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In our algebra class, we recently talked briefly about finitely generated modules, and the instructor referred to how even if a $R$-module admits a basis, we can easily have that different bases of the module are not in bijection with each other. This brought to mind a different question: For a ring $R$ and a finitely generated ideal $I$, need all minimal generators of $I$ be in bijection? That is, if $S, T \subseteq I$ are two finite sets such that $I = (S) = (T)$, and moreover such that any proper subset of $S$ or $T$ will not generate $I$ (i.e. if $S$ and $T$ are minimal), then does there exist a bijection between $S$ and $T$? I raised the question with a friend in my class who's often interested in these kinds of questions, and he brought up that any (left) ideal $I$ would carry a natural structure as a (left) $R$-module, so if there existed a finite ideal $I$ that would induce a module with non-bijective bases, then the answer would be to the negative, but I have no intuition for whether that would be possible. If the answer is in general to the negative, then is there a natural assumption that could be made about $R$ so that the conjecture does hold?

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    It's possible (although I wouldn't really call it "easy") for two finitely generated free modules over a ring to have different size bases. Rings which do this are said to lack the Invariant Basis Number property. I only know two families of examples for this. – 2017-02-16

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For a ring $R$ and a finitely generated ideal $I$, need all minimal generators of $I$ be in bijection? That is, if $S, T \subseteq I$ are two finite sets such that $I = (S) = (T)$, and moreover such that any proper subset of $S$ or $T$ will not generate $I$ (i.e. if $S$ and $T$ are minimal), then does there exist a bijection between $S$ and $T$?

This isn't even true for $\mathbb Z$. Consider $(1)$ and $(2,3)$.

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    It's as easy as $1,2,3$ `¯\_(ツ)_/¯` – 2017-02-16