# Monday April 6th

Last time: a characterization of Artinian rings, the Akizuki-Hopkins theorem

Theorem (Akizuki-Hopkins)
: $R$ is Artinian iff $R$ is Noetherian with Krull dimension 1, i.e. all primes are maximal.

Proposition (Powers of Maximal Ideals in Local Rings)
:   Suppose $(R, \mfm)$ is Noetherian and local.
    Then either

    1. $\mfm^n \gneq \mfm^{n+1}$ for all $n$, or
    2. $\mfm^n = (0)$ for some $n$.

    Moreover, (2) holds iff $R$ is Artinian.

Proof
:   If $\mfm^n = \mfm^{n+1}$ for some $n$, then by Nakayama $\mfm^n = (0)$.

    If $R$ is Artinian, then (1) can not hold, so (2) must hold.
    Conversely, if (2) holds, then $$\mfm \in \nil R = \intersect_{\spec R} \mfp,$$ but this can only happen if $\spec R = \theset{\mfm}$ is precisely one ideal.
    But then every prime ideal is maximal.

Note that Artinian rings have finitely many maximal ideals.

Theorem (Primary Decomposition)
:   Let $R$ be a nonzero Artinian ring and suppose $\maxspec R = \theset{\mfm_i}^n$.
    Then $R = \prod_{i=1}^n \mfr_i$ where $\mfr_i \definedas R_{\mfm_i}$.

Note that for $R = \ZZ$ and $N = \prod_{i=1}^n p_i^{a_i}$, this recovers $Z/N\ZZ = \prod_{i=1}^n \ZZ/p_i^{a_i}\ZZ$.

Punchline: Artinian rings split as a product of local rings.
Modules over local rings are free, and thus simple, so this gives a local-to-global principle.
Moreover, every localization is a projection onto one of the factors.

Exercise
:   For $R$ Artinian, show that every $x\in R$ is either a unit or a zero divisor.
    Thus $R$ is its own total fraction ring.

Exercise
:   For a ring $R$, TFAE

    1. $R$ is semilocal (finitely many maximal ideals).
    2. $R/\mcj(R)$ is Artinian.
    3. $R/\mcj(R)$ has finitely many ideals.

Most important theorem in the course: Hilbert's basis theorem!

Theorem (Hilbert's Basis)
: If $R$ is Noetherian, then $R[t]$ is Noetherian.

Proof
:   Toward a contradiction, let $J \normal R[t]$ that is not finitely generated.
    Construct a sequence $\theset{f_n}$  and take $J_n \definedas \gens{f_1, \cdots, f_n} \subseteq J$.
    This can be done by taking $f_0 = 0$ and $f_{n+1}$ any element of $J/J_{n}$ of minimal degree.
    Note that this ensures that $\deg f_n \leq \deg f_{n-1}$.

    Set $a_n$ to be the leading coefficient of $f_n$, and let $I = \gens{\theset{a_i}}$.
    Since $R$ is Noetherian, $I$ is finitely generated, so there is some $N \in \ZZ$ such that $I = \gens{a_1, \cdots, a_N}$.

    Thus $a_{n+1} = \sum_{i=1}^N u_i a_i$ for some $u_i \in R$.
    So set $g \definedas \sum_{i=1}^N u_i f_i t^{ \deg f_{N+1} - \deg f_i }$.
    Then $g\in J_N$ and $f_N \in J/J_n$ and $f_{N+1} - g \in J\setminus J_N$.

    Now the leading term of $g$ is $\sum u_i a_i = a_{N+1}$, and since $\deg g = \deg f_{N+1}$ where $a_{N+1}$ is also the leading term of $f_{N+1}$.
    Thus $\deg(f_{N+1} - g) < \deg f_{N+1}$, contradicting minimality.

Theorem (Formal Power Series over Noetherian is Noetherian)
: $R$ Noetherian implies that $R[[t]]$ is Noetherian.

Exercise
:   If $R$ is a ring with $R[t], R[[t]]$ both Noetherian, then $R$ is Noetherian.

    > Idea: use the fact that quotients of Noetherian rings are again Noetherian.

Corollary (Single Most Important Result!)
: If $R$ is Noetherian then every finitely generated $R\dash$algebra is Noetherian.

If such an algebra is finitely generated by $n$ generators, it's a quotient of $R[x_1, \cdots, x_n]$.

Some historical notes on the Hilbert Basis Theorem:
Given $G$ a group and $T$ a ring, we can consider actions $G \to \aut(R)$ and the ring of invariants $$T^G = \theset{t\in T \suchthat gt = t ~\forall g\in G} \subset T.$$
Note that Galois theory fits into this framework.
For a classical example, consider $T = \CC[t_1, \cdots, t_n]$.
Then $G \subset \Gl(n, \CC) \actson T$ by linear automorphisms, i.e. it acts on each $t_i$ by taking it to some linear combination of $t_j$.
Note that $G$ can be chosen to be "nice", i.e. a linear algebraic group.

Question: is $T^G$ finitely generated as a $\CC\dash$algebra?
Hilbert proved that this is true when $G$ is a linear algebraic group, and the main step was the basis theorem.
Previously, people were proving these kinds of theorems for a single group at a time, whereas this encompassed all of them simultaneously!

> Quote by Gordon: "This is not Mathematics, this is theology."

This is an early triumph of abstraction in algebra, as opposed to writing out lines upon lines of equations for single proofs.
How effective is this proof?
This doesn't necessarily lead to a good algorithm for finding a finite generating set, see computational commutative algebra.

Exercise
: For $k$ a field, $k[x, y]$ is Noetherian. Show that $k[y, xy, x^2y, \cdots]$ is not Noetherian.

Note that things work out very nicely for Noetherian rings of dimension 0 and 1, but many theorems fail in higher dimensions.

Exercise (Possibly difficult to prove)
: Show that every $k\dash$subalgebra of $k[x]$ is Noetherian.