# Tuesday, February 01

:::{.proposition title="Cayley-Hamilton"}
For $A\in \CRing, M\in \mods{A}^\fg$, $\mfa \in \Id(A)$, and $\phi: M\to \mfa M \subseteq M$, there exists $\ts{a_i}_{i\leq n} \subseteq \mfa$ such that
\[
\phi^r + a_1 \phi^{r-1} + \cdots + a_r \id = 0
.\]

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:::{.proof title="Cayley-Hamilton: Reductions"}
Reduce to showing this for $M = A^r$ a free module.
Use the diagram:

\begin{tikzcd}
	{e_i} & {A^r} && M && 0 \\
	\\
	& {\mfa^r} && {\mfa M} && 0 \\
	& {} && {a_i}
	\arrow[from=3-2, to=3-4]
	\arrow[from=3-4, to=3-6]
	\arrow[from=1-2, to=1-4]
	\arrow[from=1-4, to=1-6]
	\arrow["\phi", from=1-4, to=3-4]
	\arrow["{\tilde f}", dashed, from=1-2, to=3-2]
	\arrow[from=1-1, to=4-4]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsOSxbMSwwLCJBXnIiXSxbMywwLCJNIl0sWzUsMCwiMCJdLFszLDIsIlxcbWZhIE0iXSxbNSwyLCIwIl0sWzEsMiwiXFxtZmFeciJdLFsxLDNdLFswLDAsImVfaSJdLFszLDMsImFfaSJdLFs1LDNdLFszLDRdLFswLDFdLFsxLDJdLFsxLDMsIlxccGhpIl0sWzAsNSwiXFx0aWxkZSBmIiwwLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzcsOF1d)

Lift by sending $e_i$ to any element $a_i \in \mfa$.
Then: STS $\tilde f$ satisfies some polynomial, since $\phi$ will satisfy the same polynomial and map to zero by commutativity of the square above.
$\tilde f: A^r\to A^r$ can be written as a matrix $(a_{ij})$.
Now reduce to $\CC$: consider the map
\[
\ZZ[ \ts{ x_{ij}}_{i, j \leq r} ] &\to A \\
x_{ij} &\mapsto a_{ij}
.\]

Forming the matrix $M = (x_{ij})$ yields a commutative diagram:

\begin{tikzcd}
	{\ZZ[\ts{x_{ij}}_{i,j\leq r}]\cartpower{r}} && {A^r} \\
	\\
	{\ZZ[\ts{x_{ij}}{i,j\leq r}]\cartpower{r}} && {A^r}
	\arrow[from=1-1, to=1-3]
	\arrow["{\tilde f = (a_{ij})}", from=1-3, to=3-3]
	\arrow[from=3-1, to=3-3]
	\arrow["{M = (x_{ij})}"', from=1-1, to=3-1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXFpaW1xcdHN7eF97aWp9fV97aSxqXFxsZXEgcn1dXFxjYXJ0cG93ZXJ7cn0iXSxbMCwyLCJcXFpaW1xcdHN7eF97aWp9fXtpLGpcXGxlcSByfV1cXGNhcnRwb3dlcntyfSJdLFsyLDAsIkFeciJdLFsyLDIsIkFeciJdLFswLDJdLFsyLDMsIlxcdGlsZGUgZiA9IChhX3tpan0pIl0sWzEsM10sWzAsMSwiTSA9ICh4X3tpan0pIiwyXV0=)

Since $\ZZ[\ts{ x_{ij} }]$ is an integral domain, we can pass to the fraction field $\QQ(\ts{x_{ij}})$, where by linear algebra $M$ has a characteristic polynomial in the $x_{ij}$.
So for some homogeneous polynomials $p_i$ in the $x_{ij}$, we have
\[
\tilde f^r + p_1(x_{ij}) \tilde f^{r-1} + \cdots + p_r(x_{ij}) = 0
.\]
Now choose an arbitrary embedding $\QQ(x_{ij}) \embeds \CC$, and prove Cayley-Hamilton here using (e.g.) Jordan normal form.
:::

:::{.exercise title="Fun, everyone should have a proof!"}
Write a careful proof of Cayley-Hamilton for arbitrary fields.
:::

:::{.remark}
Useful strategy: reduce to the "universal matrix".
:::

:::{.proof title="Cayley-Hamilton for arbitrary fields, sketch"}
Recall that there exists an *adjugate* of any square matrix satisfying $M \adj(M) = \adj(M) M = \det(M) \id$.
Apply this to $M \da tI - \tilde f \in \Endo_{A[t]}(A(t)\cartpower{r})$.
Write $\adj(tI-\tilde f) = \sum B_i t^i$ with $B_i \in \Mat_{n\times n}(A)$.
We have
\[
(tI-\tilde f) \adj(tI - \tilde f) = \det(\tilde f)I
,\]
but we can't plug in $\tilde f$ here because we don't know if this lands in a *commutative* ring, so e.g. $t m t^r \neq mt^{r+1}$ doesn't necessarily hold.

Note that $B_i$ commutes with $tI - \tilde f \iff B_i$ commutes with $\tilde f$, e.g. by equating coefficients.
Write $R = Z(\tilde f)$ for the centralizer, those matrices commuting with $\tilde f$.
Then $(tI - \tilde f) \in R[t]$, which reduces us to the world of commutative rings.
Then $\ro{ \det(tI - \tilde f) }{\tilde f} = (\tilde f - \tilde f)\cdot g = 0$.
:::

:::{.exercise title="?"}
Show that if $M\in \mods{A}^\fg$ and $\mfa\in \Id(A)$,
\[
M = \mfa M \implies \exists x\cong 1\mods \mfa \text{ such that } xM = 0
.\]

> Hint: apply Cayley-Hamilton to $\id_M$.

:::


:::{.theorem title="Nakayama"}
For $M\in \mods{A}^\fg, I \in \Id(A)$ with $I \in \jacobsonrad{A}$,
\[
M = IM \implies M = 0
.\]

:::


:::{.remark}
Apply the corollary to get $x\equiv 1 \mod I$ with $xM = 0$.
But then $x \equiv 1 \mod \jacobsonrad{A}$, so $x$ is a unique and $xM = M$.

Alternatively, pick a minimal set of generators $\ts{x_i}$ of $M$, so $m = \sum a_i x_i$ with $a_i\in I$ since $M=IM$.
Since $1-a_n\in \jacobsonrad{A}$ and is a unit, so
\[
(1-a_n)x_n = \sum_{i\leq n-1} a_i x_i \implies x_n = (1-a_n)\inv \sum_{i\leq n-1}a_i x_i
.\]
$\contradiction$
:::


:::{.remark}
Notes:

- Proved last time: $A\in \Loc\Ring, M \in \mods{A}^\fg$, if $X\da \ts{x_i} \subseteq M$ with $\ts{\bar{x_i}}$ generating $M/\mfm M$, then $X$ generates $M$.
- Suppose $f\in \mods{A}(M, N)$ with $\bar{f}: M/\mfm_A M \to N/\mfm_A N$ an isomorphism -- $f$ is not necessarily an isomorphism.
  - Counterexample: $k[[x]]\to k$ is not an isomorphism in $\mods{k}$ but reduces mod $x$ to $k \mapsvia{\id} k$.

- Show that $f$ need not be injective, but is always surjective.
  - For surjectivity: use $M \mapsvia{f} N \to C \to 0$, use that $C/\mfm C =0$ to conclude $C=0$ by Nakayama.
  - For injectivity: use $0\to K\to M \mapsvia{f} N \to 0$ and try to show $K=0$.
    Not true, take $0\to \gens{x} \to k[[x]] \to k\to 0$ and apply $(\wait)\tensor_{k[[x]]} k$ to get $k \mapsvia{0} k \mapsvia{\id} k\to 0$.
- The special SES $0\to M \to M \oplus N \to N \to 0$ has a left-section $s: M \oplus N\to M$; applying $(\wait)\tensor_A S$ or $\Hom_A(S, \wait)$ actually produces a SES, since this induces left sections on the resulting sequences.
- Prove that free modules are projective.
- Prove that divisible abelian groups are injective in $\mods{\ZZ}$.
  - Prove that $\QQ$ and $\QQ/\ZZ$ are injective.
- Show that a SES splits iff it admits a right section or a left section.
- Show that $0\to I\to M\to P\to 0$ splits if either $P$ is projective or $I$ is injective.
- Show that $(\wait)\tensor_A S, \Hom_A(\wait, S), \Hom_A(S, \wait)$ are exact on SES's $0\to A\to B \to P\to 0$ with $P$ projective.

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