Minimal number of generators for $A^n$ Let $A$ be a commutative ring and $n \in \mathbb{N}$. What is the minimal number $e_A(n)$ of generators of the $A$-algebra $A^n$? Here is what I already know (I can add proofs if necessary) from a little joint work:

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*We have $e_A(n)=0$ for $n=0,1$, therefore let's exclude these trivial cases. We have $e_A(2) \leq 1$ since $A^2$ is generated by $(1,0)$.


*We have $e_A(mn) \leq e_A(m) + e_A(n)$ since $A^{mn} \simeq A^m \otimes_A A^n$ as $A$-algebras.


*The function $e_A : \mathbb{N} \to \mathbb{N}$ is non-decreasing.


*When there is a homomorphism $A \to B$, then $e_B \leq e_A$ holds pointwise. In particular we have $e_A = e_B$ when $B$ is an $A$-algebra with a section. This happens, for example, when $B$ is a polynomial ring over $A$.


*We have $e_A(n) \leq \lceil \log_2(n) \rceil$ (use 1,2,3).


*If $A$ has elements $\alpha_1,\dotsc,\alpha_n$ such that $\alpha_i - \alpha_j \in A^*$ for $i \neq j$, then $e_A(n)=1$ (use Vandermonde). In particular: If $K$ is a field with $\geq n$ elements, then $e_K(n)=1$. If $K$ is an infinite field, we therefore have $e_K=1$.


*For a finite field $\mathbb{F}_q$ we have $e_{\mathbb{F}_q}(n)=\lceil \log_q(n) \rceil$.
In particular, it follows (use 4,5,7) that $e_A = \lceil \log_2 \rceil$ when there is a    homomorphism $A \to \mathbb{F}_2$ (for example for $A=\mathbb{Z}$).


*We have $e_{A \otimes B} \leq \min(e_A,e_B)$ with equality for $A=B$. But this is not always an equality.


*We have $e_{A \times B} = \max(e_A,e_B)$.


*If $I \subseteq A$ is a nil ideal, then $e_A = e_{A/I}$. In particular, we may assume always that $A$ is reduced.


*If $A$ is a local ring with residue field $k$, then $e_A = e_k$. More generally, if $A$ has only finitely many maximal ideals $\mathfrak{m}_i$, then $e_A = \max_i e_{A/\mathfrak{m}_i}$.


*If $A=\mathrm{colim}_i A_i$ is a directed colimit, then $e_A = \min_i e_{A_i}$.
Questions. Is there any (geometric) intuition behind the number $e_A(n)$? How can we compute $e_A$ for other examples, or even for an arbitrary commutative ring? Is there always some $q \in \mathbb{N}$ such that $e_A=\lceil \log_q \rceil$?
 A: You already completely solved the question over fields, noetherian artinian rings and $\mathbb Z$. 
Let $A$ be any commutative unitary ring. Then the maximum $f_A(n)$ of all $e_k(n)$ when $k$ runs the residue fields of $A$ (at maximal ideals) satisfies clearly $e_A(n)\ge f_A(n)$ by your (4). 

Suppose $A$ is noetherian of dimension $d$, then 
  $$f_A(n) \le e_A(n) \le \max \{ d+1, f_A(n)\}.$$

Proof. Let $m=\max \{ d+1, f_A(n)\}$. We want to show that the affine space $\mathbb A^m$ over $A$ contains $n$ disjoint sections. Let $r\le n-1$ be such that $\mathbb A^m$ contains $r$ disjoint sections. We are going to show that $\mathbb A^m$ contains one more section disjoint from the previous one. This will prove the claim.
Let $T$ be the union of $r$ sections. For every residue field $k$ of $A$, $\mathbb A^m_k$ contains at least $r+1$ rational points. In particular, $T$ doesn't contain $\mathbb A^m_k(k)$. By hypothesis, we also have $\dim T=\dim A< m$. By Proposition 1.10 of this preprint, there is a section in $\mathbb A^m$ disjoint from $T$ and we are done.
Edit (Remove generalization to non-noetherian rings). 
Remark.  Let $A$ be any finite dimensional noetherian ring. If $A$ has a finite residue field, there exists $q$ such that $e_A(n)$ coincides asymptotically with $ \lceil \log_q n \rceil $. It is enough to take for $q$ the smallest cardinality of the finite residue fields of $A$. If all residue fields of $A$ are infinite, then  $e_A(n)$ is bounded hence asymptotically constant (because it is increasing). It would be interesting to decide whether these properties hold without noetherian and finite-dimensional hypothesis.
A: The paper http://msp.org/ant/2012/6-2/p03.xhtml and the references in it should be of interest to you. Please do not hesitate to contact me with any questions.
