Questions of Proposition 9.2 from Atiyah's Introduction to Commutative Algebra

Question

Proposition 9.2. Let $A$ be a Noetherian local domain of dimension one, $\mathfrak{m}$ its maximal ideal, $k = A / \mathfrak{m}$ its residue field. Then the following are equivalent:

• i) $A$ is a discrete valuation ring;
• ii) $A$ is integrally closed;
• iii) $\mathfrak{m}$ is a principal ideal;
• iv) $\dim_k( \mathfrak{m} / \mathfrak{m}^2 ) = 1$;
• v) Every non-zero ideal is a power of $\mathfrak{m}$;
• vi) There exists $x \in A$ such that every non-zero ideal is of the form $(x^k)$, $k \geq 0$.

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 $n \geq 0$. This follows from (8.6).

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iv) $\implies$ v). Let $\mathfrak{a}$ be an ideal $\neq (0), (1)$. By remark (A) we have $a \supseteq m^n$ for some $n$; from (8.8) (applied to $A / \mathfrak{m}^n$) it follows that $\mathfrak{a}$ is a power of $\mathfrak{m}$.

v) $\implies$ 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) $\implies$ 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 $v$.

I can't understand iv) $\implies$ v), and vi) $\implies$ i). Could someone explain more to me? Also, could someone explain more about Proposition 2.8 to me? I am confused about the sentence “whose images form a basis of this vector space”. Could someone explain the Proposition 2.8 more to me? (Or even the proof of Proposition 2.8, if it is necessary to understand Proposition 2.8).

Answer 1

Proposition 2.8: As explained in the paragraph before $M/\mathfrak mM$ is a $A/\mathfrak m = k$ Module and, as $k$ is a field, $M/\mathfrak mM$ is a vectorspace. So you can find a basis and because $M$ is a finite $A$-Module the vectorspace has finite dimension, so the basis is finite. Because $M/\mathfrak mM$ is a quotient of the module $M$ (take $\phi:M \to M/\mathfrak mM$, $ m \mapsto m + \mathfrak mM$). Because a quotient map is surjective each basisvector $v_i$ has to have a preimage $x_i$ (so $\phi(x_i) = v_i$).

Remark (A): If we take $\mathfrak a \not = (0)$ and ideal in $A$ then the smallest prime ideal containing it $r(\mathfrak a)\not = (0)$ is a prime ideal and because A only has one prime ideal not equal to $(0)$, because dimension of $A$ is one, which is $\mathfrak m$, because it is maximal. If you now apply 7.16 you get equivalent conditions namely that $\mathfrak a$ is $\mathfrak m$-primary and $\mathfrak m^n \subset \mathfrak a$ for some $n$.

$\bf{iv)\implies v)}$: Now we know by (A) that $\mathfrak m^n \subset \mathfrak a$ for some $n$ and $\mathfrak a \not = (0)$ some ideal. $A/\mathfrak m^n$ is by Prop. 8.6. an Artinin local ring, so the proof of Prop. 8.8. gives us by $iv)$ that $\mathfrak a$ is a power of $\mathfrak m$ in the quotient ring, but this is also the case in $A$ because all our ideals are included in each other. To see the last part we look at an element in $A$, because $\mathfrak m^n \subset \mathfrak a$ and $\mathfrak a +\mathfrak m^n \subset A/\mathfrak m^n$ is a power $r$ of $\mathfrak m +\mathfrak m^n$, this elment is of the form $a*x^r+b*x^n = (a+b*x^{n-r})*x^r$, for $x$ generator of $\mathfrak m$, this gives us that $\mathfrak a = \mathfrak m^r$ not only in the quotient but in $A$ itself.

$\bf{v)\implies vi)}$: Because $A$ is a domain $\mathfrak m^n \not =(0)$ for all $n\geq 0$, so by Prop 8.6. $\mathfrak m^n \not =\mathfrak m^{n+1}$ for all $n\geq 0$. So we can take $x \in \mathfrak m \backslash \mathfrak m^2 $ and by $v)$ $(x) = \mathfrak m^r$, but as $x \in \mathfrak m$ $r = 1$ and clearly $\mathfrak m^n = (x)^n = (x^n)$.

$\bf{vi)\implies i)}$: Again because $A$ is a domain by Prop 8.6. $(x)^n \not = (x)^{n+1}$ for all $n\geq 0$. If you now follow the same arguments as $iii)\implies iv) \implies v)$ for our specific ideal $\mathfrak m = (x)$, for any $a \in A$ nonzero $(a) = (x^k)$ for exactly one $k$. So we can define our valuation by $v(a) = k$. Because the criteria for valuation are qualitively the same as for powers you can check the well definedness by always switching to the ideals and then to the same ideal by the power of $x$, then it becomes quite clear, that everything works out.

Proposition 8.8: Let $A$ be an Artin local ring with maximal ideal $\mathfrak m$. Then following are equivalent:
i) every ideal in $A$ is principle
ii) $\mathfrak m$ is principle
iii) $dim_k(\mathfrak m/\mathfrak m^2)\leq1$

i) $\implies$ ii) is clear

As in Prop 2.8. we look at a module $M$ and we take $M = \mathfrak m$, this is an $A$-module as any ideal. Now we can proceed again as in Prop 2.8. and look at $M/\mathfrak mM = \mathfrak m/\mathfrak m^2$, this is a $A/\mathfrak m = k$-vectorspace and as such has a dimension $dim_k(\mathfrak m/\mathfrak m^2)$ this is by Prop 2.8. exactly the number of generators of $\mathfrak m$ as a module and therefor as a an ideal.
This shows ii) $\iff$ iii).

At the end of the proof we then show that $\mathfrak m = (x)$ principle implies that every ideal is principle. For that we take any ideal $\mathfrak a\not = (0),(1)$. As $\mathfrak m$ is maximal we get following inclusions $\mathfrak m^n = (0) \subset A \subset \mathfrak m$ and we can therefor,$\mathfrak a\not=(0)$, choose $r$ such that $A \subset \mathfrak m^r$ but $A \not \subset \mathfrak m^{r+1}$. Now there exists $y = a*x^n \in \mathfrak a$ such that $y \not \in \mathfrak m^{r+1}$, because of that $a \not \in (x) = \mathfrak m$ and, as $\mathfrak m$ is the maximal ideal of a local ring, $a$ is a unit. So we get by the ideal property of $\mathfrak a$ that $a^{-1}*y = a^{-1}*a*x^r = x^r \in \mathfrak a$, so we have $\mathfrak a = (x^r)$ and $\mathfrak a$ is a principle ideal.

A general remark on the proof: It is presented as being a "normal" equivalence proof, but in the proof itself in a way you often use previous parts as well, as similar properties appear in the referenced proposition. Especially $iii),iv),v)$ are parallel to Prop. 8.8., so this (and 2.8.) is where most things you were confused about actually happen. So you often repeat similar arguments, for reasons of brevity those repeated arguments are for the most part omitted in the book, as it is meant to be very brief.

Answer 2

Prop2.8(Nakayama's lemma)

Let $(A,\frak{m})$ be a local ring, and denote the residue field $k=A/\frak{m}$. $M$ be a finitely generated $A$-module. Then we can consider $M/{\frak m}M$ is $A/{\frak m}=k$ vector space. Now we denote canonical map $\phi:M\to M/{\frak m}M$.

In this case, for $x_1,\dots,x_n\in M$ if $\phi(x_1),\dots,\phi(x_n)$ generate $M/{\frak m}M$ as $k$-vector space, then $x_1,\dots, x_n$ generate $M$ as $A$-module.