# Monday February 3rd

Some module topics from Chapter 8.

## Noetherian and Artinian Modules

Definition (Noetherian Posets)
: A poset $(X, \leq)$ is said to satisfy the **ACC** or to be **Noetherian** iff there does not exist an infinite sequence (a chain) $\theset{x_n}$ with strict inequalities $x_1 < x_2 < \cdots$.
  Equivalently, every weakly ascending chain $x_1 \leq x_2 \leq \cdots$ eventually stabilizes, i.e. there exists an $N$ such that $x_N = x_{N+1} = \cdots$.

Definition (Artinian Posets)
: Similarly, a poset satisfies the **DCC** or is **Artinian** iff there does not exist an infinite decreasing sequence $x_1 > x_2 > \cdots$.

Definition (Order Dual)
: For $(X, \leq)$, define the **order dual** $(X\dual, \leq)$ where $x\leq y \in X\dual \iff y\leq x \in X$.

Proposition (Noetherian/Artinian Duality)
: $X$ is Noetherian iff $X\dual$ is Artinian.

Lemma
: The ACC holds iff every nonempty subset has a maximum (and similarly the DCC with minimums).

Proof 
: Otherwise use AOC to pick elements $x_i$; if $x_i$ isn't the maximum then there is some $x_{i+1} > x_i$, and this yields an infinite ascending chain iff no maximum.

Let $M$ be an $R\dash$module, and define $\mathrm{Sub}_R M = \theset{(R\dash\text{submodules of } M, \leq)}$.

Lemma
: $M$ is Noetherian $\iff$ every submodule $N\leq M$ is finitely generated.

Proof
: Apply the DCC.

Exercise
:   Let $M' \subset M$ and $q: M \to M/M'$ and $N_1 \subset N_2 \subset M$ such that

    - $N_1 \intersect M' = N_2 = M'$, and
    - $q(N_1) = q(N_2)$.

    Then $N_1 = N_2$.

Proposition (2 out of 3 Property for Noetherian/Artinian)
: If $0 \to M' \to M \to M'' \to 0$ is exact and $M$ is Noetherian (resp. Artinian) then $M', M''$ are both Noetherian (resp. Artinian).

Proof
: Note that $\mathrm{Sub}_R {M'}, \mathrm{Sub}_R {M''} \injects \mathrm{Sub}_{R} M$ in an order-preserving manner.
  If we then have $N_1 \subset N_2 \subset \cdots$ with $N_i \leq M$ submodules of $M$, we can consider $N_n = \frac{N_n + M'}{M'}$, which is weakly increasing in $M'$. 

  > Note: this is how we push forward into quotients.

  Thus this chain stabilizes, so for $i, j \gg 0$ we have $N_i + M' = N_j + M'$.
  So then $N_i \intersect M' = N_j \intersect M'$, and by the exercise, $N_i = N_j$ for all $i, j \gg 0$.

Corollary
: $R$ is Noetherian (resp. Artinian) iff every finitely-generated $R\dash$module is Noetherian (resp. Artinian)

Proof
: $\implies$:
  Suppose $R$ is Noetherian.
  Note that $0 \to R \to R^2 \to R \to 0$ since $R^2$ is an extension of $R$ by $R$.
  Thus $R^2$ is Noetherian, and inductively $R^n$ is a Noetherian $R\dash$module. \newline\newline
  If $M$ is a finitely-generated $R\dash$module, it is a quotient of a finitely-generated free $R\dash$module, and in particular $0\to K \to R^n \to M \to 0$ is exact.
  So $M$ is Noetherian, by the previous proposition (middle of a SES Noetherian $\implies$ ends are Noetherian).

## Tensor Products

Motivation from Representation Theory:
For $G$ finite, $H\leq G$, and $\rho: G \to V$ a finite-dimensional $\CC\dash$representation, this data is equivalent to a $\CC[G]\dash$module structure on $V$.
If $W$ is a representation on $H$, then $\ind_H^G W$ is a representation of $G$ given by $V = \ind_H^G W = W \tensor_{\CC[H]} \CC[G]$.

Definition (Tensor Product)
:   Let $M, N$ be $R\dash$modules, then the **tensor product** $M\tensor_R N$ is an object characterized up to canonical isomorphism by the following universal property:
    If $P$ is an $R\dash$module and $\Phi: M\cross N \to P$ is any bilinear map, then there exists a unique lift such that the following diagram commutes:

    \begin{center}
    \begin{tikzcd}
    M \tensor_R N \arrow[rrdd, "\exists! \psi", dotted] &  &   \\
                                                        &  &   \\
    M\times N \arrow[rr, "\Phi"] \arrow[uu, "\iota"]    &  & P
    \end{tikzcd}
    \end{center}

    where $\iota: M \cross N \to M \tensor_R N$ is $R\dash$bilinear and for all $(m, n) \in M\cross N$, we denote $m\tensor n \definedas \iota(m, n)$. 

By dimension counting in the finite-dimensional case of vector spaces, it's clear that $\iota$ need not be surjective.
In general, elements in $M \tensor_R N$ are *finite sums* of simple tensors, not just simple tensors, i.e. $M\tensor_R N = \generators{ \iota(m, n)  }$.

Proof (existence)
: Let $F$ be the free $R\dash$module on $M\cross N$ with basis $\theset{(m, n) \suchthat m\in M,~ n\in N}$.
  Mod out by the following relations: for all $m, m_1, M_2 \in M$ and for all $n, n_1, n_2 \in N$ and all $r\in R$,

  - $(m_1 + m_2) \tensor n - m_1\tensor n - m_2 \tensor n$
  - $m \tensor (n_1 + n_2) - m\tensor n_1 - m\tensor n_2$
  - $r(m\tensor n) - (rm)\tensor n$
  - $r(m\tensor n) - m\tensor (rn)$

  Let $\mathcal{R}$ be the ideal generated by these relations, the define $M \tensor_R N = F/\mathcal{R}$ by $(m, n) \mapsto (m, n) + \mathcal{R}$.
  Then (straightforward check) the universal mapping property holds.

How do we work with tensor products?
Namely, how do we even know whether an arbitrary element is zero or not in this complicated quotient.

- To show $m\tensor n = 0$, use bilinear relations (reduce to relations above)
- To show $m\tensor n\neq 0$, find an $R\dash$module and a bilinear map $\psi: M\tensor_R N \to P$ such that $\im(m\tensor n) \neq 0$.
- To show $M\tensor_R N \cong X$, show that $X$ satisfies the universal property.

Exercise
: $R\tensor_R M \equiv M$ by $(r, m) \mapsto r\cdot m$, with $\cdot$ the $R\dash$module action on $M$.
  Let $P$ be arbitrary, let $\phi: R\cross M \to P$ be arbitrary, and define $\psi: M \to P$ by $m \mapsto \phi(1, m)$.

Exercise
: $\ZZ/m\ZZ \tensor_\ZZ \ZZ/n\ZZ \equiv \ZZ/\gcd(n, m) \ZZ$.
Show that every element is both $n\dash$torsion and $m\dash$torsion.

Proposition (Existence of Base Change)
: For $M$ and $R\dash$module and $f: R\to S$, we can create an $S\dash$module $S \tensor_R M$ by *base change*.

Definitely the most important concept thus far!