# Tuesday, April 19

## Completion

:::{.remark}
Recall that $\cocolim_{i\in I} M_i\in \amod$ is the universal $A\dash$module living above all of the $M_i$, such that for any other $N$ above the $M_i$ there is a morphism $N\to \cocolim_i M_i$.
This can be realized as $\cocolim M_i = \ts{ (m_i)_{i\in I} \in \prod_{i\in I} M_i \st \phi(m_i) = m_{i+1}\, \forall i }$.

> Note that the inverse limit has a mapping **in** property, as does $\prod$.
> Limits can be constructed out of equalizers and products: 
\[
\cocolim F \cong \eq\qty{ \prod_{X \in \cat C} F(X) \to \prod_{f,Y\in \cat{C}(X, Y) } F(Y) }
.\]

:::

:::{.exercise title="?"}
Show that this satisfies the correct universal property.
:::

:::{.exercise title="?"}
Show the following:
\[
\cocolim_i k[t]/\gens{t^i} & \cong k \fps{t} \\
\cocolim(\cdots \to \ZZ/p^2\ZZ \to \ZZ/p\ZZ \to 0) &\cong \ZZpadic \\
\cocolim_n \ZZ/n!\ZZ &\cong \hat{\ZZ} = \prod_p \ZZpadic
.\]

:::

:::{.exercise title="?"}
Show that the functor $\cocolim_i(\wait)$ is left-exact and $\colim_i(\wait)$ is exact, i.e. if $0\to (M_i)\to (N_i) \to (L_i)\to 0$ is a SES of inverse systems, then $0\to \cocolim_i M_i \to \cocolim_i N_i \to\cocolim_i L_i$.
:::

:::{.solution}
Hint: for injectivity, use the product definition to realize $f: \cocolim_i M_i \to \cocolim_i N_i$ as $f=\prod_i f_i: M_i\to N_i$.

:::

:::{.remark}
Show that $\tau_{\geq 2}\RR \cocolim_i(\wait) = 0$, so there is always a 6-term exact sequence.
:::

:::{.theorem title="Mittag-Leffler, sufficient conditions for $\lim^1$ vanishing"}
If $M_i \to M_{i+1}$ is surjective, then $\lim^1 = 0$ and $\cocolim N_i \surjects \cocolim L_i$.
:::

:::{.exercise title="?"}
Prove this!

Hints:

- Given $(n_i)\in \cocolim N_i$, we want to produce $(l_i)$ with $(g_i(n_i)) = \ell_i$.
- Lift $\ell_0$ to $n_0$ and induct on $i$:
- Choose an approximate lift $\tilde n_i$ of $\ell_i$.
- Use commutativity of the diagrams of inverse systems to correct this choice, using the surjectivity assumption.

:::

:::{.definition title="Completion"}
Recall that given a sequence of submodules $M_i$ of a module $M$, so $M \geq M_1 \geq M_2 \geq \cdots$, one can form an inverse system $\cdots \to M/M_2\to M/M_1\to 0$.
The **completion** of $M$ is $\cocolim_i M/M_i$.
:::

:::{.remark}
Topological interpretation: define a subset $U \subseteq M$ iff for each $m\in U$ there is some $M_i$ such that $m+ M_i \subseteq U$; equivalently $\ts{M_i}$ forms a basis of neighborhoods of zero.
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:::{.exercise title="?"}
Show that $\cocolim M/M_i$ is Hausdorff iff $\intersect_i M_i = \ts{0}$ (i.e. $\hat M$ is separated).
:::

:::{.solution}
Hints:

$\implies$: 
If $m\neq 0\in \intersect M_i$, show every open containing zero contains $m$.

$\impliedby$:
Pick $m_1\neq m_2$ and find $M_i$ with $(m_1 + M_i) \intersect (m_2 + M_i) = \emptyset$.

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:::{.remark}
Call a sequence $(m_i)$ **Cauchy** iff $m_n - m_{n'}\in M_i$ for all $n, n' > N_i$.
Note that one can define a metric this way and take the Cauchy completion, defined as $\hat{M}$, which is canonically isomorphic to $\hat{M}$.
:::

:::{.exercise title="?"}
Produce the isomorphism from Cauchy sequences modulo equivalence to $\hat{M}$.
:::

:::{.solution}
For $\hat{M} \to M/M_i$, take $N_i \gg 0$ and project (i.e. send the sequence to its limit).
For $\cocolim M/M_i \to \hat{M}$, send $(m_i)\mapsto (\tilde m_i)$ by choosing arbitrary lifts.
:::

:::{.exercise title="?"}
Show that if $0\to M_1\to M_2\to M_3\to 0$ and $M_2$ admits a filtration, then there is a SES of the completions at *induced* filtrations $0\to \hat M_1\to \hat M_2\to \hat M_3\to 0$.
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:::{.solution}
Start with the SES
\[
0 \to {M_1\over M_1 \intersect \Fil_i M_2} \to {M_2\over \Fil_i M_2} \to {M_3\over f(\Fil_i M_2)} \to 0
,\]
and letting $i$ vary yields a SES of inverse systems.
By Mittag-Leffler, the limits are exact.
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:::{.exercise title="?"}
Show that completion is idempotent, so $\hat{\hat{M}} \cong \hat{M}$.
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:::{.definition title="adic filtration"}
For $A\in \CRing, M\in \amod, \alpha\in \Id(A)$, there is an $\alpha\dash$adic filtration $\Fil_i M \da \alpha^i M$, and the corresponding completion $\hat{M}$ is the **$\alpha\dash$adic completion**.
:::

:::{.example title="?"}
Some examples:

- $k[t], \gens{t}$
- $\ZZ, \gens{p}$
- $k[x,y], \gens{x,y}$

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:::{.slogan}
Completions are inverse limits of Artin rings and are local, and thus mediate between the two.
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:::{.exercise title="?"}
Show that if $A\in \CRing, m\in \mspec A$, then $\hat{A}$ completed with respect to $m$ is a local ring.
:::