In a Noetherian ring, any given generating set is equivalent to some of its finite subsets

Suppose S \subseteq R is a subset and (S) a finitely generated ideal. Then there exists a finite subset A \subseteq S such that (S) = (A).

Let (S) \subseteq R be a finitely generated ideal; say (S) = (B) for some finite set B = \{b_i\}_{i=0}^n \subseteq R. Note that each b_i \in B has the form b_i = \sum_{j_i=0}^{n_i} r_{i,j_i}a_{i,j_i} for some r_{i,j_i} \in R and a_{i,j_i} \in S. Let A = \{a_{i,j_i} \ |\ 0 \leq i \leq n, 0 \leq j_i \leq n_i \}; note that A is finite. Moreover, we have (S) = (B) \subseteq (A) \subseteq (S), so that (S) = (A).

In particular, if R is a Noetherian ring, then every generating set may be “cut down” to one of its finite subsets.

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