 0 \text{ and } I_1 + I_2 = R} \\
&\iff
\correspond{\text{Idempotents } e \neq 0, 1}
.\end{align*}
Thus a ring can be decomposed as a product iff it contains nontrivial idempotents.
Definition (Connected Rings)
: $R$ is connected iff there do not exists nonzero $R_1, R_2$ such that $R \cong R_1 \cross R_2$ iff $R$ does not contain an idempotents $e\neq 0, 1$.
Exercise
: Show that $R$ is connected iff $\spec(R)$ is connected as a topological space.
Note:
Not every ring is a finite product of connected rings.
## Modules
For $(M, +)$ a commutative group, we want an action $R\actson M$ for $R$ a ring.
Recall that $\endo(M)$ for a group is a (potentially noncommutative) ring.
An $R\dash$module structure is a ring homomorphism $R \to \endo(M)$.
Equivalently, it is a function $R\cross M \to M$ with $rs(x) = r(sx)$, $r(x+y) = rx + ry$, and $1\cdot x = x$ for all $x\in M$.
> Note that this defines a left $R\dash$module, but right/left modules coincide for commutative rings.
Exercise
: Let $M$ be an $R\dash$module and for $m\in M$ define $$\ann(m) = \theset{r\in R \suchthat xm = 0} \normal R,$$ show this is in fact an ideal.
Note: skipped chapter on Galois connections, i.e. some binary relation on a pair of sets. This is an instance of such a connection, where $x\sim m \iff xm = 0$.
Exercise
: For any subset $S \subset M$, define
$$
\ann(S) \definedas \theset{x\in R\suchthat xm = 0 ~\forall m\in S}
.$$
Show that $\ann(S) = \intersect_{m\in S} \ann(m)$ and
$$
\ann(M) = \theset{x\in R \suchthat xM = 0} = \ker(R \to \endo(M))
.$$
Definition (Faithful Modules)
: $M$ is **faithful** iff $\ann(M) = 0$ iff $R \injects \endo(M)$ is an injection.
Exercise
: Any $M$ is naturally a faithful $R/\ann(M) \dash$module.