The Stacks project

Proof. Setting $M = N/K$ we find that part (1) follows from part (2). Let $0 \to K \to N \to M \to 0$ be as in (2). For each $n$ we get the short exact sequence

By Lemma 10.87.1 we obtain the exact sequence

By the Artin-Rees Lemma 10.51.2 we may choose $c$ such that $I^ nK \subset I^ n N \cap K \subset I^{n-c} K$ for $n \geq c$. Hence $K^\wedge = \mathop{\mathrm{lim}}\nolimits K/I^ nK = \mathop{\mathrm{lim}}\nolimits K/(I^ nN \cap K)$ and we conclude that (2) is true.

Let $M$ be as in (3) and let $0 \to K \to R^{\oplus t} \to M \to 0$ be a presentation of $M$. We get a commutative diagram

The top row is exact, see Section 10.39. The bottom row is exact by part (2). By Lemma 10.96.1 the vertical arrows are surjective. The middle vertical arrow is an isomorphism. We conclude (3) holds by the Snake Lemma 10.4.1. $\square$

Proof. Consider $J \otimes _ R R^\wedge \to R \otimes _ R R^\wedge = R^\wedge$ where $J$ is an arbitrary ideal of $R$. According to Lemma 10.97.1 this is identified with $J^\wedge \to R^\wedge$ and $J^\wedge \to R^\wedge$ is injective. Part (1) follows from Lemma 10.39.5. Part (2) is a reformulation of Lemma 10.97.1 part (2). $\square$

Proof. By Lemma 10.97.2 it is flat. The composition $R \to R^\wedge \to R/I$ where the last map is the projection map $R^\wedge \to R/I$ shows that any maximal ideal of $R$ is in the image of $\mathop{\mathrm{Spec}}(R^\wedge ) \to \mathop{\mathrm{Spec}}(R)$. Hence the map is faithfully flat by Lemma 10.39.15. $\square$

Proof. This is a special case of Lemma 10.96.3 because $I$ is a finitely generated ideal. $\square$

Proof. By Lemma 10.96.3 we see that $R^\wedge$ is $I$-adically complete. Hence it is also $IR^\wedge$-adically complete. Since $R^\wedge /IR^\wedge = R/I$ is Noetherian we see that after replacing $R$ by $R^\wedge$ we may in addition to assumptions (1) and (2) assume that also $R$ is $I$-adically complete.

Let $f_1, \ldots , f_ t$ be generators of $I$. Then there is a surjection of rings $R/I[T_1, \ldots , T_ t] \to \bigoplus I^ n/I^{n + 1}$ mapping $T_ i$ to the element $\overline{f}_ i \in I/I^2$. Hence $\bigoplus I^ n/I^{n + 1}$ is a Noetherian ring. Let $J \subset R$ be an ideal. Consider the ideal

Let $\overline{g}_1, \ldots , \overline{g}_ m$ be generators of this ideal. We may choose $\overline{g}_ j$ to be a homogeneous element of degree $d_ j$ and we may pick $g_ j \in J \cap I^{d_ j}$ mapping to $\overline{g}_ j \in J \cap I^{d_ j}/J \cap I^{d_ j + 1}$. We claim that $g_1, \ldots , g_ m$ generate $J$.

Let $x \in J \cap I^ n$. There exist $a_ j \in I^{\max (0, n - d_ j)}$ such that $x - \sum a_ j g_ j \in J \cap I^{n + 1}$. The reason is that $J \cap I^ n/J \cap I^{n + 1}$ is equal to $\sum \overline{g}_ j I^{n - d_ j}/I^{n - d_ j + 1}$ by our choice of $g_1, \ldots , g_ m$. Hence starting with $x \in J$ we can find a sequence of vectors $(a_{1, n}, \ldots , a_{m, n})_{n \geq 0}$ with $a_{j, n} \in I^{\max (0, n - d_ j)}$ such that

Setting $A_ j = \sum _{n \geq 0} a_{j, n}$ we see that $x = \sum A_ j g_ j$ as $R$ is complete. Hence $J$ is finitely generated and we win. $\square$

Proof. This is a consequence of Lemma 10.97.5. It can also be seen directly as follows. Choose generators $f_1, \ldots , f_ n$ of $I$. Consider the map

This is a well defined and surjective ring map (details omitted). Since $R[[x_1, \ldots , x_ n]]$ is Noetherian (see Lemma 10.31.2) we win. $\square$

Proof. We have $\mathfrak mS \subset \mathfrak n$ because $R \to S$ is a local ring map. The assumption $\dim _{\kappa (\mathfrak m)} S/\mathfrak mS < \infty$ implies that $S/\mathfrak mS$ is an Artinian ring, see Lemma 10.53.2. Hence has dimension $0$, see Lemma 10.60.5, hence $\mathfrak n = \sqrt{\mathfrak mS}$. This and the fact that $\mathfrak n$ is finitely generated implies that $\mathfrak n^ t \subset \mathfrak mS$ for some $t \geq 1$. By Lemma 10.96.9 we see that $S^\wedge$ can be identified with the $\mathfrak m$-adic completion of $S$. As $\mathfrak m$ is finitely generated we see from Lemma 10.96.3 that $S^\wedge$ and $R^\wedge$ are $\mathfrak m$-adically complete. At this point we may apply Lemma 10.96.12 to $S^\wedge$ as an $R^\wedge$-module to conclude. $\square$

Proof. The first equality follows from Lemma 10.97.1. We may replace $R$ by the localization $R_\mathfrak p$ and $S$ by $S_\mathfrak p = S \otimes _ R R_\mathfrak p$. Hence we may assume that $R$ is a local Noetherian ring and that $\mathfrak p = \mathfrak m$ is its maximal ideal. The $\mathfrak q_ iS_{\mathfrak q_ i}$-adic completion $S_{\mathfrak q_ i}^\wedge$ is equal to the $\mathfrak m$-adic completion by Lemma 10.97.7. For every $n \geq 1$ prime ideals of $S/\mathfrak m^ nS$ are in 1-to-1 correspondence with the maximal ideals $\mathfrak q_1, \ldots , \mathfrak q_ m$ of $S$ (by going up for $S$ over $R$, see Lemma 10.36.22). Hence $S/\mathfrak m^ nS = \prod S_{\mathfrak q_ i}/\mathfrak m^ nS_{\mathfrak q_ i}$ by Lemma 10.53.6 (using for example Proposition 10.60.7 to see that $S/\mathfrak m^ nS$ is Artinian). Hence the $\mathfrak m$-adic completion $S^\wedge$ of $S$ is equal to $\prod S_{\mathfrak q_ i}^\wedge$. Finally, we have $R^\wedge \otimes _ R S = S^\wedge$ by Lemma 10.97.1. $\square$

Proof. As $M$ is flat, each of the sequences

is short exact, see Lemma 10.39.12 and the sequence $0 \to K^\wedge \to P^\wedge \to M^\wedge \to 0$ is a short exact sequence, see Lemma 10.96.1. It suffices to show that we can find splittings $s_ n : M/I^ nM \to P/I^ nP$ such that $s_{n + 1} \bmod I^ n = s_ n$. We will construct these $s_ n$ by induction on $n$. Pick any splitting $s_1$, which exists as $M/IM$ is a projective $R/I$-module. Assume given $s_ n$ for some $n > 0$. Set $P_{n + 1} = \{ x \in P \mid x \bmod I^ nP \in \mathop{\mathrm{Im}}(s_ n)\}$. The map $\pi : P_{n + 1}/I^{n + 1}P_{n + 1} \to M/I^{n + 1}M$ is surjective (details omitted). As $M/I^{n + 1}M$ is projective as a $R/I^{n + 1}$-module by Lemma 10.77.7 we may choose a section $t : M/I^{n + 1}M \to P_{n + 1}/I^{n + 1}P_{n + 1}$ of $\pi$. Setting $s_{n + 1}$ equal to the composition of $t$ with the canonical map $P_{n + 1}/I^{n + 1}P_{n + 1} \to P/I^{n + 1}P$ works. $\square$

Proof. Let $B$ be the $(I + J)$-adic completion of $A$. By Lemma 10.97.2 $B/IB$ is the $J$-adic completion of $A/I$ hence isomorphic to $A/I$ by assumption. Moreover $B$ is $I$-adically complete by Lemma 10.96.8. Hence $B$ is a finite $A$-module by Lemma 10.96.12. By Nakayama's lemma (Lemma 10.20.1 using $I$ is in the Jacobson radical of $A$ by Lemma 10.96.6) we find that $A \to B$ is surjective. The map $A \to B$ is flat by Lemma 10.97.2. The image of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ contains $V(I)$ and as $I$ is contained in the Jacobson radical of $A$ we find $A \to B$ is faithfully flat (Lemma 10.39.16). Thus $A \to B$ is injective. Thus $A$ is complete with respect to $I + J$, hence a fortiori complete with respect to $J$. $\square$