Transfinite induction (Atiyah and Macdonald)

Question

Exercise 17 in chapter 4 of Introduction to Commutative Algebra by Atiyah and Macdonald hints for you to use transfinite induction to complete the proof.

I have not come across transfinite induction before but the Wolfram page seems to make it fairly simple- it's looks exactly like strong induction but just on a well-ordered set $S$ instead of $\mathbb{N}$.

My main issue however is with identifying what $S$ should be in this case.

I tried something like " if $a_k \neq (1) \forall k \leq n$ then $q_{n+1}$ exists and $a=\cap_{i=1}^{n+1} q_i \cap (a_{n+1} + (x_{n+1}))$" but I don't see why this is any different from normal induction nor how it uses the observation that $a_{n-1} \subset a_n$.

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Full text of the exercise:

17. Let $A$ be a ring with the following property.
(L1) For every ideal $\mathfrak a\ne(1)$ in $A$ and every prime ideal $\mathfrak p$, there exists $x\notin\mathfrak p$ such that $S_{\mathfrak p}(\mathfrak a) = (\mathfrak a:x)$, where $S_{\mathfrak p}=A - \mathfrak p$.
Then every ideal in $A$ is an intersection of (possibly infinitely many) primary ideals.

[Let $\mathfrak a$ be an ideal $\ne (1)$ in $A$ and let $\mathfrak p_1$ be a minimal element of the set of prime ideals containing $\mathfrak a$. Then $\mathfrak q_1=S_{\mathfrak p_1}(\mathfrak a)$ is $\mathfrak p$-primary (by Exercise 11), and $\mathfrak q_1= (\mathfrak a:x)$ for some $x\notin \mathfrak p_1$. Show that $\mathfrak a= \mathfrak q_1 \cap (\mathfrak a+(x))$.

Now let $\mathfrak a_1$ be a maximal element of the set of ideals $\mathfrak b \supseteq \mathfrak a$ such that $\mathfrak q_1\cap \mathfrak b = \mathfrak a$, and choose $\mathfrak a_1$ so that $x\in \mathfrak a_1$, and therefore $\mathfrak a_1\not\subseteq \mathfrak p_1$. Repeat the construction starting with $\mathfrak a_1$, and so on. At the $n$th stage we have $\mathfrak a= \mathfrak q_1 \cap \dots \cap \mathfrak q_n \cap \mathfrak a_n$ where the $\mathfrak q_t$ are primary ideals, $\mathfrak a_n$ is maximal among the ideals $\mathfrak b$ containing $\mathfrak a_{n-1}=\mathfrak a_n \cap \mathfrak q_n$, such that $\mathfrak a=\mathfrak q_1\cap\cdots\cap\mathfrak q_n\cap\mathfrak b$, and $\mathfrak a_n \not\subseteq \mathfrak p_n$. If at any stage we have $\mathfrak a_n=(1)$, the process stops, and $\mathfrak a$ is a finite intersection of primary ideals. If not, continue by transfinite induction, observing that each $\mathfrak a_n$ strictly contains $\mathfrak a_{n-1}$.]

Answer 1

Here is a sketch of a solution.

We try to construct $\mathfrak a_\alpha,\mathfrak p_\alpha,\mathfrak q_\alpha$ by induction on the ordinal $\alpha$.

As $\mathfrak a_\alpha$ will increase strictly with $\alpha$, the ideal $\mathfrak a_\alpha$ cannot exist if the cardinality of $\alpha$ exceeds that of $A$. It will be clear that $\mathfrak a_\alpha$ can be defined whenever $\mathfrak a_\beta\ne(1)$ for all $\beta < \alpha$. This will imply $\mathfrak a_\alpha=(1)$ for some $\alpha$.

The ideals $\mathfrak a_\alpha,\mathfrak p_\alpha,\mathfrak q_\alpha$ having already been constructed in the hint of the book for finite ordinals $\alpha$, we can assume from now on $\alpha\ge\omega$. (For a finite ordinal $\alpha$ we define $\mathfrak a_\alpha$ as being the ideal denoted by $\mathfrak a_{\alpha+1}$ in the book, and similarly for $\mathfrak p_\alpha$ and $\mathfrak q_\alpha$, so that we start our transfinite induction at stage $0$ instead of $1$.)

Let $\alpha$ be an infinite ordinal and assume that $\mathfrak a_\beta,\mathfrak p_\beta,\mathfrak q_\beta$ have been defined for $\beta < \alpha$.

When $\alpha$ has a predecessor, the definition of $\mathfrak a_\alpha,\mathfrak p_\alpha,\mathfrak q_\alpha$ follows easily from the hint given in the book.

If $\alpha$ is a limit ordinal, we set $$ \mathfrak a_\alpha:=\sum_{\beta < \alpha}\mathfrak a_\beta, $$ $\mathfrak p_\alpha$ is a minimal prime ideal of $\mathfrak a_\alpha$, and $\mathfrak q_\alpha:=S_{\mathfrak p_\alpha}(\mathfrak a)$.

EDIT 1. I forgot to state the important fact that we have $$ \mathfrak a=\mathfrak a_\alpha\cap\bigcap_{\beta\le\alpha}\mathfrak q_\beta $$ whenever the right-hand side is defined.

EDIT 2. The following lemma is implicitly used:

Let $|X|$ denote the cardinality of the set $X$. Let $(X_\beta)$ be a strictly increasing sequence of sets indexed by the ordinals $\beta$ such that $\beta\le\alpha$, where $\alpha$ is a given ordinal. Then we have $|X_\beta|\ge|\beta|$ for all $\beta\le\alpha$.

We prove $|X_\beta|\ge|\beta|$ by induction. The case where $\beta$ is $0$ or a successor is straightforward and left to the reader. If $\beta$ is a limit ordinal, we have $$ |X_\beta|\ge\left|\ \bigcup_{\gamma < \beta}\ \left(X_{\gamma+1}\setminus X_\gamma\right)\ \right|=\sum_{\gamma < \beta}\left|X_{\gamma+1}\setminus X_\gamma\right|\ge\sum_{\gamma < \beta}1=|\beta|. $$ EDIT 3. The main point, I think, is to understand the structure of the argument. I would summarize it as follows:

We assume by contradiction

(a) we have $\mathfrak a_\alpha\ne(1)$ whenever $\mathfrak a_\alpha$ is defined.

We show that (a) implies

(b) $\mathfrak a_\alpha$ is defined for all $\alpha$.

But in Edit 2 we proved

(c) $\mathfrak a_\alpha$ is not defined for all $\alpha$ large enough.

Clearly (c) contradicts (b), and thus (a). The conclusion is that (a) is false, i.e. there is an $\alpha$ for which $\mathfrak a_\alpha$ is defined and coincides with $(1)$.