# Wednesday February 5th

Recall that if $M, N$ are $R\dash$modules then there is a map 

\begin{align*}
M\cross N &\mapsvia{\Phi} M\tensor_R N \\
(r, m) &\mapsto r\tensor m
\end{align*}

which is universal wrt the property that any bilinear map $\phi: M \cross N \to A$ factors through $\Phi$ uniquely.

We have a notion of *pullback*, where if $i:R \to S$ and $N$ is an $S\dash$module then $i^*N$ is an $R$ module with action given by composition $R \mapsvia{i} S \mapsvia{\cdot} \endo_\ZZ(N)$.

Dually, we have a notion of *base change*, where for $M$ an $R\dash$module we can form $i_* M \definedas S\tensor_R M$ an $S\dash$module where $s(\sum s_i\tensor m_i) = \sum ss_i \tensor m_i$.

An $R\dash$algebra is $i:R\to S$ a ring morphism, where algebra morphisms $f: S_1 \to S_2$ are given by commutative diagrams

\begin{center}
\begin{tikzcd}
R \arrow[rr, "i_1"] \arrow[rrdd, "i_2"] &  & S_1 \arrow[dd, "f"] \\
&  &                     \\
&  & S_2                
\end{tikzcd}
\end{center}

For $S, T$ $R\dash$algebras, the tensor product $S \tensor_R T$ is an $R\dash$algebra with $(s_1 \tensor m_1) \cdot (s_2 \tensor m_2) = s_1s_2 \tensor m_1 m_2$.
Note that the tensor product satisfies the universal property of the direct sum or coproduct:

\begin{center}
\begin{tikzcd}
&  & t \arrow[rrd, dotted]                              &  &                                              &  &   \\
&  & T \arrow[rrd] \arrow[rrrrd, "\psi", bend left=49]  &  & 1\tensor t                                   &  &   \\
R \arrow[rrd] \arrow[rru] &  &                                                    &  & S \tensor_R T \arrow[rr, "\exists!", dashed] &  & W \\
&  & S \arrow[rru] \arrow[rrrru, "\phi", bend right=49] &  & s\tensor 1                                   &  &   \\
&  & s \arrow[rru, dotted]                              &  &                                              &  &  
\end{tikzcd}
\end{center}

Exercise (Important)
:   Verify the following identities

    *One:*
    Let $M$ be an $R\dash$module and $N$ an $S\dash$module with $\iota: R\to S$.
    $\hom_R(M, \iota^* N) = \hom_S(\iota_* M, N) = \hom_S(S\tensor_R M, N)$.
    What's the map? $s\tensor m \mapsto sf(m)$.

    *Two:*
    For $P$ and $R\dash$module and $M, N$ $S\dash$modules, we have $M\tensor_X(i^*N \tensor_R P) = i^*(M\tensor_S N) \tensor_R P$.
    So for $N = S$, then $M \tensor_S (S \tensor_R P) = M\tensor_R P$.

    **Three (Good Exercise! Very important!):**
    For $M$ an $R\dash$module and $I \normal R$, we have $IM \subset_R M$.
    Show that we can identify the base change as $R/I \tensor_R M = M/I$.

    > Hint: Show that the RHS satisfies the appropriate universal property.

    *Four:*

    - $(\oplus M_i) \tensor_R N = \oplus (M_i \tensor_R N)$.
    - The tensor product of free modules is free.
    - If $F$ is a free $R\dash$module and we base change with $\iota: R\to S$ then $S \tensor_R F$ is a free $S\dash$module.


Definition (Invariant Basis Number)
: Let $R$ be a ring, then $R$ satisfies the *invariant basis number property* (IBN) iff any two bases for a free left $R\dash$module have the same cardinality.

Definition (Rank Condition)
: $R$ satisfies the *rank condition* iff whenever there exists a $q: R^m \surjects R^n$, $n\leq m$.

Definition (Strong Rank Condition)
: $R$ satisfies the *strong rank condition* iff whenever $q: R^m \injects R^n$ then $n\leq m$.

Facts
: If $R$ is commutative or (left)-Noetherian, then strong rank condition $\implies$ rank condition $\implies$ IBN.

Note: this is not obvious, since if $R$ is not Noetherian there are submodules that aren't finitely generated but can still have bounded rank.

Exercise (Non-Commutative)
: Let $k$ be a field and $V$ an infinite dimensional $k\dash$vector space, i.e. $V \cong V \oplus V$.
  Let $R \definedas \endo_k(V)$; then $R$ does not satisfy the IBN.

Proposition (Commutative Rings have Invariant Basis Number)
: If $R$ is nonzero and commutative then $R$ satisfies IBN.

Proof
: Suppose there exist $I, J$ such that  $\oplus_{i\in I} R \cong_R \oplus_{j\in J} R$.
  We want to show that $\abs{I} = \abs{J}$.
  Since $R\neq 0$, there is a maximal ideal $\mfm \in \maxspec(R)$.
  Since $R/\mfm$ is a field, we base change to it to obtain $R/\mfm \tensor_R (\bigoplus_{i\in I} R) = \bigoplus_{i\in I} R/\mfm$.
  We know this equals $R/\mfm \tensor_R (\bigoplus_{j\in J} R) = \oplus_{j\in J} R/\mfm$.
  So $I, J$ are bases of isomorphic vector spaces and thus $\abs{I} = \abs{J}$ by linear algebra.

Definition (Noetherian and Artinian Modules)
: A module $M$ is *Noetherian* iff ACC on submodules, and *Artinian* iff DCC on submodules.

Exercise
: If $R = k$ is a field and $V$ is a $k\dash$vector space, then $V$ is Noetherian iff Artinian iff infinite-dimensional.

Exercise
: If $R= \ZZ$, $R$ is Noetherian but not Artinian.
  Find a $\ZZ\dash$module that is Artinian but not Noetherian.

> Try all $2^n$ possibilities for adjectives!

Exercise
: If $R$ is finite, it is both Artinian and Noetherian, and moreover has only finitely many ideals.

Artinian is much stronger, and implies Noetherian? 
Converse iff every ideal is maximal.
The only Artinian integral domains are fields. Very small class of rings.
It's not true that Artinian alone implies finitely many ideals.

Exercise (8.29 in Notes)
: Let $I = (x^2, xy, y^2) = (xy)^2 \normal \CC[x, y]$ and take $R = \CC[x, y]/I$.

  a. Show that a $\CC\dash$basis for $R$ is given by $\theset{1 + I, x + I, y + I}$.

  b. Deduce that $R$ is Noetherian and Artinian.

  c. Show proper ideals of $R$ are precisely the $\CC\dash$subspaces of $\generators{x, y} + I$.

  d. Deduce that $\RR$ has continuum many ideals.