## 87.2 Two categories

Let $A$ be a ring and let $I \subset A$ be an ideal. In this section ${}^\wedge $ will mean $I$-adic completion. Set $A_ n = A/I^ n$ so that the $I$-adic completion of $A$ is $A^\wedge = \mathop{\mathrm{lim}}\nolimits A_ n$. Let $\mathcal{C}$ be the category

87.2.0.1
\begin{equation} \label{restricted-equation-C} \mathcal{C} = \left\{ \begin{matrix} \text{inverse systems }\ldots \to B_3 \to B_2 \to B_1
\\ \text{where }B_ n\text{ is a finite type }A_ n\text{-algebra,}
\\ B_{n + 1} \to B_ n\text{ is an }A_{n + 1}\text{-algebra map}
\\ \text{which induces }B_{n + 1}/I^ nB_{n + 1} \cong B_ n
\end{matrix} \right\} \end{equation}

Morphisms in $\mathcal{C}$ are given by systems of homomorphisms. Let $\mathcal{C}'$ be the category

87.2.0.2
\begin{equation} \label{restricted-equation-C-prime} \mathcal{C}' = \left\{ \begin{matrix} A\text{-algebras }B\text{ which are }I\text{-adically complete}
\\ \text{such that }B/IB\text{ is of finite type over }A/I
\end{matrix} \right\} \end{equation}

Morphisms in $\mathcal{C}'$ are $A$-algebra maps. There is a functor

87.2.0.3
\begin{equation} \label{restricted-equation-from-complete-to-systems} \mathcal{C}' \longrightarrow \mathcal{C},\quad B \longmapsto (B/I^ nB) \end{equation}

Indeed, since $B/IB$ is of finite type over $A/I$ the ring maps $A_ n = A/I^ n \to B/I^ nB$ are of finite type by Algebra, Lemma 10.126.8.

Lemma 87.2.1. Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. The functor

\[ \mathcal{C} \longrightarrow \mathcal{C}',\quad (B_ n) \longmapsto B = \mathop{\mathrm{lim}}\nolimits B_ n \]

is a quasi-inverse to (87.2.0.3). The completions $A[x_1, \ldots , x_ r]^\wedge $ are in $\mathcal{C}'$ and any object of $\mathcal{C}'$ is of the form

\[ B = A[x_1, \ldots , x_ r]^\wedge / J \]

for some ideal $J \subset A[x_1, \ldots , x_ r]^\wedge $.

**Proof.**
Let $(B_ n)$ be an object of $\mathcal{C}$. By Algebra, Lemma 10.98.2 we see that $B = \mathop{\mathrm{lim}}\nolimits B_ n$ is $I$-adically complete and $B/I^ nB = B_ n$. Hence we see that $B$ is an object of $\mathcal{C}'$ and that we can recover the object $(B_ n)$ by taking the quotients. Conversely, if $B$ is an object of $\mathcal{C}'$, then $B = \mathop{\mathrm{lim}}\nolimits B/I^ nB$ by assumption. Thus $B \mapsto (B/I^ nB)$ is a quasi-inverse to the functor of the lemma.

Since $A[x_1, \ldots , x_ r]^\wedge = \mathop{\mathrm{lim}}\nolimits A_ n[x_1, \ldots , x_ r]$ it is an object of $\mathcal{C}'$ by the first statement of the lemma. Finally, let $B$ be an object of $\mathcal{C}'$. Choose $b_1, \ldots , b_ r \in B$ whose images in $B/IB$ generate $B/IB$ as an algebra over $A/I$. Since $B$ is $I$-adically complete, the $A$-algebra map $A[x_1, \ldots , x_ r] \to B$, $x_ i \mapsto b_ i$ extends to an $A$-algebra map $A[x_1, \ldots , x_ r]^\wedge \to B$. To finish the proof we have to show this map is surjective which follows from Algebra, Lemma 10.96.1 as our map $A[x_1, \ldots , x_ r] \to B$ is surjective modulo $I$ and as $B = B^\wedge $.
$\square$

We warn the reader that, in case $A$ is not Noetherian, the quotient of an object of $\mathcal{C}'$ may not be an object of $\mathcal{C}'$. See Examples, Lemma 109.8.1. Next we show this does not happen when $A$ is Noetherian.

reference
Lemma 87.2.2. Let $A$ be a Noetherian ring and let $I \subset A$ be an ideal. Then

every object of the category $\mathcal{C}'$ (87.2.0.2) is Noetherian,

if $B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}')$ and $J \subset B$ is an ideal, then $B/J$ is an object of $\mathcal{C}'$,

for a finite type $A$-algebra $C$ the $I$-adic completion $C^\wedge $ is in $\mathcal{C}'$,

in particular the completion $A[x_1, \ldots , x_ r]^\wedge $ is in $\mathcal{C}'$.

**Proof.**
Part (4) follows from Algebra, Lemma 10.97.6 as $A[x_1, \ldots , x_ r]$ is Noetherian (Algebra, Lemma 10.31.1). To see (1) by Lemma 87.2.1 we reduce to the case of the completion of the polynomial ring which we just proved. Part (2) follows from Algebra, Lemma 10.97.1 which tells us that ever finite $B$-module is $IB$-adically complete. Part (3) follows in the same manner as part (4).
$\square$

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