## 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.

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

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

2. 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. $\square$

Remark 86.2.3 (Base change). Let $\varphi : A_1 \to A_2$ be a ring map and let $I_ i \subset A_ i$ be ideals such that $\varphi (I_1^ c) \subset I_2$ for some $c \geq 1$. This induces ring maps $A_{1, cn} = A_1/I_1^{cn} \to A_2/I_2^ n = A_{2, n}$ for all $n \geq 1$. Let $\mathcal{C}_ i$ be the category (86.2.0.1) for $(A_ i, I_ i)$. There is a base change functor

86.2.3.1
\begin{equation} \label{restricted-equation-base-change-systems} \mathcal{C}_1 \longrightarrow \mathcal{C}_2,\quad (B_ n) \longmapsto (B_{cn} \otimes _{A_{1, cn}} A_{2, n}) \end{equation}

Let $\mathcal{C}_ i'$ be the category (86.2.0.2) for $(A_ i, I_ i)$. If $I_2$ is finitely generated, then there is a base change functor

86.2.3.2
\begin{equation} \label{restricted-equation-base-change-complete} \mathcal{C}_1' \longrightarrow \mathcal{C}_2',\quad B \longmapsto (B \otimes _{A_1} A_2)^\wedge \end{equation}

because in this case the completion is complete (Algebra, Lemma 10.95.3). If both $I_1$ and $I_2$ are finitely generated, then the two base change functors agree via the functors (86.2.0.3) which are equivalences by Lemma 86.2.1.

Remark 86.2.4 (Base change by closed immersion). Let $A$ be a Noetherian ring and $I \subset A$ an ideal. Let $\mathfrak a \subset A$ be an ideal. Denote $\bar A = A/\mathfrak a$. Let $\bar I \subset \bar A$ be an ideal such that $I^ c \bar A \subset \bar I$ and $\bar I^ d \subset I\bar A$ for some $c, d \geq 1$. In this case the base change functor (86.2.3.2) for $(A, I)$ to $(\bar A, \bar I)$ is given by $B \mapsto \bar B = B/\mathfrak aB$. Namely, we have

86.2.4.1
\begin{equation} \label{restricted-equation-base-change-to-closed} \bar B = (B \otimes _ A \bar A)^\wedge = (B/\mathfrak a B)^\wedge = B/\mathfrak a B \end{equation}

the last equality because any finite $B$-module is $I$-adically complete by Algebra, Lemma 10.96.1 and if annihilated by $\mathfrak a$ also $\bar I$-adically complete by Algebra, Lemma 10.95.9.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).