MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $e_A=\lceil \log_q \rceil$?

share|cite|improve this question
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 $e_{\mathbb Z}(3)\ge 2$. – Qing Liu Nov 17 '12 at 11:15
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 ${\bf A}^e_S$. – François Brunault Nov 17 '12 at 23:33
@François, I guess you mean $e_S(n)$. – Qing Liu Nov 18 '12 at 0:04
@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). – François Brunault Nov 18 '12 at 0:31
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 $A$. – François Brunault Nov 19 '12 at 14:01

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.

share|cite|improve this answer
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. – Qing Liu Nov 18 '12 at 22:45
What can be said about the eventual value $\lim_{n \to \infty} e_A(n)$ when $A$ has only infinite residue fields? – Martin Brandenburg Nov 19 '12 at 11:15
I don't know, even when $A$ is noetherian (of infinite dimension). – Qing Liu Nov 19 '12 at 12:48
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 $>1$? – Martin Brandenburg Nov 19 '12 at 15:30
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. – Qing Liu Nov 20 '12 at 14:07

The paper and the references in it should be of interest to you. Please do not hesitate to contact me with any questions.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.