# Thursday, March 03

:::{.remark}
Next topic: Artin rings.
The general way we reduce the study of arbitrary rings:

\begin{tikzcd}
	\CRing \\
	&& {\text{Checking on covers by }\spec A} \\
	\Loc\CRing \\
	&& {\Loc\CRing^{\text{complete}}} \\
	\Art\CRing \\
	&& {} \\
	\Field
	\arrow[squiggly, from=1-1, to=3-1]
	\arrow[squiggly, from=3-1, to=5-1]
	\arrow[squiggly, from=3-1, to=4-3]
	\arrow[squiggly, from=4-3, to=5-1]
	\arrow[squiggly, from=5-1, to=7-1]
	\arrow[squiggly, from=1-1, to=2-3]
	\arrow[squiggly, from=2-3, to=3-1]
\end{tikzcd}

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


- Recall that Artin rings are defined by the DCC condition on ideals.
- The **length** of a module is the maximal length of a strictly increasing filtration.

:::

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

Show the following:

- $\CC[t]/\gens{t^n}$ is Artin.
- $C_{p^n}$ is Artin.
- $\ZZ$ is Noetherian but not Artin.
- Any finite product of fields is Artin.
- An Artin domain is a field.
- Prime implies maximal in any Artin ring.

  > Hint: quotient by the prime, and use that any element $a$ satisfies $a^n = ba^{n+1}$ for some $n$ and $b$ to produce an inverse.

- $\spec A = \ts{\pt}$ for a local Artin ring.
- Artin rings $A$ have finite length in $\mods{A}$.
- Quotients of Artin rings by their Jacobson radicals are products of fields.
- $\spec A = \Disjoint_i \ts{\pt_i}$ is a disjoint union of points, and is Hausdorff.
  - The only Noetherian rings with Hausdorff spectra are Artin.
:::

:::{.theorem title="?"}
Artin rings are Noetherian.
:::

:::{.proof title="?"}
Use that

- Any module has a maximal proper submodule.
- Choose $\mfa_i \subseteq A$ simple, so $A/\mfa_i$ is Artinian, to produce an increasing chain $0 \subseteq \mfa_1 \subseteq \mfa_2 \cdots$ where $\mfa_i/\mfa_{i+1}$ is simple for all $i$.
- Enumerate maximal ideals $\mfm_i$ and produce a chain $\mfm_1 \contains \mfm_2 \contains \cdots$ and take colon ideals.
- Reduce to this case by showing every chain refines.

Steps:

- Make a descending filtration with semisimple associated graded, whose filtration is finite.
- Use Jordan-Holder, every such sequence has the same length.
- Refine an arbitrary filtration to one in which the quotients are simple.

:::

:::{.corollary title="?"}
$\size \mspec A < \infty$ for $A\in \Art\CRing$.
:::

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

- If not, take a decreasing chain of $\mfm_1 \contains \mfm_1\mfm_2 \cdots$, stabilize, use that the $\mfm_i$ are finitely-generated and apply Nakayama.
- If $I$ is defined as the product at the minimal stabilized step, $I \subseteq \jacobsonrad{A}$.
- Without loss of generality, assume $\jacobsonrad{A} = 0$, so $I=0$ since $\mspec A/\jacobsonrad{A} \cong \mspec A$.
- $A/\mfm_1\cdots \mfm_n = \prod A/\mfm_i$ is a product of fields
- Corollary: $n> N$, $\mfm_N$ will have empty support.

:::