Ring Theory – Zero Divisors and Prime Ideals

commutative-algebraring-theory

I've looked at the following similar questions but none of them solved my specific doubt:

I am doing exercise 14 in chapter 1 of Atiyah-MacDonald. I have successfully completed the first part:

In a ring $A$, let $\Sigma$ be the set of all ideals in which every element is a zero-divisor. Show that the set $\Sigma$ has maximal elements and that every maximal element of $\Sigma$ is a prime ideal.

What I'm struggling with is the conclusion from the above:

Hence the set of zero-divisors in $A$ is a union of prime ideals.

Note that I'm interested in specifically how to conclude this from the part above, and not in some other way.

Here is what I've tried:

I claim that the set $D$ of zero-divisors of $A$ is: $$D = \bigcup \{ \text{maximal elements of }\Sigma \}$$
which will give the result (this seems to be the "obvious" suggestion from the way the exercise is stated).

The inclusion $\supseteq$ is clear, since all elements of members of $\Sigma$ are zero divisors by construction.

I'm struggling with the other inclusion, $\subseteq$.
Let $x \in D$, and let $y$ be nonzero in $A$ such that $xy = 0$.
Ideally, what I'd like to say is that $(x) \in \Sigma$ and therefore $(x)$ must be included in a maximal element of $\Sigma$, and we're done.
However, Zorn's lemma just gives us some maximal element of $\Sigma$, it doesn't say anything about where that maximal element is in the poset.
For all I know, $(x)$ could be living not under some maximal element:

(...) <- infinite chain with no maximal element
  |
  .
  |
  .
  |   .  <- a maximal element of \Sigma
 (x) /
  | /
  .

To prove this, I tried the following.
Apply the result from the first part to $A/(x)$.
Then we get a prime ideal $\overline{\mathfrak{p}}$ of $A/(x)$ whose elements are zero divisors and which is maximal in this regard.
By the correspondence theorem, this corresponds to a prime ideal $\mathfrak{p}$ of $A$ containing $x$.
To finish the proof, I need to show that $\mathfrak{p}$ is maximal in $\Sigma$.

First, we need to show that $\mathfrak{p} \in \Sigma$.
Let $a \in \mathfrak{p}$, the goal is to show that $a$ is a zero divisor.
Then $\overline{a} \in \overline{\mathfrak{p}}$ is a zero divisor in $A/(x)$, so $\overline{a}\overline{b} = \overline{0}$ for some $\overline{b} \ne 0$, i.e. $b \notin (x)$.
Thus $ab \in (x)$, so $ab = \lambda x$ for some $\lambda \in A$.
Then $aby = \lambda x y = 0$.
But in order to conclude that $a$ is a zero divisor, I would need that $by \ne 0$, which is not necessarily true as illustrated by the following example: take $A = k[\alpha, \beta, \gamma]/(\alpha\beta, \beta\gamma, \alpha\gamma – \alpha)$, $x = \alpha$, $y = \beta$, $a = \alpha$, $b = \gamma$.

How to proceed?

I think I could modify the Zorn's lemma argument directly (let $\Sigma_x$ be the elements of $\Sigma$ containing $x$…) to get that $x$ is contained in a maximal element of $\Sigma$, but the way the exercise is written seems to imply that the final conclusion is "direct" from the first part, but I can't figure it out, which is bugging me.

Best Answer

I guess a partially satisfactory answer is that the following corollary of Zorn's lemma seems to be true:

If a poset $(P, \le)$ satisfies the hypotheses for Zorn's lemma ($P$ nonempty and every chain has an upper bound), then for any $x \in P$, $x$ lies under a maximal element of $P$.

Consider the poset $(P_x = \{y \in P \mid x \le y\}, \le)$, the claim is that $P_x$ has a maximal element. Indeed, $P_x$ is nonempty as it contains $x$. Also, given a nonempty chain $\mathcal{C} \subseteq P_x \subseteq P$, it is also a chain in $P$, and so by hypothesis it has an upper bound $u$ in $P$; but since $\mathcal{C}$ is nonempty and $\mathcal{C} \subseteq P_x$, we have some $c \in \mathcal{C}$ with $x \le c \le u$, and so $x \le u$, ergo $u \in P_x$. Therefore $(P_x, \le)$ satisfies the hypotheses of Zorn's lemma, so it has a maximal element $m \in P_x$ (so we have $x \le m$). To finish the proof, we need to show that $m$ is still maximal in $P$. Suppose there is a $y \in P$ with $m \le y$. By transitivity, $y \in P_x$. By maximality of $m$ in $P_x$, we conclude $y = m$.

Therefore, if Atiyah & MacDonald are taking this fact for granted implicitly, it is a direct application of the first part in some way, but I don't know for sure if this is what they intended.

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