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Let $I$ be an ideal of commutative ring $A$ with unity. Does $I$ have a minimal generating set?

At times, I am able to compute what they are for specific example, but it seems like it is true in general with some existence proof (I guess it is true for non-commutative rings with some adjectives such as "left" or "right").

I have thought about Zorn's lemma but generating sets are very likely to be incomparable (nor the existence of lower bound of each chain was not clear) and the intersection of all generating sets may be too small to generate an ideal. Sounds not too difficult, but it seems not clear what to do.

user26857
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user123454321
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2 Answers2

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Consider the ring: $$\mathbb Q\left[\{x^{1/n}\}_{n\in\mathbb Z^+}\right]$$

The maximal ideal consisting of the expressions with constant term zero, has no minimal set of generators - you can always remove any element of a set of generators and still have a set of generators.

Proving this is not entirely trivial. It is clear that the obvious set of generators does not have a minimal generating subset, so anything Zorn-like will fail.

Thomas Andrews
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Here is a slight modification of Thomas's example that is easy to prove. Let $R$ be a quasilocal ring with a nonzero idempotent maximal ideal $M$, e.g., the ring in Thomas's solution localized at his maximal ideal. Then $M$ has no minimal generating subset. Suppose to the contrary that it did have such a subset $\{a_i\}$. Then we have an expression $a_j = \sum_i x_ia_i$ with each $x_i \in M$. Solving for $a_j$ (note that $1-x_j$ is a unit), we see that it can be written as an $R$-linear combination of the other $a_i$'s, a contradiction.

Jason Juett
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    More generally, by the same argument, any nonzero ideal I of a ring R with I = J(R)I does not have a minimal generating set. – Jason Juett Aug 24 '15 at 22:33
  • you should mention in your hypothesis that $M$ is not principal .... if $M$ is principal, then you have no contradiction ... btw, do you know of any other example of such a ring other than the example in Thomas's solution ? – user521337 Feb 05 '19 at 23:04
  • Actually the hypotheses imply that $M$ is not finitely generated. (By Nakayama's Lemma or the argument of the proof, a quasilocal ring cannot have any nonzero idempotent finitely generated ideals.) – Jason Juett Feb 05 '19 at 23:49
  • I follow the convention of using "quasilocal" to mean the ring has a unique maximal ideal, without implying anything about whether the ring is Noetherian or not. I personally use "local" to mean "quasilocal plus Noetherian". – Jason Juett Feb 05 '19 at 23:56
  • The method of constructing examples by localizing monoid domains seems to be the easiest to describe and prove, but certainly that's not the only way to get a quasilocal ring with an idempotent maximal ideal. (Note that my example above is technically a slight modification of Thomas's idea.) Here is another pretty similar method. Consider $\mathbb{Q}[[\mathbb{Q}^+]]$, a "generalized power series ring" in the sense of Ribenboim. It is quasilocal with an idempotent maximal ideal. Or pick any valuation domain with a non-principal maximal ideal. – Jason Juett Feb 05 '19 at 23:58
  • Let us continue this discussion in chat. – Jason Juett Feb 06 '19 at 00:03