## 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.