Lemma 10.97.5. Let $I$ be an ideal of a ring $R$. Assume

$R/I$ is a Noetherian ring,

$I$ is finitely generated.

Then the completion $R^\wedge $ of $R$ with respect to $I$ is a Noetherian ring complete with respect to $IR^\wedge $.

Lemma 10.97.5. Let $I$ be an ideal of a ring $R$. Assume

$R/I$ is a Noetherian ring,

$I$ is finitely generated.

Then the completion $R^\wedge $ of $R$ with respect to $I$ is a Noetherian ring complete with respect to $IR^\wedge $.

**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

\[ \bigoplus J \cap I^ n/J \cap I^{n + 1} \subset \bigoplus I^ n/I^{n + 1}. \]

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

\[ x = \sum \nolimits _{n = 0, \ldots , N} \sum \nolimits _{j = 1, \ldots , m} a_{j, n} g_ j \bmod I^{N + 1} \]

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. $\square$

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