Minimum number of generators of a module and its 'local dimension'

Question

I am struggling to prove that a certain module cannot be generated by too few elements. I have came up with these questions while tackling the problem.

Let $A$ be a ring, and $M$ a finitely generated $A$-module. Define

• $mg_A(M) = \min\{n: \exists \ m_1, \ldots, m_n: M = \langle m_1, \ldots,m_n\rangle_A\}$ its minimum number of generators;
• $\displaystyle lg_A(M) = \max_{\mathfrak{p} \in Spec(A)} mg_{A_{\mathfrak{p} }}( M_{\mathfrak{p}}) $ the maximum of the 'local' number of generators. Note that by Nakayama $$ mg_{A_{\mathfrak{p} }}( M_{\mathfrak{p}})= \dim_{k(A/\mathfrak{p}) } (M_{\mathfrak{p}}/\mathfrak{p}M_{\mathfrak{p}})$$ where $k(A/\mathfrak{p}) = A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$ is the quotient field of $A_{\mathfrak{p}}$. This is much easier to calculate.

It is easy to prove that $mg_A(M) \ge lg_A(M)$, because a set of generators for $M$ also generates $M_{\mathfrak{p}}$. I am interested in understanding for which rings the other inequality holds. We will call $A$ locally ranked if $lg_A(M) = mg_A(M)$ holds for every module $M$. I list for you what I have discovered and what I would like to discover:

1. If $A$ is local, then it is (trivially) locally ranked.

2. (product) If $A,B$ are locally ranked, then so it is $A \times B$. This essentially comes from the fact that an $A \times B$ module is of the form $M_A \times M_B$, and $$mg_{A \times B} (M_A \times M_B ) = \max(mg_A(M_A),mg_B(M_B))$$

3. Using (1), (2), we get that an artinian ring, which is a product of local rings, is locally ranked.

4. If $A$ is PID, then using the canonical (Smith?) form for A-modules it is easy to prove that A is locally ranked. In particular, $k[X]$ is locally ranked if $k$ is a field.

5. (quotient) If A is locally ranked and $I$ an ideal, then so is $A/I$. In fact an $A/I$ module is an $A$ module $M$ such that $IM=0$. If $\mathfrak{p}$ realizes the $mg_A(M)$, then $I \subseteq \mathfrak{p}$ . Otherwise there would exist $i \in I\setminus \mathfrak{p}$, an invertible element of $A_{\mathfrak{p}}$ such that $iM_{\mathfrak{p}}=0$; so $M_{\mathfrak{p}}=0$ and then $M=0$ because it is generated by $0=mg_{A_{\mathfrak{p}}}(M_{\mathfrak{p}})$ elements.

   Essentially, I don't know how to raise the (Krull) dimension of A. Here are my three questions:

6. If $A,B$ are locally ranked $k$-algebras, is $A \otimes_k B$ so?

7. If $A$ is locally ranked, is $A[x]$ so?

8. Is $\mathbb{Z}[x]$ locally ranked?

(1-5) were really 'natural'; these seem much harder to me. In particular, note that (4)+(6) or (7) give th result for $k[x_1, \ldots, x_n]$, and together with (5) give any finitely generated k-algebra. This can be useful for counterexamples.

Thank you for the aid.

Answer 1

Note that if $A$ is locally ranked and $M$ is a finitely generated projective $A$-module of constant rank, then $M$ must be free. Indeed, if $M$ has rank $n$, then $M$ is locally free on $n$ generators everywhere, so since $A$ is locally ranked, $M$ is generated by $n$ elements. These $n$ elements then freely generate $M$ locally, and hence freely generate $M$ globally.

This gives lots of examples of rings that are not locally ranked. For instance, if $A$ is a Dedekind domain but not a PID, then $A$ is not locally ranked (since any nonprincipal ideal is projective but not free). This gives counterexamples to your (6), (7), and (8). For instance, $\mathbb{Z}[\sqrt{-5}]$ is a quotient of $\mathbb{Z}[x]$ which is a Dedekind domain but not a PID, so $\mathbb{Z}[x]$ cannot be locally ranked. Similarly, the coordinate ring of any smooth plane curve of positive genus is a quotient of $k[x,y]$ which is a Dedekind domain but not a PID, so $k[x,y]$ cannot be locally ranked.