# Tuesday, April 05

## Hilbert dimension

:::{.remark}
Preliminary definitions of dimension:

- For $M\in \amod^\fg$, find a *good filtration* $\ts{\Fil_i M}$ where $\dim_k M_i$ is eventually a polynomial $p$ in $i$, so define $\dim M \da \deg p$.
- For $A\slice k$ a finitely-generated domain, define $\dim A \da \trdeg_k A$ as the transcendence degree.
- Krull dimension: $\dim A = n$ iff the longest chain of prime ideals $\mfp_0 \subsetneq \mfp_1 \subsetneq \cdots \subsetneq \mfp_n$ (note that $n$ is the number of *inclusions*!)

When they're all defined, they all agree.
:::

:::{.definition title="Good filtrations"}
Let $A$ be a filtered ring and $M\in \amod^\fg$.
A filtration $\complex\Fil M$ is a **good filtration** iff 

- The pair $(M, \complex \Fil M)$ is a filtered module, i.e. $A_i M_j \subseteq M_{i+j}$,
- The more contentful condition: $\gr(M) \in \mods{\gr(A)}^\fg$.

:::

:::{.exercise title="?"}
Show that every such $M$ admits a good filtration.
:::

:::{.solution}
Use that $A^n \mapsvia{f}  M$ and $A^n$ has a good filtration, and take its image.
Show that $M_i \da f(A_i^n)$ is good using that $\gr(A)^n \surjects \gr(M)$, which generally won't stay surjective but will in this case because $M$ receives the induced filtration.
:::

:::{.theorem title="Artin-Rees lemma (extremely important for any arguments involving filtrations!!)"}
Suppose $\gr A$ is Noetherian and let $(M, M_i)$ be an $A\dash$module with a good filtration and let $N\leq M$.
Then the induced filtration $N_i \da N \intersect M_i$ is a good filtration.
:::

:::{.proof title="?"}
Since $N\injects M$ and we're taking an induced filtration, $\gr(N) \injects \gr(M)$ remains injective.
Since $\gr(M)$ is finitely-generated over $\gr(A)$ which is Noetherian, $\gr(N)$ is finitely-generated.
:::

:::{.theorem title="?"}
Let $A$ be a filtered ring, $M\in \amod^\fg$, and let $\complex F, \complex G$ be two good filtrations on $M$.
Then there exists a $k$ such that
\[
F_{i-k} M \subseteq G_i M \subseteq F_{i+k} M
.\]
:::

:::{.remark}
There is a notion of a topology on $M$ induced by a filtration, and this theorem says all good filtrations induce the same topology.
We'll need the following to prove this theorem:
:::

:::{.lemma title="?"}
Let $(M, F)$ be a module with a good filtration.
Then there exists some $n$ and $i_0$ such that 
\[
i\geq i_0 \implies F_{i+n} \subseteq A_{i+i_0} F_n
.\]
:::

:::{.proof title="of theorem, assuming lemma"}
Choose $n$ as in the lemma, choose $m$ such that $F_n \subseteq G_{n+m}$, then
\[
F_{i+n} \subseteq A_{i+i_0} F_n \subseteq A_{i+ i_0} G_{n+m } \subseteq G_{i+n+m}
.\]
Now run the same argument on $G$.
:::

:::{.proof title="of lemma"}
Since $\gr(M)$ is finitely-generated over $\gr(A)$, take a finite set of homogeneous generators $m_i$.
Choose $n, i_0$ such that $n-i_0 < \deg m_i < n$, the claim is that these work.
Induct on $i$: suppose $F_{i+n} \subseteq A_{i+i_0} F_n$ and we WTS this still holds when $i\mapsto i+1$.
Letting $m\in F_{i+n+1}$, if $m\in F_{i+n}$ we're done.
Otherwise $\bar m\neq 0 \in \gr_{i+n+1} M$, so $\bar m = \sum \bar c_i m_i$ and picking lifts, $\bar m - \sum c_i m_i \in F_{i+n}$.
:::

:::{.theorem title="?"}
Let $A \da \kxn$, $(M, M_i)\in \amod^\fg$ have a good filtration, and let $\Phi(i) \da \dim_k M_i$.
Then

- There exists a polynomial $p$ with $\deg p \leq n$ such that $\Phi(i) = p(i)$ for $i\gg 0$.
- The degree and leading coefficient of $p$ are independent of the choice of good filtration

We then define 
\[
\dim_A M = \deg p
.\]
:::

:::{.exercise title="Method of finite differences"}
Let $\Phi: \ZZ\to \ZZ$ such that $\Phi(i+1) - \Phi(i)$ is eventually polynomial of degree $d' \leq n-1$.
Show that $\Phi$ is eventually polynomial of some bounded degree $d\leq m$.
:::

:::{.solution}
Write $\Phi(i) = C + \sum_{i\geq 1} \Phi(i) - \Phi(i-1) = C + \sum_{i\geq 1} q(i)$ for some polynomial $q$ with $\deg q \leq n-1$.
So it's ETS $\sum_{0\leq n\leq i}n^{a}$ polynomial in $i$ for $a\geq 0$.
:::

:::{.proof title="of theorem, part 1"}
Proceed by induction on the number of variables.
Since $\dim_k(M_i) = \sum_{i} \dim_k(\gr_i M)$ where $\gr_i M = M_i/M_{i+1}$, it's ETS $\dim M_i - \dim M_{i-1} = \dim \gr_i M$.
An awesome maneuver: take a regrading to get a LES
\[
0 \to K\da \ker f \to \gr M \mapsvia{f} \Sigma \gr M \to C\da \coker f\to 0
.\]

Note that $K, C\in k[x_1,\cdots, x_{n-1}]$ has fewer variables, and by alternating additivity in exact sequences,
\[
\dim \gr_i K - \dim \gr_i M + \dim \gr_{i+1} M - \dim \gr_i C = 0
.\]
Thus
\[
\dim \gr_{i+1} M - \dim \gr_{i} M = \dim \gr_i C - \dim \gr_i K
,\]
and since the RHS involves good filtrations, these are eventually polynomial, thus so is the LHS.
:::

:::{.proof title="of theorem, part 2"}
Let $F, G$ be good filtrations with $F_{i-k} \subseteq G_{i} \subseteq F_{i+k}$ with $\Phi_F(i) = \dim F_i, \Phi_G(i) = \dim G_i$.
Then 
\[
\Phi_F(i-k) \leq \Phi_G(i) \leq \Phi_G(i+k)
,\]
and if $\Phi_G, \Phi_G$ have different degrees or the same degrees but different leading terms, it would violate this inequality for large $i$.
:::

## Properties of dimension

:::{.exercise title="?"}
Let $M \in \amod^\fg$.

- Show that if $N \leq M$ then $\dim N \leq \dim M$.
- Show that if $M \surjects N$ then $\dim N\leq \dim M$.
:::

:::{.solution}
\envlist

- Restrict the good filtration on $M$, this is good by Artin-Rees.
- Take an induced good filtration. Use that quotients of finitely-generated are finitely-generated and $\gr(A)^n \surjects \gr(M) \surjects \gr(N)$.

:::

:::{.exercise title="?"}
Show that if $0\to M_1\to M_2\to M_3\to 0$ is a SES in $\amod$ for $A=\kxn$, then 
\[
\dim M_2 = \max( \dim M_1, \dim M_3)
.\]
:::

:::{.solution}
Choose good filtrations $F_i, M_i, G_i$ on $M_1, M_2, M_3$ respectively, so $\dim M_i = \dim F_i + \dim G_i$.
The RHS involves polynomials in $i$ with non-negative leading terms, and their sum has the max of the 2 degrees and again has a nonnegative leading term.
:::

:::{.exercise title="?"}
Show that if $A=\kxn$ then $\dim_A A = n$ and $\dim_A A^n = n$.
:::