Can every ideal have a minimal generating set?

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.

Solution 1:

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.

Solution 2:

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.