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Following are part of the proof of proposition 9.2 from Atiyah that I still not understand could anyone explain more to me? Or suggest me some other reference to read to help me to understand the proof?

Proof. Before we start going the rounds, we make two remarks:
(A) If $\mathfrak{a}$ is an ideal $\neq 0,(1)$, then $\mathfrak{a}$ is $\mathfrak{m}$-primary and $\mathfrak{a}\supseteq\mathfrak{m}^n$ for some n. For $r(\mathfrak{a})=\mathfrak{m}$, since $\mathfrak{m}$ is the only non-zero prime ideal; now use (7.16).

(B) $\mathfrak{m}^n\neq \mathfrak{m}^{n+1}$ for all $n\ge 0$. This follows from (8.6).

$v)\Rightarrow vi)$. By remark (B), $\mathfrak{m}\neq\mathfrak{m}^2$ , hence there exists $x\in\mathfrak{m},x\notin\mathfrak{m}^2$. But $(x) = \mathfrak{m}^r$ by hypothesis, hence $r = 1$, $(x) =\mathfrak{m}$, $(x^k) = \mathfrak{m}^k$.

$vi)\Rightarrow i)$. Clearly $(x) = \mathfrak{m}$, hence $(x^k)\neq(x^{k+1})$ by remark (B). Hence if $a$ is any non-zero element of $A$, we have $(a) = (x^k)$ for exactly one value of $k$. Define $v(a) = k$ and extend $v$ to $K^*$ by defining $v(ab^{-1}) = v(a) - v(b)$. Check that v is well-defined and is a discrete valuation, and that $A$ is the valuation ring of $v$.

Question. For remark (B), why $\mathfrak{m}^n\neq \mathfrak{m}^{n+1}$ for all $n\ge 0$ this is true, not $\mathfrak{m}^n=0$ is true? (8.6 says that one of them will be true)

Please explain more about the proof. I can't understand it.

Reference: Questions of the proof of proposition 9.2 from commutative algebra by Atiyah

Arturo Magidin
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Judy Judy
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    $A$ is a Noetherian local domain and $\mathfrak{m}<A$. How can it be that $\mathfrak{m}^n=0$? – kabenyuk Dec 02 '22 at 14:47
  • Sorry, I still can't get it. Local domain to me it just means it's a ID with only one maximal ideal, and $m^n$ is a subset of m, so I still don't know where is wrong？ Could you explain more to me？ – Judy Judy Dec 03 '22 at 08:32

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This is a long commentary, but not an answer to the question. Although it may be an answer :)

Here it seems worthwhile to mention the terminology in this and other books on commutative algebra.

  1. A ring with no zero-divisors $\neq0$ (and in which $1\neq0$) is called an integral domain.

  2. In some cases the word 'integral' is omitted e.g. (the pages are from the book by M. F. Atiyah and I. G. Macdonald, Introduction to Commutative Algebra):

  • principal ideal domain (see p.5);

  • unique factorization domain (see p.50);

  • integrally closed domain (see p.64);

  • Dedekind domain (see p.93);

  • Noetherian domain (see p.93);

  • local domain (see p.97).

In all these cases the term 'domain' means 'integral domain'.

So if $\mathfrak{m}^n=0$, then it follows that $x_1x_2\ldots x_n=0$ for any $x_i\in\mathfrak{m}$ and hence $\mathfrak{m}=0$ since $A$ is an integral domain. By the way, note that $\mathfrak{m}^n$ consists of all possible sums of products of the form $x_1x_2\ldots x_n$, where $x_i$ runs through all elements of $\mathfrak{m}$.

kabenyuk
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