Lemma 90.4.8. Let $\ldots \to A_3 \to A_2 \to A_1$ be a sequence of surjective ring maps in $\mathcal{C}_\Lambda $. If $\dim _ k (\mathfrak m_{A_ n}/\mathfrak m_{A_ n}^2)$ is bounded, then $S = \mathop{\mathrm{lim}}\nolimits A_ n$ is an object in $\widehat{\mathcal{C}}_\Lambda $ and the ideals $I_ n = \mathop{\mathrm{Ker}}(S \to A_ n)$ define the $\mathfrak m_ S$-adic topology on $S$.

**Proof.**
We will use freely that the maps $S \to A_ n$ are surjective for all $n$. Note that the maps $\mathfrak m_{A_{n + 1}}/\mathfrak m_{A_{n + 1}}^2 \to \mathfrak m_{A_ n}/\mathfrak m_{A_ n}^2$ are surjective, see Lemma 90.4.2. Hence for $n$ sufficiently large the dimension $\dim _ k (\mathfrak m_{A_ n}/\mathfrak m_{A_ n}^2)$ stabilizes to an integer, say $r$. Thus we can find $x_1, \ldots , x_ r \in \mathfrak m_ S$ whose images in $A_ n$ generate $\mathfrak m_{A_ n}$. Moreover, pick $y_1, \ldots , y_ t \in S$ whose images in $k$ generate $k$ over $\Lambda $. Then we get a ring map $P = \Lambda [z_1, \ldots , z_{r + t}] \to S$, $z_ i \mapsto x_ i$ and $z_{r + j} \mapsto y_ j$ such that the composition $P \to S \to A_ n$ is surjective for all $n$. Let $\mathfrak m \subset P$ be the kernel of $P \to k$. Let $R = P^\wedge $ be the $\mathfrak m$-adic completion of $P$; this is an object of $\widehat{\mathcal{C}}_\Lambda $. Since we still have the compatible system of (surjective) maps $R \to A_ n$ we get a map $R \to S$. Set $J_ n = \mathop{\mathrm{Ker}}(R \to A_ n)$. Set $J = \bigcap J_ n$. By Lemma 90.4.7 we see that $R/J = \mathop{\mathrm{lim}}\nolimits R/J_ n = \mathop{\mathrm{lim}}\nolimits A_ n = S$ and that the ideals $J_ n/J = I_ n$ define the $\mathfrak m$-adic topology. (Note that for each $n$ we have $\mathfrak m_ R^{N_ n} \subset J_ n$ for some $N_ n$ and not necessarily $N_ n = n$, so a renumbering of the ideals $J_ n$ may be necessary before applying the lemma.)
$\square$

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