\documentclass[12pt]{amsart}
\title{Lecture 7}
\author{}

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\newtheorem*{rem}{Remark}
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\newtheorem*{ex}{Example}

\begin{document}
\maketitle

\PP{7.0}{Review}
%----------------
Recall that last time we defined direct and inverse limits.
Given a poset $(I,\leq)$, a direct (resp. inverse) system, 
valued in a category $\CalC$ can be quickly
defined as an object of the functor category
$\Cov{\mathbf{I}}{\CalC}$ (resp. $\Contra{\mathbf{I}}{\CalC}$).
Here, $\mathbf{I}$ is the category defined using
the poset $(I,\leq)$ as in \S 6.2.\\

Thus, given a direct system $\{(X_i),(\psi_{ji}:X_i\to X_j)\}$,
its direct limit is the unique object representing the
following covariant functor $\CalC\to\Sets$:
\[
Y\mapsto \{(f_i:X_i\to Y) : f_j\circ\psi_{ji} = f_i \text{ for
every } i\leq j\}.
\]

Similary, for an inverse system $\{(X_i),(\varphi_{ij}:X_j\to X_i)\}$,
its inverse limit is the unique object representing the
following contravariant functor $\CalC\to\Sets$:
\[
Y\mapsto \{(g_i:Y\to X_i) : \varphi_{ij}\circ g_j=g_i \text{ for
every } i\leq j\}.
\]

Last time we saw some examples of the direct limit. Today, we will
see another such example where the {\em right directedness} hypothesis
is imposed on the poset $(I,\leq)$. Then, we will go over
some examples of the inverse limit.\\

\PP{7.1}{Example involving right directed posets}
%----------------------------------------------
Assume that $(I,\leq)$ is a right directed poset. Recall
that this means that for every $i,j\in I$, there exists $k\in I$
such that $i\leq k$ and $j\leq k$.\\

Let $\mathbf{CommRings}$ be the category of unital, commutative
rings, where the morphisms are assumed to be unital as well.
Let $\{(A_i)_{i\in I}, (\psi_{ji}:A_i\to A_j)_{i\leq j}\}$
be a direct system, valued in $\mathbf{CommRings}$.\\

\begin{thm}
The direct limit $A=\Dlim{(I,\leq)} A_i$ exists in the category
$\mathbf{CommRings}$. Moreover, if each $A_i$ is an integral
domain {\footnote{Recall that a commutative ring $A$ is an integral domain
if $ab=0$ implies either $a=0$ or $b=0$.}} (resp. a field)
then so is $A$.
\end{thm}

Compare this result with Example V of \S 5.4. The statement
would be obviously false if the right directedness is not
assumed, for instance, for a discrete poset (i.e, any two
distinct elements are non-comparable).\\

\begin{pf}
We will construct $A$, as a set, by imposing an equivalence
relation on the disjoint union of $A_i's$. Namely,
\[
A := \left(\bigsqcup_{i\in I} A_i\right)/\sim\ , \text{ where}
\]
$a\in A_i$ and $b\in A_j$ are {\em equivalent} iff there exists
$k$, $i\leq k$ and $j\leq k$, such that $\psi_{ki}(a) = \psi_{kj}(b)$
in $A_k$. I am going to leave it for you to check that this
is indeed an equivalence relation.\\

We first have to give $A$ a structure of a ring, then check that
it is unital and commutative. The canonical maps $A_i\to A$
are the natural inclusion followed by the natural surjection,
i.e, $a\in A_i$ goes to the equivalence class containing $a$.
We will have to make sure that these are (unital) ring homomorphisms,
and that they satisfy the universal property of direct limits.\\

So, given two elements $\alpha,\beta\in A$, choose representatives
$a\in \alpha$, $b\in \beta$. Let $i,j\in I$ be such that $a\in A_i$
and $b\in A_j$. Pick $k\in I$ such that $i\leq k$ and $j\leq k$,
and define the sum and product as follows:
\[
\alpha+\beta := \text{Equivalence class of }
\psi_{ki}(a) + \psi_{kj}(b).
\]
\[
\alpha\cdot\beta := \text{Equivalence class of } \psi_{ki}(a)\psi_{kj}(b) \in A_k.
\]

We now prove that these operations are well--defined. Let us
show it for the product (the proof for the sum is exactly the same
- replace multiplication by addition in the following argument).\\

Assume that
we pick some representatives other than $a,b$, say $a'\in A_{i'}$ and
$b'\in A_{j'}$ from $\alpha$ and $\beta$ respectively, and
take $k'\geq i',j'$ to get an element $\psi_{k'i'}(a')\psi_{k'j'}(b')$
in $A_{k'}$. We have to show that this element belongs to the
same equivalence class as $\psi_{ki}(a)\psi_{kj}(b)$. See
the picture below:

\[
\xy
(0,40)*{a\in A_i};
(0,0)*{b\in A_j};
(35,20)*{A_k\ni \psi_{ki}(a)\psi_{kj}(b)};
{\ar^{\psi_{ki}} (5,38); (18,22)};
{\ar_{\psi_{kj}} (5,2); (18,18)};
(130,40)*{A_{i'}\ni a'};
(130,0)*{A_{j'}\ni b'};
(85,20)*{\psi_{k'i'}(a')\psi_{k'j'}(b')\in A_{k'}};
{\ar_{\psi_{k'i'}} (123,38); (103,22)};
{\ar^{\psi_{k'j'}} (123,2); (103,18)};
(58,22)*{\stackrel{?}{\sim}};
\endxy
\]

Since $a\sim a'$, there exists an element $i''\geq i,i'$ for
which $\psi_{i''i}(a) = \psi_{i''i'}(a')$ (similarly for $b\sim b'$, we
denote such an index by $j''$).
Let us choose an index $\ell\in I$ larger than any index from the finite
set $\{i'',j'',k,k'\}$ (exists by the right--directedness hypothesis). 
It suffices to prove the following:\\

\noindent {\bf Claim.} $\ds \psi_{\ell k}(\psi_{ki}(a)\psi_{kj}(b))
= \psi_{\ell k'}(\psi_{k'i'}(a')\psi_{k'j'}(b'))$.\\

{\em Proof of the claim.} We have the following computation,
using the transitivity property of $\psi$ morphisms, and the
fact that these are homomorphisms of rings:\\

\begin{align*}
\psi_{\ell k}(\psi_{ki}(a)\psi_{kj}(b))
&= \psi_{\ell i}(a)\psi_{\ell j}(b) = 
\psi_{\ell i''}(\psi_{i''i}(a))\psi_{\ell j''}(\psi_{j'' j}(b)) \\
&= \psi_{\ell i''}(\psi_{i''i'}(a'))
\psi_{\ell j''}(\psi_{j'' j'}(b')) = \psi_{\ell i'}(a')\psi_{\ell j'}(b')\\
&= \psi_{\ell k'}(\psi_{k'i'}(a'))\psi_{\ell k'}(\psi_{k'j'}(b'))
= \psi_{\ell k'}(\psi_{k'i'}(a')\psi_{k'j'}(b')).
\end{align*}

Thus the claim is established.\\

The role of $0\in A$ is played by the equivalence class containing
$0\in A_i$ (for any $i\in I$), and that of $1\in A$ by the class
containing $1\in A_i$ (again, for any $i\in I$). It remains to
check that $A$ is indeed a unital, commutative ring. The following
observation follows immediately from the right--directedness assumption:\\

\noindent {\bf Observation.} For a finite number of equivalence
classes, $\{\alpha_1,\ldots,\alpha_n\}$, there exists $k\in I$
and representatives $a_1\in\alpha_1,\ldots,a_n\in\alpha_n$
such that $a_1,\ldots,a_n\in A_k$.\\

Since axioms for $A$ to be a unital, commutative ring only
involve finitely many elements, and they hold for $A_k$, they
will continue to hold for $A$. For instance, let us write down
how the proof of associativity will proceed. Let $\alpha,\beta,\gamma$
be three elements of $A$. Pick representatives $a,b,c$ from
these equivalence classes, which all belong to the same $A_k$
(for some $k\in I$). Since in $A_k$, $(ab)c = a(bc)$, it
follows that $(\alpha\beta)\gamma = \alpha(\beta\gamma)$.\\

Same argument as above goes into showing that, if each
$A_i$ is an integral domain (resp. a field), then so is $A$.\\

Finally, let us assume that $B\in\mathbf{CommRings}$,
and we are given morphisms $f_i : A_i\to B$, for every $i\in I$,
satisfying $f_j\circ \psi_{ji} = f_i$ for every $i\leq j$.
Consider the unique (set) map we get $\ds \wt{f} : \bigsqcup_{i\in I}
A_i\to B$, whose restriction to $A_i$ is $f_i$.\\

If $a\in A_i$ and $b\in A_j$ are equivalent, i.e, there exists
$k\geq i,j$ such that $\psi_{ki}(a) = \psi_{kj}(b)$, then
$f_i(a) = f_k(\psi_{ki}(a)) = f_k(\psi_{kj}(b)) = f_j(b)$.
Hence, $\wt{f}$ factors through the equivalence relation
to give $f:A\to B$. It is obvious (left as an easy exercise)
to see that $f$ is a ring homomorphism, and is the unique
one which makes the following diagram commute, for every
$i\leq j$:
\[
\xy
(0,20)*{A_i}="a"; (20,20)*{A}="b"; (20,0)*{B}="c";
{\ar "a"; "b"};
{\ar^{f} "b"; "c"};
{\ar_{f_i} "a"; "c"};
\endxy
\]
where, recall that $A_i\to A$ is given by the following composition
$\ds A_i \subset \bigsqcup_{j\in I} A_j \twoheadrightarrow A$.
\end{pf}

\PP{7.2}{Example of inverse limit of groups}
%------------------------------------------
Now, let us focus on inverse limits. As direct limits are contructed,
in practice, as ``quotients" of direct sums, inverse limits will
be constructed as ``subobjects" of direct product. The following
lemma illustrates this point. We are {\bf not} assuming any
directedness hypothesis here.\\

\begin{lem}
Inverse limits exist in the category of groups.
\end{lem}

\begin{pf}
Let $(I,\leq)$ be a poset, and consider an inverse system
of groups:
\[
\left\{ (G_i)_{i\in I}; (\varphi_{ij}:G_j\to G_i)_{i\leq j} \right\}
\]
Recall that this means: (i) $\varphi_{ii}=\Id_{G_i}$ ($\forall i\in I$),
and (ii) $\varphi_{ij}\varphi_{jk}=\varphi_{ik}$ ($\forall i\leq j\leq k$).\\

Consider $\ds \wt{G} = \prod_{i\in I} G_i$ together with canonical
projections $\wt{p}_i : \wt{G}\to G_i$. Let $G$ be the following
subset of $\wt{G}$:
\[
G = \{(x_i\in G_i)_{i\in I} : \varphi_{ij}(x_j) = x_i \text{ for
every } i\leq j\}.
\]

Note that $G$ is a subgroup of $\wt{G}$. This is because, if $e_i\in G_i$
is the neutral element, then $e=(e_i)_{i\in I}$ is in $G$ which
is neutral with respect to componentwise multiplication.
Moreover if $(x_i)_{i\in I}$ and $(y_i)_{i\in I}$ are two
elements of $G$, then $(x_i^{-1}y_i)_{i\in I}$ is also in $G$, because:
\[
\text{for every } i\leq j,\ 
\varphi_{ij}(x_j^{-1}y_j) = \varphi_{ij}(x_j)^{-1}\varphi_{ij}(y_j)
= x_i^{-1}y_i.
\]

Now let $p_i : G\to G_i$ be the restriction of $\wt{p}_i$ to
$G\subset \wt{G}$. We have: $\varphi_{ij} \circ p_j = p_i$,
for every $i\leq j$, because, once evaluated on a tuple $(x_k)_{k\in I}$,
this becomes $\varphi_{ij}(x_j) = x_i$. This holds by definition
of $G$.\\

Finally, let $H$ be another group, and let $g_i:H\to G_i$ be
group homomorphisms, for which $\varphi_{ij}\circ g_j = g_i$.
Thus we get a unique group homomorphism $\wt{g}:H\to\wt{G}$,
given by $x\mapsto (g_i(x))_{i\in I}$. All we need to check
is that $\Im(g)\subset G$. That is, for every $i\leq j$,
$\varphi_{ij}(g_j(x)) = g_i(x)$. As it was assumed for the group homomorphisms
$(g_i)_{i\in I}$, we are done.
\end{pf}

\begin{rem}
The same proof goes over to prove that inverse limits exist
in the category of rings, abelian groups, modules over a ring,
vector spaces and so on.
\end{rem}

\PP{7.3}{A concrete example}
%--------------------------
Consider $I = \N$ with its natural (total) order. For each
$n\in \N$, let $A_n = \C[x]/(x^{n+1})$, an object of the category
$\mathbf{CommRings}$. For $m<n$, we take the following ring
homomorphism:
\[
\varphi_{mn} : \C[x]/(x^{n+1}) \to \C[x]/(x^{m+1}),\qquad
\sum_{i=0}^n a_ix^i \mapsto \sum_{j=0}^m a_jx^j.
\]
I am going to leave it as an exercise for you to prove
the following (the limit is taken in the category
$\mathbf{CommRings}$):

\[
\Ilim{n\in\N} \C[x]/(x^{n+1}) = \C[\![x]\!],
\]
the ring of formal power series in a variable $x$.\\

\PP{7.4}{Completion of a ring with respect to an ideal}
%------------------------------------------------------
Another application of inverse limits is to define completion
of a ring with respect to an ideal. This generalizes the example
presented in the previous paragraph, where the ring under
consideration was $\C[x]$ and the ideal was $(x)\subset\C[x]$.\\

So, let $A\in\mathbf{CommRings}$, and let $\mathfrak{a}\subset A$
be an ideal. Recall that this means $\mathfrak{a}$ is an abelian
subgroup of $A$ (under addition), and for every $x\in A$, $y\in\mathfrak{a}$,
we have $xy\in \mathfrak{a}$.\\

Consider the following inverse system in $\mathbf{CommRings}$,
with respect to $I=\N$.
\[
A_n = A/\mathfrak{a}^{n+1},\qquad
\pi_{mn}:A_n\to A_m,\ \forall\ m\leq n.
\]
Here, $\pi_{mn}$ is the natural projection.\\

{\em The completion of $A$ with respect to the ideal $\mathfrak{a}$}
is defined as the inverse limit of this inverse system of rings.
\[
\widehat{A} := \Ilim{n\in\N} A/\mathfrak{a}^{n+1}.
\]
Note that there are natural ring homomorphisms $A\to A_n$,
for every $n\in \N$, which give rise to a ring homomorphism
(by the universal property of inverse limits) $A\to \widehat{A}$.
For instance, in the set up of the previous paragraph,
this is just the usual inclusion of polynomials into power series
$\C[x]\subset \C[\![x]\!]$.\\

For instance, let us take $A=\Z$ and $\mathfrak{a}=(p)$, for a prime
number $p\in\Z_{\geq 2}$. The resulting inverse limit is denoted
by $\widehat{\Z}_p$, and is called {\em the ring of $p$--adic
integers}:
\[
\widehat{\Z}_p := \Ilim{n\in\N} \Z/(p^{n+1}).
\]
\end{document}
