# Tuesday, January 18

:::{.remark}
A reference for pictures: Mumford's red book.
Note the typo in A&M problem 10: it is false for the zero ring.

Recall some definitions:

- \(R\dash\)modules, what are some examples?
- Submodules and quotient modules.
- The submodule generated by a subset.
- A morphism of modules and the \(R\dash\)module structure on $\Hom_R(M, N)$, $(rf)(x) \da r\cdot f(x)$.
  - This makes $\rmod$ into a category enriched over itself.
- $\im f, \ker f, \coker f$.
- $\prod M_i, \bigoplus M_i$
:::

:::{.exercise title="Module structure on quotients"}
Show that if $M\leq N \in \mods{R}$, then the following action makes $M/N$ into an \(R\dash\)module:
\[
r\cdot [x] \da [rx]
,\]
i.e. that if $[x] = [y]$ then $r\cdot [x] = r\cdot [y]$.
:::

:::{.exercise title="Universal properties of kernels and cokernels"}
Show the universal properties of kernels and cokernels: given $f: M\to N$ and $Q\in \rmod$, 
\[
\Hom_R(Q, \ker f) &= \ts{ s\in \Hom_R(Q, M) \st f\circ s = 0} \\
\Hom_R(\coker f, Q) &= \ts{t\in \Hom_R(N, Q) \st t\circ f = 0 }
.\]
:::

:::{.exercise title="Direct sum and product coincide for finite index sets"}
Show that if $I$ is a finite indexing set, there is an isomorphism 
\[ 
\bigoplus _{i\in I} M_i \iso \prod_{i\in I} M_i
, \] and that
\[
\Hom_R\qty{ T, \prod M_i} &\cong \prod_{i\in I} \Hom_R\qty{T, M_i} \\
\Hom_R\qty{ \bigoplus M_i, T} &\cong \bigoplus \Hom_R\qty{ M_i, T}
.\]

Show that by Yoneda, this satisfies the universal property.
:::

:::{.solution}
Sketch:

- Define projections $\pi_j: \prod_i M_i \to M_j$.
- Send $f\in \Hom(T, \prod_i M_i)$ to $\pi_j \circ f\in \prod_i \Hom(T, M_i)$
- For the other direction, given $(f_i) \in \prod_i \Hom(M_i, T)$, send $(f_i)$ to $\sum f_i$.
:::

:::{.exercise title="Modules have products and coproducts"}
Show that $\prod$ is a categorical product and $\bigoplus$ is a categorical coproduct.
What are the product and coproduct in $\Top$?

:::

:::{.definition title="Colon ideals and annihilators"}
Given $N \leq M \in \rmod$, define the **colon ideal**
\[
(N: M) \da \ts{a\in R\st aM \subseteq N}
\]
and the **annihilator** of $M$ 
\[
\Ann(M) = (0: M) = \ts{a\in R \st aM = 0}
.\]
:::

:::{.example title="Annihilators"}
Some annihilators:

- $C_n \in \mods{\ZZ}$, so $\Ann(C_n) = n\ZZ = \gens{n}$.
- Again in $\mods{\ZZ}$, $\Ann(C_n \oplus C_m) = n\ZZ \intersect m\ZZ = \lcm(m, n) \ZZ$.

:::

:::{.remark}
An $R\dash$module is free iff $R \cong \bigoplus _{i\in I} R$, where $I$ can be infinite but (importantly) we need the direct sum instead of the direct product.
Note that generally $\prod_{i\in I} R$ may not be free as an $R\dash$module. 

A module is finitely generated if there exists a generating set $X \da \ts{x_i}_{i\leq n} \subseteq M$ such that any submodule containing $X$ is all of $M$, or equivalently $x\in M \implies x = \sum r_i x_i$ for some $r_i \in R$.
:::

:::{.exercise title="Finitely generated iff surjective image of a free module"}
Show that $M$ is finitely-generated iff there is a surjective morphism $R^n \to M$ for some $n\in \ZZ_{\geq 0}$.
:::

:::{.solution}
Sketch:

- finitely-generated $\implies$ surjection:
  - Take $e_i = \ts{0,\cdots, 1,\cdots, 0} \in R^n$ and define $f(e_i) \da x_i$ 
  - Use the universal property of the direct sum.
- $\impliedby$:
  - Show that the $f(e_i)$ generate $M$: by surjectivity, $m = f(x) = f(\sum r_i e_i) = \sum r_i f(e_i)$.

:::

:::{.remark}
Thus every $M\in \mods{R}$ is the quotient of a free module: find a surjection $f: F\to M$, so $\im f = M$, then use that $\im f \cong /\ker f$. 
For example, one can take \( F \da \bigoplus _{m\in M} R \to M \).

Recall

- The definition of exact sequences
- $0\to A \mapsvia{f} B$ is exact iff $f$ is injective,
- $A \mapsvia{f} B \to 0$ iff $f$ is surjective,
- $0\to A \mapsvia{f} B \to 0$ iff $f$ is an isomorphism,
- $0\to A\to B\to C\to 0$ is called a short exact sequence.

:::

:::{.exercise title="Direct sums are exact"}
Check that $(\wait) \oplus M$ is exact.
:::

:::{.exercise title="Left exactness of hom"}
Show that $\Hom(N, \wait)$ and $\Hom(\wait, N)$ are both left-exact for any $N\in\mods{R}$, where $0 \to A \mapsvia{f} B \mapsvia{g} C\to 0$ is sent to $0 \to \Hom(N, A) \mapsvia{f\circ \wait} \cdots$.
Give an example of when right-exactness fails.

> Hint: try $0\to \ZZ \mapsvia{\cdot 2} \ZZ \to C_2 \to 0$ and apply $\Hom_\ZZ(\wait, C_2)$.

:::

:::{.remark}
Recall that $\Hom_{\mods{\ZZ}}(\ZZ, \wait) \cong \id$ and $\Hom_{\mods{\ZZ}}(C_n, \wait) \cong (\wait)[n]$ picks out the $n\dash$torsion.
:::