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In a Noetherian ring, every ideal is finitely generated. Suppose an ideal $I$ in a Noetherian ring $R$ is generated by a set of generators $S$. If $S$ is infinite, does it have a finite subset that generates $I$.

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up vote 10 down vote accepted

Suppose $G$ is an infinite set generating your ideal $I$ such that no finite subset also generates $I$. Let $g_0\in G$. Since $g_0$ does not generate $I$, there is a $g_1\in G$ such that $(g_0)\subsetneq(g_0,g_1)$. Since $(g_0,g_1)\neq I$, there is a $g_2\in G$ such that $(g_0,g_1)\subsetneq(g_0,g_1,g_2)$. Etc.

This way you construct a strictly increasing sequence of ideals, and such a thing does not exist because the ring is noetherian.

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Yes. Let $R$ be the noetherian ring, let $I$ be the ideal in question, and let $\{s_\lambda\}$ be the given (infinite) set of generators, and let $a_1,\dots,a_n$ be the finite set of generators that generate $I$. Since the $s_\lambda$ generate $I$, each $a_i$ can be written as a finite linear combination of the $s_\lambda$: $$ a_i=\sum_{j=1}^{k_i} r_{ij}s_{\lambda_{ij}} $$ Then the total number of the $s_\lambda$ that are used to represent all the $a_i$ is finite, and these generate the whole ideal.

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Let $S$ be the set that generates $I$. Let $\mathcal{I}$ be the collection of all ideals that are generated by finite subsets of $S$. Since the ring is noetherian and this set is nonempty, it contains a maximal element $J$, which is generated by some finite subset $S_0$ of $S$. By maximality of $J$, for every $s\in S$ we have that $J\subseteq \langle S_0,s\rangle$, hence $\langle S_0,s\rangle= J$. Thus, for every $s\in S$, $s\in J$, so $I=\langle S\rangle\subseteq J\subseteq I$, proving equality. Thus, $I$ is generated by a finite subset of $S$.

Added. I am using the following equivalence:

Theorem. Let $R$ be a ring; the following are equivalent:

  1. Every left (resp. right, two-sided) ideal of $R$ is finitely generated.
  2. Every ascending chain of left (resp. right, two-sided) ideals of $R$ stabilizes.
  3. Every nonempty family of left (resp. right, two-sided) ideals of $R$ has a maximal element under the partial order given by inclusion.

Some of the implications use the Axiom of Choice.

(1)$\Rightarrow$(2) is Choice-free: take the union of the chain; it is an ideal, so it is finitely generated; there is an index that contains all generators of the union, and that's where the chain stabilizes.

(2)$\Rightarrow$(1) I think uses at least some Choice: take a non-finitely generated $I$. Pick $a_1\in I$; then pick $a_2\in I-\langle a_1\rangle$, etc, to construct an infinite ascending chain.

(1)$\Rightarrow$(3) uses at least some Choice: one uses (1) to prove the hypotheses of Zorn's Lemma hold in family (via (2)).

(3)$\Rightarrow$(1) is Choice-free: take the family of finitely generated ideals contained in a given $I$, and proceed as I did above.

(3)$\Rightarrow$(2) is Choice-free: take the elements of the chain as a family.

(2)$\Rightarrow$(3) generally uses at least some Choice: use (2) to establish that the hypothesis of Zorn's Lemma hold, but may require only Countable Choice.

But most ring theorists I know are perfectly happy to work with the Axiom of Choice.

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If $I=(y_1,y_2,...,y_k)$ is a finitely-generated ideal, and $I$ is also generated by a set $X$, then each $y_i$ is a linear combination of finitely many elements of $X$ - that is, there is a finite set $X_i\subset X$ such that $y_i=\sum_{x\in X_i} a_x x$ for some set of $a_x\in R$.

But then $\cup X_i$ is a finite subset of $X$ which generates each of the $y_i$, and hence all of $I$.

Since all ideals in a Noetherian ring are finitely generated, we are done.

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