2
$\begingroup$

In my lecture notes, I found the following proposition. I believe that it is incorrect.

Proposition (Nakayama for Local Rings). Let $(R, \mathfrak{m})$ be a commutative local ring with residue field $\mathcal{K} = R/\mathfrak{m}$, and let $p: M \to M/\mathfrak{m}M$ be the canonical projection. Then $\{p(x_1), \ldots, p(x_n)\}$ is a basis of the $\mathcal{K}$-vector space $M/\mathfrak{m}M$ if and only if $\{x_1, \ldots, x_n\}$ is a minimal set of generators for $M$.

Indeed, let $(R, \mathfrak{m})$ be a local ring such that $\mathfrak{m} \neq \{0\}$. Let $a \in \mathfrak{m} \setminus \{0\}$. Take $M = R$ and $x_1 = 1, x_2 = 1 + a$. Then $\{x_1, x_2\}$ is clearly not a minimal generating set for $M$. However, $\{p(x_1), p(x_2)\} = \{1, 1\} = \{1\}$ is a basis for $M/\mathfrak{m}M = R/\mathfrak{m}$.

Is my counterexample correct? If so, how can the claim be salvaged?

$\endgroup$
1
  • 1
    $\begingroup$ Interestingly, the statement is correct without the set braces! $\endgroup$
    – walkar
    Commented Nov 15 at 14:16

1 Answer 1

Reset to default
4
$\begingroup$

Yes. We should assume that the $p(x_i)$ are pairwise distinct, and surely $M$ must be finitely generated.

$\endgroup$
5
  • 1
    $\begingroup$ Do you need to assume finitely generated in the statement? As worded, the basis/putative minimal generating set are both required to be finite, so it says nothing about infinitely generated modules. $\endgroup$
    – walkar
    Commented Nov 15 at 14:17
  • 1
    $\begingroup$ Yes we need that for $\implies$ to use Nakayama. $\endgroup$
    – Martin Brandenburg
    Commented Nov 15 at 14:19
  • $\begingroup$ @MartinBrandenburg Interesting, do you have a counterexample where $M$ is infinitely generated but $M/\mathfrak{m}M$ is finite-dimensional? $\endgroup$
    – Smiley1000
    Commented Nov 15 at 14:41
  • 1
    $\begingroup$ Yes see here, where even $M/\mathfrak{m}M=0$. $\endgroup$
    – Martin Brandenburg
    Commented Nov 15 at 14:48
  • $\begingroup$ @MartinBrandenburg Thanks, so for example $\mathbb{Q}$ as a $\mathbb{Z}_{(p)}$-module is not finitely generated (in particular nonzero), but the $\mathbb{Z}_{(p)}/p\mathbb{Z}_{(p)} = \mathbb{F}_p$-vector space $\mathbb{Q}/p\mathbb{Q}$ is zero. $\endgroup$
    – Smiley1000
    Commented Nov 16 at 21:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .