# Thursday, February 17

:::{.exercise title="?"}
Prove the correspondence theorem between $\Id(A)$ and $\Id(S\inv A)$.

\begin{tikzcd}
	A && {S\inv A} \\
	\\
	{\iota\inv(I)} && I
	\arrow["\iota", from=1-1, to=1-3]
	\arrow[hook, from=3-3, to=1-3]
	\arrow[from=3-1, to=3-3]
	\arrow[from=3-1, to=1-1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJBIl0sWzIsMCwiU1xcaW52IEEiXSxbMCwyLCJcXGlvdGFcXGludihJKSJdLFsyLDIsIkkiXSxbMCwxLCJcXGlvdGEiXSxbMywxLCIiLDIseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFsyLDNdLFsyLDBdXQ==)

Use that

- $\iota\inv I\in \Id(A)$
- $\iota(\iota\inv (I))$ generates $I$.
- ${a\over s} = {1\over s}{a\over 1}$ and ${a\over 1}$ is in the image of the lower map.

:::

:::{.exercise title="?"}
Show that the localization of any Noetherian ring is again Noetherian.
:::

:::{.exercise title="?"}
Say $M\in \rmod^\fp$ iff there is an exact sequence $R^a\to R^b\to M\to 0$.
Show that if $R$ is Noetherian, $\rmod^\fp = \rmod^{\fg}$.
:::

:::{.definition title="Integrality"}
For $f\in \CRing(A, B)$ and $b\in B$ define
\[
f_b: A[x] &\to B \\
x&\mapsto b \\
\sum a_i x^j &\mapsto \sum f(a_i) b^i
.\]
The element $b$ is **integral over $A$** iff $\im f_b \in \mods{A}^\fg$.
The ring $B$ is **integral over $A$** iff every $b\in B$ is integral over $A$ as above.
:::

:::{.exercise title="?"}
Show that $\ZZ\invert{2} \subseteq \QQ$ is not integral.
:::

:::{.exercise title="?"}
Show TFAE:

- $b$ is integral over $A$
- $b$ satisfies a monic polynomial over $A$, so $b^n + \sum_{i = 1}^n a_i b^{n-i} = 0$.

:::

:::{.solution}
Hints:

- Write $\im f_b = I\da \gens{1,b,\cdots, b^{n-1}}$, suppose $b^n\in I$, and show $b^{n+1} \in I$ by expanding as a sum.
- Set $M_i = \gens{1,b, \cdots, b^i}$ and use that these stabilize to conclude $b^n \in M_{n-1}$ for some $n$.

:::

:::{.warnings}
$B$ Noetherian over $A$ does not imply $B$ is Noetherian!
Consider $\bar{\ZZ}$, the algebraic integers, which is integral over $\ZZ$.
Note that $\gens{2^{1\over n}}$ is not finitely-generated.
:::

# Toward the Nullstellensatz

:::{.remark}
Idea: there is a dictionary between $\kxn$ and $\AA^n\slice k$ for $k=\kbar$:

- Points in $\AA^n$ correspond to maps $\kxn\to k$ given by $x_i\mapsto c_i\in k$.
- Rings $R$ correspond to their prime spectra $\spec R$.
- Kernels correspond to maximal ideals
- Ideals $I\in \Id(R)$ correspond to $V(I) =\ts{\mfm \st I \subseteq \mfm}$
- Ideals $I\in \Id(\kxn)$ correspond to $V(I) = \ts{\vector c\in k\cartpower{n} \st f(\vector c) = 0 \, \forall f\in I}$. 

:::

:::{.theorem title="Nullstellensatz V1"}
Any $\mfm \in \mspec \kxn$ is given by $\ker \pi_p$ for some morphism
\[
\pi_p: \kxn \to k \\
x_i &\mapsto c_i
\]
for some $p = \tv{c_1,\cdots, c_n} \in k\cartpower{n}$.
Equivalently, $\mfm = \gens{x_1-c_1,\cdots, x_n - c_n}$.
:::

:::{.theorem title="Nullstellensatz"}
If $f\in \kxn$ satisfies $\ro{f}{V(I)} = 0$ for a fixed $I\in \Id(\kxn)$, then $f^n\in I$ for some $n>0$.
I.e. there is a bijective correspondence $I\mapstofrom V(I)$ for radical ideals $\sqrt{I} = I$.
:::

:::{.example title="?"}
Necessity of conditions:

- Why $k=\kbar$ is needed: take $\gens{x^-2}\in \QQ[x]$, this is a maximal ideal but $V(I) = 0$.

- Why the radical condition is needed: take $I = \gens{x^{10} }$ so $V(I) = \ts{0} \subseteq k$, but $\ro{x}{V(I)} = 0$.

:::

:::{.corollary title="?"}
For $k=\kbar$ and $R\in \kalg^\fg$,
\[
\mspec R \mapstofrom \Hom_{\kalg}(R, k)
.\]
:::

:::{.corollary title="?"}
$V(f) = k\cartpower{n} = V(0)$ iff $f\in \nilrad{R}$.
:::

:::{.corollary title="?"}
Let $R\in \kalg^\fg$, so $R = \kxn/I$ for some $n$, and let $V(I) \subseteq k\cartpower{n}$.
Given $J \subseteq R$ radical, $V(J) \subseteq V(I)$, and $f\in R$ vanishes on $V(J)$ iff $f^n\in J$ for some $n>0$.
:::

:::{.theorem title="Maximal idealansatz"}
For $k=\kbar$ and $A\in \kalg^\fg$ with $\mfm \in \mspec A$, 
\[
A/\mfm \cong k
.\]
:::

:::{.theorem title="Maximal idealansatz 2"}
For $k$ an arbitrary field and $A\in \kalg^\fg$ with $\mfm \in \mspec A$, 
\[
A/\mfm
\text{ is a finite extension of } k
.\]
:::

:::{.theorem title="EEKS / proto Zariski's lemma"}
If $k \subseteq F$ fields with $F\in \kalg^\fg$, then $F\slice k$ is a finite extension of fields.
:::

:::{.corollary title="?"}
If $R\in \zalg^\fg$ and $\mfm \in \mspec R$, then $R/\mfm$ is finite.
:::

:::{.proof title="?"}
Check $\mfm \intersect \ZZ = \mfp \in \spec \ZZ$ and $R/\mfm$ is a finitely generated extension of $\FF_p$ and hence finite.
:::

:::{.lemma title="Zariski"}
If $A\in \CRing^\Noeth$ and $A \subseteq B \subseteq C$ with

- $C\in \algs{A}^\fg$ and
- $C\in \mods{B}^\fg$,

then $B\in \algs{A}^\fg$.
:::

:::{.proof title="?"}
Sketch:

- Choose $x_1,\cdots, x_n$ generating $C$ as an $A\dash$algebra
- Choose $y_1,\cdots, y_m$ generating $C$ as a $B\dash$module.
- Write $x_i = \sum_j b_{ij} y_j$ with $b_{ij}\in B$
- Write $x_i x_j = \sum_k b_{ijk} y_k$ with $b_{ijk}\in B$
- Let $B_0 \subseteq B$ be the $A\dash$algebra generated by the $b_{ij}$ and $b_{ijk}$.
- Observe that $B_0$ is Noetherian by the Hilbert basis theorem since it's finitely generated as an $A\dash$algebra.
- For any $c\in C$, write $c = \sum_i b_i y_i$ since the $y_i$ generate $C$ as a $B_0$ module.
- Idea: can rewrite in terms of lower degree monomials?
- Since each $b_i = \sum a_{I}^i x^I$, we have $c = \sum_i \sum_I a_I^i x^I y_i$, which is a polynomial in $\sum b_{ij} y_j$.
- Then $y_{i} \cdots y_{i_k} = \sum p_s y_s$ where the $p_s$ are polynomials in the $b_{ijk}$?
- Since $B_0$ is Noetherian and $B \subseteq C$, $B$ is finitely generated as a $B_0$ module and thus as a $B_0$ algebra.
- Since $A \subseteq B_0 \subseteq B$ and $B_0$ is finitely-generated as an $A$ algebra and $B$ is finitely-generated as a $B_0$ algebra, we have that $B$ is finitely-generated as an $A$ algebra.

:::