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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:
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 $e_A=\lceil \log_q \rceil$?
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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:
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 $A^{mn} \otimes simeq A^m \otimes_A A^n$ 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 $e_A = \max_i e_{A/\mathfrak{m}_i}$.
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 $e_A=\lceil \log_q \rceil$?
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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:
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 $A^{mn} \otimes A^m \otimes_A A^n$ 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 $e_A = \max_i e_{A/\mathfrak{m}_i}$.
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 $e_A=\lceil \log_q \rceil$?
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