The Stacks project

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
\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
\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
\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 ( 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.


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

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

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

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

  4. 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). $\square$

Remark 87.2.3 (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 ( for $(A_ i, I_ i)$. There is a base change functor
\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 ( for $(A_ i, I_ i)$. If $I_2$ is finitely generated, then there is a base change functor
\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 10.96.3). If both $I_1$ and $I_2$ are finitely generated, then the two base change functors agree via the functors ( which are equivalences by Lemma 87.2.1.

Remark 87.2.4 (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 ( for $(A, I)$ to $(\bar A, \bar I)$ is given by $B \mapsto \bar B = B/\mathfrak aB$. Namely, we have
\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 10.97.1 and if annihilated by $\mathfrak a$ also $\bar I$-adically complete by Algebra, Lemma 10.96.9.

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