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:

  1. 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 $(1,0)$.

  2. 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.

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

  4. 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 $A$.

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

  6. 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 $e_K=1$.

  7. 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 $A=\mathbb{Z}$).

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

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

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

  11. 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 $e_A = \max_i e_{A/\mathfrak{m}_i}$.

  12. 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 $e_A=\lceil \log_q \rceil$?

  • 1
    $\begingroup$ A related number is, given $m\ge 1$, how many disjoint sections you can put in the affine space $\mathbb A^m$ over $A$. The maximal number is the greastest $n$ such that $e_A(n)\le m$. For exemple, in $\mathbb A^1_{\mathbb Z}$, you can't put three disjoint sections. So $e_{\mathbb Z}(3)\ge 2$. $\endgroup$
    – Qing Liu
    Nov 17, 2012 at 11:15
  • 2
    $\begingroup$ The number $e_A(n)$ makes sense for any scheme $S$ : it is the smallest integer $e$ such that there is a closed immersion from $S \sqcup \ldots \sqcup S$ ($n$ times) into the affine space ${\bf A}^e_S$. $\endgroup$ Nov 17, 2012 at 23:33
  • $\begingroup$ @François, I guess you mean $e_S(n)$. $\endgroup$
    – Qing Liu
    Nov 18, 2012 at 0:04
  • $\begingroup$ @Qing Liu : Yes, I should have written $e_S(n)$. There is also the related number $n_S(e)$ which one can define as in your first comment (fixing the dimension of the affine space and trying to put the maximal number of disjoint sections). $\endgroup$ Nov 18, 2012 at 0:31
  • $\begingroup$ Write $A$ as an inductive limit of finitely generated algebras $A_0$ over ${\bf Z}$. Then $e_A = \min{A_0} e_{A_0}$ pointwise. So in some sense one is reduced to the case of f.g. algebras. Don't know if this helps : it is not clear how to deduce a result purely in terms of $A$. $\endgroup$ Nov 19, 2012 at 14:01

2 Answers 2


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.

  • $\begingroup$ There is gap in my proof in this part. Enlarging $A_0$ can introduce maximal ideals that I can't control. So I will delete the part concerning non-necessarily noetherian rings. $\endgroup$
    – Qing Liu
    Nov 18, 2012 at 22:45
  • $\begingroup$ What can be said about the eventual value $\lim_{n \to \infty} e_A(n)$ when $A$ has only infinite residue fields? $\endgroup$ Nov 19, 2012 at 11:15
  • $\begingroup$ I don't know, even when $A$ is noetherian (of infinite dimension). $\endgroup$
    – Qing Liu
    Nov 19, 2012 at 12:48
  • $\begingroup$ Let's assume that $A$ is noetherian of finite dimension $d$. Then your result just says that $e_A(n)$ is eventually constant with some value $< d$. How can we determine this, say, when $A$ is given by generators and relations over a field? And can you give an example where this value is $>1$? $\endgroup$ Nov 19, 2012 at 15:30
  • $\begingroup$ If $A$ has only infinite residue fields and contains a field $k$, then probably $A$ contains also an infinite field ? Anyway I think this is not the hardest case. Staying with the hypothesis on the residue fields, if we want $e_A(n)=1$ for all $n\ge 2$, then this means that for any $n$, there are $n$ units in $A$ whose differences are all units. This is true for example if $A$ is the ring of all algebraic integers thanks to cyclotomic units $(\xi^n-1)/(\xi -1)$. So far I don't have an example with $e_A(n)>1$. But this is an interesting question. $\endgroup$
    – Qing Liu
    Nov 20, 2012 at 14:07

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.


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