2
$\begingroup$

In the chapter on Artinian rings in "Introduction to Commutative Algebra" by Atiyah and MacDonald, we have:

Proposition 8.6. Let $(A,\mathfrak{m})$ be a local Noetherian ring. Then exactly one of the following holds:

  1. $\frak{m}^n\neq\frak{m}^{n+1}$ for all $n \in \mathbb{N}$;

  2. $\frak{m}$ is a nilpotent ideal, in which case, $A$ is Artinian.

Proposition 8.8. Let $(A,\mathfrak{m})$ be a local Artinian ring. Then the following are equivalent:

  1. $A$ is a principal ideal ring;

  2. $\frak{m}$ is principal;

  3. $\dim_K(\frak{m}/\frak{m}^2)\leq 1$ (where $K=A/\frak{m}$ is the residue field).

In the proof of 8.8, A&M quickly boil down to the case:

$$\dim_K(\mathfrak{m}/\mathfrak{m}^2)= 1 \Rightarrow A \text{ is a principal ideal ring.}$$

A&M begin by explaining that $\mathfrak{m}$ is nilpotent, which is because $\mathfrak{m}$ equals the Jacobson radical of $A$ (as $A$ is a local ring) and the Jacobson radical of an Artinian ring is nilpotent (as A&M prove earlier).

However, doesn't this also follow from 8.6 as Artinian rings are Noetherian (which A&M prove in Proposition 8.5)?

$\endgroup$
0

1 Answer 1

1
$\begingroup$

It follows as long as you know that an Artinian ring is Noetherian; I assume they're trying to avoid that fact here.

$\endgroup$
4
  • $\begingroup$ I suppose so, although, they proved this as proposition 8.5. Thank you for your response! :) $\endgroup$ Commented Nov 25, 2014 at 19:39
  • $\begingroup$ I have now improved the question $\endgroup$ Commented Nov 25, 2014 at 19:41
  • $\begingroup$ Well, I think I already answered this: yes, it does follow, as soon as you point out that of course case 1 of proposition 8.6 doesn't hold for $A$ artinian. $\endgroup$ Commented Nov 25, 2014 at 19:44
  • $\begingroup$ I know, I was just embedding your answer in the question for anyone else who might read it :) $\endgroup$ Commented Nov 25, 2014 at 19:48

You must log in to answer this question.

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