# Thursday, January 27

:::{.remark}
Nakayama: called a lemma, but arguably the most important statement in commutative algebra!
Recall that $\jacobsonrad{A} = \intersect_{\mfm\in \mspec A} \mfm$, and being zero mod every $\mfm\in \mspec A$ means being zero mod $\jacobsonrad{A}$.
:::

:::{.theorem title="Nakayama's Lemma"}
Let $A \in \CRing$ with $I \subseteq \jacobsonrad{I}$ and let $M\in \mods{A}^\fg$.
Then
\[
M=IM \implies M=0
.\]

:::

:::{.example title="?"}
This reduces statements about local rings to statements about fields.
Commonly used examples:

- $I = \nilrad{R}$.
- $A\in \Loc\Ring$ with $I = \mfm_A$ its maximal ideal.

:::

:::{.corollary title="?"}
If $A$ is local and $M\in \mods{A}^\fg$,
\[
M/\mfm_A M = 0 \iff M = 0
.\]
:::

:::{.remark}
$M$ yields a sheaf (or vector bundle) on $\spec A$, so think of $m\in M$ as a function on $\spec A$ in the following way: if $m\in M$,
\[
m: \mfp \mapsto m \mod \mfp \in M/\mfp M
.\]

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![](figures/2022-01-27_12-57-28.png)

:::

:::{.proposition title="Equivalent formulation of Nakayama"}
If $n\in N\in \mods{A}^\fg$ and $n\equiv 0 \mod \mfm$ for all $\mfm\in \mspec A$, then $M=0$.
:::

:::{.exercise title="?"}
Show that if $A \in \Loc\Ring$ and $M\in \mods{A}^\fg$ with $\ts{x_i}_{i\leq n} \subseteq M$,
then
\[
\gens{x_1,\cdots, x_n} = M \iff \gens{\bar{x_1}, \cdots, \bar{x_n}} = M/\mfm M, \qquad \bar{x_i} \da x_i \mod \mfm
.\]
:::

:::{.solution}
$\implies$:
If $\bar y\in M/\mfm M$, lift to $y\in M$ to write $y = \sum c_i x_i$ and thus $\bar y = \sum c_i \bar{x_i}$.

$\impliedby$:
Present $M$ by $A^n \mapsvia{f} M \to C =\coker f \to 0$ where $f(e_i) = x_i$ -- we want to show $C = 0$.
Reduce mod $\mfm$ by applying $(\wait)\tensor_A A/\mfm$ to get 
\[
(A/\mfm)^n \mapsvia{\bar f} M/\mfm M \to C/\mfm C\to 0
.\]
This is surjective so $C/\mfm C = 0$.
By Nakayama, $C=0$.

:::

:::{.example title="?"}
Let the local ring $R = k\powerseries{t}$ with $\mfm_R = \gens{t}$, and let $M = R\sumpower{3}$.
Consider
\[
\tv{1+t^3 + t^5, t+t^2, t^3}, \tv{t^{22}, 1+t^9, t^{10} + t^{10^{10}}}, \tv{1+ t^{10^{10^{10}}}, 1+t^5 + t^9, 1+t^7 + t^{11} }
,\]
which reduced $\mod t$ yields
\[
\tv{1,0,0}, \tv{0,1,0}, \tv{1,1,1}
.\]
So the original elements generate $M$.
:::

:::{.remark}
Next goal: proving Nakayama. 
We'll need a version of Cayley-Hamilton.
Recall the definition of multiplicative subsets and localization, along with its universal property.

:::

:::{.example title="of multiplicative sets $S$"}
Examples:

- For any element $f$, $S\da \ts{1, f, f^2,\cdots}$
- For $A$ an integral domain, $S \da A\smz$
- All nonzero zero divisors.

:::

:::{.exercise title="?"}
Prove $S\inv A$ exists and is unique for $A\in \CRing$.
Give an example where $s'a = sb$ is not sufficient.
:::

:::{.remark}

Remarks on localization:

- ${a\over s} = {b\over s'} \iff s'' s' a - s'' s b$ for some $s''\in S$ -- this is needed because it will hold in $S\inv A$, since $s''\in (S\inv A)\units$ for any $s''\in S$ by construction.
- ${a\over s}=0$ iff $a$ is annihilated by an element of $S$.
- Producing the actual map for the universal property: if $f:A\to B$ sends $S$ to invertible elements,

\begin{tikzcd}
	A && B & {f(a)f(s)\inv} \\
	\\
	{S\inv A} \\
	{{a\over s}}
	\arrow["\iota", from=1-1, to=3-1]
	\arrow["f", from=1-1, to=1-3]
	\arrow[from=3-1, to=1-3]
	\arrow[dashed, maps to, from=4-1, to=1-4]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMCwwLCJBIl0sWzIsMCwiQiJdLFswLDIsIlNcXGludiBBIl0sWzAsMywie2FcXG92ZXIgc30iXSxbMywwLCJmKGEpZihzKVxcaW52Il0sWzAsMiwiXFxpb3RhIl0sWzAsMSwiZiJdLFsyLDFdLFszLDQsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Im1hcHMgdG8ifSwiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d)

- $A_f \da S\inv A$ for $S \da \ts{1,f,f^2,\cdots}$.
  For $A = \ZZ$ and $f=p$ a prime, $\ZZ_f = \ZZ\invert{p} \subseteq \QQ$ are fractions whose denominator is a power of $p$.
- For $A$ an integral domain and $S=A\smz$ yields $S\inv A = \ff(A)$ the fraction field.
- For $\mfp \in \spec A$, $S\da A\sm\mfp$, then $A_\mfp \da S\inv A$ is localization at a prime ideal
- $\ZZ_{\gens{p}} = \ZZ\invert{\ell}_{\ell\neq p \text{ prime}}$, fractions in $\QQ$ with denominators not divisible by $p$.
:::

:::{.exercise title="?"}
Some exercises:

- Use the universal property to show $(\ZZ/15\ZZ)_5 \cong \ZZ/3\ZZ$.
- Show that for $M\in \mods{A}$, $S\inv M$ exists and is unique.
- Show that $S\inv A \actson S\inv M$.
- Show that $S\inv M = M\tensor_A S\inv A$ using the universal property. 
  Use 
  \[
  M = M\tensor_A A \mapsvia{\id \tensor S\inv(\wait)} M\tensor_A S\inv A
  \]
  where $m\mapsto m\tensor 1$
  For $f:M\to N$ where $s$ acts invertibly on $N$, produce a map $M\times S\inv A\to N$ where $(m, a/s)\mapsto as\inv f(m)$ where $s\inv$ is the inverse of the action $s: N\to N$.
- Show that $(\wait)\tensor_A S\inv A$ is left exact and thus exact.
  - Injectivity: use that ${m\over s}\mapsto 0 \iff s'f(m) = 0$ for some $s'\in S$.
- Show that $S\inv A \in \mods{A}^\flat$.
:::