## 86.2 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

86.2.0.1
\begin{equation} \label{restricted-equation-C} \mathcal{C} = \left\{ \begin{matrix} \text{systems }(B_ n, B_{n + 1} \to B_ n)_{n \in \mathbf{N}}\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

86.2.0.2
\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

86.2.0.3
\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 (apply Algebra, Lemma 10.19.1 to a ring map $A/I^ n[x_1, \ldots , x_ r] \to B/I^ nB$ such that the images of $x_1, \ldots , x_ r$ generate $B/IB$ over $A/I$).

Lemma 86.2.1. 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 (86.2.0.3). 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 10.97.2 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 10.95.1 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 108.7.1. Next we show this does not happen when $A$ is Noetherian.

reference
Lemma 86.2.2. Let $A$ be a Noetherian ring and let $I \subset A$ be an ideal. Then

every object of the category $\mathcal{C}'$, in particular the completion $A[x_1, \ldots , x_ r]^\wedge $, is Noetherian,

if $B$ is an object of $\mathcal{C}'$ and $J \subset B$ is an ideal, then $B/J$ is an object of $\mathcal{C}'$.

**Proof.**
To see (1) by Lemma 86.2.1 we reduce to the case of the completion of the polynomial ring. This case follows from Algebra, Lemma 10.96.6 as $A[x_1, \ldots , x_ r]$ is Noetherian (Algebra, Lemma 10.30.1). Part (2) follows from Algebra, Lemma 10.96.1 which tells us that ever finite $B$-module is $IB$-adically complete.
$\square$

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