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
Morphisms in $\mathcal{C}$ are given by systems of homomorphisms. Let $\mathcal{C}'$ be the category
Morphisms in $\mathcal{C}'$ are $A$-algebra maps. There is a functor
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 88.2.1. Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. The functor is a quasi-inverse to (88.2.0.3). The completions $A[x_1, \ldots , x_ r]^\wedge $ are in $\mathcal{C}'$ and any object of $\mathcal{C}'$ is of the form for some ideal $J \subset A[x_1, \ldots , x_ r]^\wedge $.
\[ \mathcal{C} \longrightarrow \mathcal{C}',\quad (B_ n) \longmapsto B = \mathop{\mathrm{lim}}\nolimits B_ n \]
\[ B = A[x_1, \ldots , x_ r]^\wedge / J \]
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 110.8.1. Next we show this does not happen when $A$ is Noetherian.
Lemma 88.2.2. Let $A$ be a Noetherian ring and let $I \subset A$ be an ideal. Then
every object of the category $\mathcal{C}'$ (88.2.0.2) 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 88.2.1 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 88.2.3 (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 (88.2.0.1) for $(A_ i, I_ i)$. There is a base change functor Let $\mathcal{C}_ i'$ be the category (88.2.0.2) for $(A_ i, I_ i)$. If $I_2$ is finitely generated, then there is a base change functor 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 (88.2.0.3) which are equivalences by Lemma 88.2.1.
88.2.3.1
\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}
88.2.3.2
\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}
Remark 88.2.4 (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 (88.2.3.2) for $(A, I)$ to $(\bar A, \bar I)$ is given by $B \mapsto \bar B = B/\mathfrak aB$. Namely, we have 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.
88.2.4.1
\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}
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