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
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
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