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I am currently working on Exercise 11.4 in Atiyah & Macdonald's text, Introduction to Commutative Algebra. The exercise is as follows:

An example of a Noetherian domain of infinite dimension (Nagata). Let $k$ be a field and let $A = k[x_1,x_2,...,x_n,...]$ be a polynomial ring over $k$ in a countable infinite set of indeterminates. Let $m_1,m_2,...$ be an increasing sequence of positive integers such that $m_{i+1} - m_i > m_i - m_{i-1}$ for all $i > 1$. Let $p_i = (x_{m_i + 1},...,x_{m_{i + 1}})$ and let $S$ be the complement in $A$ of the union of the ideals $p_i$.

Each $p_i$ is a prime ideal and therefore the set $S$ is multiplicatively closed. The ring $S^{-1}A$ is Noetherian by Chapter 7, Exercise 9. Each $S^{-1}p_i$ has height equal to $m_{i+1} - m_i$, hence dim $S^{-1}A = \infty$.

For reference, Exercise 9 of Chapter 7 states the following:

Let $A$ be a ring such that (1) for each maximal ideal $m$ of $A$, the local ring $A_m$ is Noetherian; (2) for each $x \neq 0$ in $A$, the set of maximal ideals of $A$ which contain $x$ is finite. Show that $A$ is Noetherian.

Working on Exercise 11.4, I was able to show that each $p_i$ is a prime ideal and that $S$ is multiplicatively closed. However, I'm struggling with the following:

  • How can I use the result of Exercise 9 of Chapter 7 to show that $S^{-1}A$ is Noetherian? It's not clear to me what the maximal ideals of $S^{-1}A$ are. If $S$ was the complement of a single prime ideal of $A$, then we'd know that $S^{-1}A$ was a local ring with just one maximal ideal.
  • I'm not sure how to show that the height of each $S^{-1}p_i$ is equal to $m_{i+1} - m_i$. I'd like to consider the chain $(x_{m_{i+1}}) \subset (x_{m_{i+1}}, x_{m_{i+2}}) \subset \cdots \subset p_i$ of prime ideals in $A$ with length $m_{i+1} - m_i$. If I localize each object in this chain, do I still get a chain of prime ideals with the same length? If so, that only shows that the height of each $S^{-1}p_i$ is bounded below by $m_{i+1} - m_i$, not equal to $m_{i+1} - m_i$. I believe this is still enough to conclude that dim $S^{-1}A = \infty$, but nevertheless, it's not the desired claim in the exercise.

Thank you!

michiganbiker898
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1 Answers1

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Update: I did some googling. And found this: answer.

This should answer all your questions.

To tackle the 2nd point:

Note that there is a one-one correspondence between the prime ideals of $A$ disjoint from $S$ and the prime ideals of $S^{-1}A$ and the correspondence is inclusion preserving. This tells you that the chain you consider is indeed a maximal one.

Evans Gambit
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