This question is motivated by the following problem from a problem set on ring theory:
Prove that, if $R$ is a Noetherian commutative ring, and $I$ is its ideal, $R/I$ is a again Noetherian.
My first thought was: "of course, let $a_{1}, ..., a_{n}$ be a minimal generating set of $I$. Let $b_{1}, ..., b_{n'}$ be a minimal generating set of some other ideal $J$. Since the ring is commutative, $I\cap J$ is again an ideal, which is, again, finitely generated, since the ring is Noetherian, so there is some common set of generators $c_{1},...,c_{k}$ that belong to both $I$ and $J$ and generate $I\cap J$, and that set can be extended to both a generating set of $I$, $c_{1},\cdots, c_{k}, \alpha_{k+1}, \cdots, \alpha_{l}$ and a generating set of $J$, $c_{1}, \cdots, c_{k}, \beta_{k+1}, \cdots, \beta_{m}$, in such a way that no linear combination of $\alpha_{i}$'s and $c_{i}$'s with coefficients in $R$ gives an element $\beta_{j}$. Then $\pi(\beta_{1}), \cdots, \pi(\beta_{l})$ is a generating set of $\pi(J)$, where $\pi$ is the canonical projection $\pi: R \to R/I$."
If generating sets were working like bases, and ideals like subspaces, this would prove the statement, since $\alpha_{i}$ are "linearly independent" from $\beta_{j}$, and projections don't create new subspaces. In this case, I realized that in some of the previous problem sets there was a task to prove that, if $\phi: R_{1}\to R_{2}$ is a surjective morphism of rings, there is a bijection between ideals of $R_{2}$ and ideals of $R_{1}$ containing the kernel of $\phi$, which in this case gives a bijection between ideals of $R/I$ and ideals of $R$ containing $I$, so there is no need for considering the part containing $\alpha_{i}$, and the projections of $\beta_{i}$ indeed form generating sets in all ideals of $R/I$, proving $R/I$ to be Noetherian.
But this made me wondering, what properties of bases generalize to generating sets of ideals? Is it true, for instance, that, if $I \subset J$ and $I, J$ are finitely generated ideals, a minimal generating set of $I$ has smaller size than a minimal generating set of $J$? Does a minimal generating set of an ideal have to have a fixed number of elements, and does it then make any sense to talk of dimension of an ideal? Is it always true that a minimal generating set of a subideal can be extended to a minimal generating set of the bigger ideal? If two ideals, $I, J$ have linearly independent minimal generating sets, i.e. no element of one is a linear combination of another, and vice versa, is it true that $\pi(I)\cong I$ when we factor by $J$? What are, in general, the useful properties of bases of linear spaces that generalize to generating sets, and what properties do not?
Edit: Of course, I mean the ideals not coinciding with the ring itself, since for a ring $R$ $\langle 1_{R}\rangle = R$ and the answers to almost all questions are obviously negative.
No: $\langle X^2, Y^2 \rangle \subsetneq \langle 1 \rangle = R[X, Y]$
It should help to look here, here, here, and here
If by 'extend' here one means 'taking unions while preserving linear independence', then no:
$\langle X^2 \rangle \subsetneq \langle X \rangle \subsetneq R[X, Y]$