# Tuesday, March 01

:::{.remark}
Recall the definition of the localization of an $R\in \CRing^\unital$ at a submonoid $S \leq (M, \cdot)$, written $R\localize{S}$.
Similarly for $M\in \rmod$, one can form $M\localize{S}$, and $(\wait)\localize{S}$ is a functor where the induced map on $M \mapsvia{f} N$ is $f_S(m/s) \da f(m)/s$.
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:::{.proposition title="?"}
For $I\in \Id(R)$, let $j(I) \da \ts{a\in R\st a/s\in I \text{ for some } s\in S}$ which is again an ideal in $R$.
Then

1. $j(I)_S = I$,
2. $I_S = R_S \iff I$ contains an element of $S$.

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:::{.proof title="of 2"}
$\impliedby$:
$I_S \subseteq R_S$ is clear.
Let $x/t\in R_S$ and $s\in I \intersect S$, then ${sx\over st} = {x\over t}\in I_S$.

$\implies$:
Write $1=i/s$ to produce $t\in s$ with $t(s-i) = 0$.
Then $z=ts \in S$ and $z=it\in I$ so $z \in I \intersect S$.
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:::{.proposition title="?"}
Let $P\in \spec R$ with $S \intersect p = \emptyset$, then $j(P_S) = P$.
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:::{.proof title="?"}
$\supseteq$:
Clear.

$\subseteq$:
Let $a\in j(P_S)$, so $a/s=p/t$ for $s,t\in S, p\in P$ and $\exists u\in S$ such that $u(at-sp)=0\in P$, so $uat - usp\in P$ where $usp\in P$.
Thus $uat\in P \implies a(ut)\in P\implies a\in P$, since $ut\in S$ and $ut\not\in P$.
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:::{.proposition title="?"}
There is an order-preserving correspondence
\[
\ts{p\in \spec R \st p \intersect S = \emptyset} &\mapstofrom \spec R\localize{S} \\
P &\mapsto P\localize{S} \\
j(P') &\mapsfrom P'
.\]
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:::{.proof title="?"}
We need to show

1. $P\localize{S} \in \spec R\localize{S}$ is actually prime.
2. If $P'\in \spec R\localize{S}$ then $j(P')\in \spec R$ with $j(P') \intersect S = \emptyset$.

For one:
\[
{x\over t}, {y\over t} \in P_S 
&\implies {xy\over st} \in P_S \\
&\implies xy \in j(P_S) = P \\
&\implies x\in P \text{ or } y\in P \\
&\implies x/s\in P \text{ or } y/s\in P
.\]

For two:
\[
xy\in j(P') 
&\implies {xy\over s}\in P' \\
&\implies {x\over 1}{y\over s}\in P' \\
&\implies {x\over 1}\in P' \text{ or } {y\over s}\in P' \\
&\implies {x}\in P' \text{ or } {y}\in P' \\
.\]

If $x\in j(P') \intersect S$ then ${x\over t}\in P'$ so ${t\over x}{x\over t}\in P'$. $\contradiction$

One can then check that these two maps compose to the identity.
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:::{.exercise title="?"}
Show that if $p\in \spec R$ then $R_p \in \Loc\Ring$ is local.
Use that the image of $p$ in $R_p$ is $P_p = R_p\sm R_p\units$, making it maximal and unique.
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:::{.exercise title="?"}
Show that

1. $M=0 \iff M_S = 0$ for all $S$,
2. $M=0 \iff M_p = 0\, \forall p\in \mspec R$,
3. $M=0 \iff M_p = 0\, \forall p\in \spec R$, noting that this is a stronger condition than maximal.

For (2), use that $\Ann_R(x)$ is a proper ideal and thus contained in a maximal, and show by contradiction that $x/1\neq 0\in M_p$.

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:::{.exercise title="?"}
Show that if $f\in \rmod(M, N)$ then 

- $f$ injective (resp. surjective) $\implies f_S$ injective (resp. surjective)
- If $f_p$ is injective for all $p\in \spec R$, then $f$ is injective (resp. surjective)
- If $M$ is flat then $M_S$ is flat
- If $M_p$ is flat for all $p$ then $M$ is flat.

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:::{.remark}
Recall that for $A \subseteq R$, $V(A) \da \ts{p\in \spec R\st p\contains A}$.
Letting $I(A)$ be the ideal generated by $A$, then check that $V(I(A)) = V(A)$ and $V(I) = V(\sqrt I)$.
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:::{.exercise title="?"}
Check that defining closed sets as $\ts{V(A) \st A \subseteq R}$ forms the basis for a topology on $\spec R$, and $V(p) \intersect V(q) = V(pq)$.
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:::{.remark}
Next time: generic points, idempotents, irreducible sets.
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