# Thursday, April 14

:::{.remark}
Recall that if $A\in \kalg$, the algebraic differentials are constructed as 
\[
\Omega_{A\slice k} = \Free\ts{dA\st a\in A}/ \gens{d(ab) = ad(b) + d(a) b}
.\]

Note that $\Hom_{\amod}( \Omega_{A\slice k}, M) \cong \Der_{\kmod}(A, M)$, where derivations are importantly not $A\dash$linear!
This is meant to emulate a cotangent bundle.

There is a SES
\[
0\to I\to A\tensor_k A \mapsvia{m} A\to 0
.\]
:::

:::{.example title="?"}
For $A=k[x]$, check that $x,y\mapsto x$ and $I = \gens{x-y}$, and moreover $\Omega_{k[x]\slice k} \cong I/I^2 = \gens{x-y}/\gens{x^2+xy+y^2}$.
:::

:::{.slogan}
The cotangent bundle is the conormal bundles of the diagonal:

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\end{tikzpicture}

:::

:::{.exercise title="?"}
Show that $\Omega_{A\slice k} \cong I/I^2$.
:::

:::{.solution}
Hints: a map \( \Omega_{A\slice k} \mapsvia{f} I/I^2 \) is equivalent to a derivation \( A\to I/I^2 \).
Show that if $\psi(a) \da a\tensor 1 - 1\tensor a\in I$, then $m(\psi(a)) = 0$ and $\psi$ is a derivation, i.e. $\psi(ab) - a\psi(b) - \psi(b)a \in I^2$.
Check
\[
ab \otimes 1 -1 \otimes ab - a(b \otimes 1 - 1 \otimes b) - b(a \otimes 1 - 1 \otimes a)
&= ab \otimes 1 - 1 \otimes ab - ab \otimes 1 + a \otimes b - ba \otimes 1 + b \otimes a \\
&= -1 \otimes ab + a \otimes b - ab \otimes 1 + b \otimes a \\
&= -(a \otimes 1 - 1 \otimes a)(b \otimes 1 - 1 \otimes b) \in I^2
.\]
For an inverse, take 
\[
g: I/I^2 &\to \Omega_{A\slice k} \\
\sum a_i \otimes b_i &\mapsto \sum b_i d(a_i)
\]
where $\sum a_i b_i = 0$ and check 

- $g(I)^2=0$
- $f \circ g = \id, g \circ f = \id$.

:::

:::{.exercise title="?"}
Show that $(1 \circ x - x \circ 1) \to d1 + 1dx$ where $x-y = -\dx$.
:::

:::{.exercise title="General algebra"}
Show that $M\to N\to L\to 0\in \amod$ is exact iff $0\to [L, S]_A \to [N, S]_A \to [M, S]_A$ is exact for all $S\in \amod$.
:::

:::{.solution}
Hint: $\impliedby$ is the nontrivial direction. Toward a contradiction take $S\da \coker(A\to L)$.
:::

:::{.exercise title="?"}
Show that if $A,B\in \kalg$ and $0\to I\to A \mapsvia{f} B\to 0$ is exact then there is an exact sequence

\begin{tikzcd}
	{I/I^2} && {\Omega_{A/\slice k}} && {\Omega_{B/\slice k}} && 0 && {\in \mods{B}} \\
	x && dx \\
	&& da && {d(f(a))}
	\arrow[from=1-1, to=1-3]
	\arrow[from=1-3, to=1-5]
	\arrow[from=1-5, to=1-7]
	\arrow[maps to, from=2-1, to=2-3]
	\arrow[maps to, from=3-3, to=3-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsOSxbMCwwLCJJL0leMiJdLFsyLDAsIlxcT21lZ2Ffe0EvXFxzbGljZSBrfSJdLFs0LDAsIlxcT21lZ2Ffe0IvXFxzbGljZSBrfSJdLFs2LDAsIjAiXSxbOCwwLCJcXGluIFxcbW9kc3tBfSJdLFswLDEsIngiXSxbMiwxLCJkeCJdLFsyLDIsImRhIl0sWzQsMiwiZChmKGEpKSJdLFswLDFdLFsxLDJdLFsyLDNdLFs1LDYsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Im1hcHMgdG8ifX19XSxbNyw4LCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJtYXBzIHRvIn19fV1d)

Idea: $\spec B\injects \spec A$ is like an embedded submanifold, and $I/I^2$ is the conormal bundle.

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\end{tikzpicture}

:::

:::{.solution}
Identify

\begin{tikzcd}
	0 && {[\Omega_{B\slice k}, S]_B} && {[\Omega_{A\slice k}\tensor_A B, S]_B \cong [\Omega_{A\slice k}, S]_A} && {[I/I^2, S]_B} \\
	\\
	0 && {\Der_k[B,S]} && {\Der_k[A,S]} && {[I/I^2, S]_B}
	\arrow["\cong", from=1-3, to=3-3]
	\arrow["\cong", from=1-5, to=3-5]
	\arrow["\cong", from=1-7, to=3-7]
	\arrow[from=3-3, to=3-5]
	\arrow[from=3-5, to=3-7]
	\arrow[from=1-3, to=1-5]
	\arrow[from=1-5, to=1-7]
	\arrow["{?}"{description}, from=1-1, to=1-3]
	\arrow["\therefore", from=3-1, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbMCwwLCIwIl0sWzIsMCwiW1xcT21lZ2Ffe0JcXHNsaWNlIGt9LCBTXV9CIl0sWzQsMCwiW1xcT21lZ2Ffe0FcXHNsaWNlIGt9XFx0ZW5zb3JfQSBCLCBTXV9CIFxcY29uZyBbXFxPbWVnYV97QVxcc2xpY2Uga30sIFNdX0EiXSxbNiwwLCJbSS9JXjIsIFNdX0IiXSxbMiwyLCJcXERlcl9rW0IsU10iXSxbNCwyLCJcXERlcl9rW0EsU10iXSxbNiwyLCJbSS9JXjIsIFNdX0IiXSxbMCwyLCIwIl0sWzEsNCwiXFxjb25nIl0sWzIsNSwiXFxjb25nIl0sWzMsNiwiXFxjb25nIl0sWzQsNV0sWzUsNl0sWzEsMl0sWzIsM10sWzAsMSwiPyIsMV0sWzcsNCwiXFx0aGVyZWZvcmUiXV0=)


Check:

- $\psi \in \Der_k(B, S)$ with $\psi \circ f$ surjective implies $\psi = 0$ when $f$ is surjective.
- $\psi \in \Der_k(A, S)$ with $\rho\psi{I} = 0$ implies $\psi\in \im(\Der_K(B, S) \to \Der_k(A, S))$

:::

:::{.definition title="?"}
For $f\in \CRing(A, B)$, $\Omega_{B/A}$ is the unique object in $\mods{B}$ such that 
\[
[\Omega_{B/A}, M] = \Der_A(B, M)
.\]
Explicitly, 
\[
\Omega_{B/A} = \Free\ts{b\in B}/\gens{d(b_1 b_2) = b_1 d(b_2) + d(b_1) b_2, da \st a\in A, b_i\in B }
.\]
:::

:::{.exercise title="?"}
Show that for $A,B\in \kalg$, there is a SES
\[
\Omega_{A\slice{k}}\tensor_{A} B \to \Omega_{B\slice{k}} \to \Omega_{B/A} \to 0 \qquad \in \mods{B}
.\]
:::

:::{.remark}
If $A\in \kalg^\fg$ and $k=\kbar$ and $\mfm\in \mspec A$, then $0\to \mfm \to A\to k \to 0$ (by EEKS) and there is a SES
\[
0\to \mfm/\mfm^2\to \Omega_{A\slice{k}} \tensor_A k\to \Omega_{k\slice{k}} = 0\to 0
.\]
:::

:::{.definition title="Tangent/cotangent spaces"}
Define $\T\dual A \da \mfm/\mfm^2$ and $\T A \da (\mfm/\mfm^2)\dual \da [\mfm/\mfm^2, k]_{k}$.
:::

:::{.exercise title="?"}
Let $A = k[x]$ and $\mfm=\gens{x}$, then for $f\in \mfm$ define $\tau\in \T\dual A$, so $\tau(\bar f)\in k$, by $\tau = \dd{}{x}$.
Then $\bar{f}\to f'(0)$.

Similarly for $A=\kxn$ and $\mfm = \gens{x_1,\cdots, x_n}$, check $\T_0 = \spanof_\CC\gens{\dd{}{x_1}, \cdots, \dd{}{x_n}}$.
:::

:::{.theorem title="?"}
For $k=\kbar$ and $A\in \kalg^\fg$ a domain,

- For any $m\in \mspec A$,
\[
\dim_k \Omega_{A\slice k}\tensor_k A/m = \dim_k m/m^2 \geq \dim A
.\]

- There exists a nonempty open $U \subseteq \spec A$ such that for some $m\in U$ this is an equality, so
\[
\dim_k m/m^2 = \dim_{\ff(A)} \ff(A) \tensor_A \Omega_{A\slice k}
.\]


:::

:::{.remark}
What goes into a proof: 

- Find $\kxn \injects A$ with $n=\dim A$ making $A\in \mods{\kxn}^\fg$.
- Check $\Omega_{A/\kxn}$ is torsion.

:::


:::{.definition title="Smooth points"}
A point $m\in \mspec A$ is **smooth** if
\[
\rank \T\dual A = \dim A
.\]

:::


:::{.example title="?"}
A non-example: $\krulldim A = 0$ for $A=k[x]/x^2$, since $A$ is Artin.
Check $\Omega_{A\slice k} = A\dx/\gens{d(x)^2=0} \cong {k[x]/x^2 \dx \over 2x\dx }$, which is $k\dx$ if $\chr k \neq 2$ and $k[x]/x^2 \dx$ if $\chr k = 2$.
:::


:::{.example title="?"}
For $A= k[x,y]/\gens{y^2-x^3}$ with $\chr k = 0$, check
\[
\Omega_{A\slice k} = {A\dx \oplus A\dy \over d(y^2-x^3)} = {A\dx \oplus A\dy\over dy\dy - 3x^2\dx }
.\]

Given $m\in \mspec A$, 
\[
\dim_{A/m} \Omega_{A/k} \tensor_A A/m = 
\begin{cases}
1 & m\not\in \gens{x,y} 
\\
2 & m\in \gens{x,y} .
\end{cases}
\]

Note that this records the nodal singularity:

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\end{tikzpicture}


:::


:::{.example title="?"}
Over $k = \FF_p\functionfield{t}$ and $A = \FF_p\functionfield{t^{1\over p}} = k[x]/\gens{x^p-t}$, check $\dim A = 0$ and
\[
\Omega_{A\slice k} = {A\dx \over d(x^p - t)} = {A\dx \over d(x^p)} = A \dx
.\]

So the algebraic differentials detect when a field fails to be separable.
:::