For context, I am learning commutative algebra from Kaplansky's book. Kaplansky writes, "We note that, given any ideal $J$ disjoint from a multiplicatively closed set $S$, we can by Zorn's lemma expand $J$ to an ideal $I$ maximal with respect to disjointness from $S$. Thus we have a method of constructing prime ideals." This seems like an important result for solving the exercises, but this is all Kaplansky states about it. I am confused since I have never used Zorn's lemma before in my undergraduate classes. Can you explain step-by-step what Kaplansky means here?

  • $\begingroup$ This is essentially the proof that every proper ideal is contained in a maximal ideal. Here you can find a detailed proof of that result. Your question can be done either by passing to the localization or by immitating that proof. $\endgroup$
    – WhatsUp
    Dec 17, 2021 at 20:44
  • $\begingroup$ Just to explain the last quoted remark: Once you have the ideal by Zorn's lemma, you can show that it is prime directly, using the fact that $S$ is multiplicatively closed. $\endgroup$ Dec 17, 2021 at 21:13
  • $\begingroup$ Another example of Zorn's lemma that you can practice with is the proof of Proposition 1.8 in Atiyah-Macdonald: The nilradical is the intersection of all prime ideals. $\endgroup$ Dec 17, 2021 at 21:15

1 Answer 1


Zorn’s Lemma states that a nonempty partially ordered set where every chain has an upper bound within the set, necessarily has a maximal element. It is equivalent to the Axiom of Choice.

The set of ideals disjoint from S is nonempty and partially ordered by inclusion. In this partial order, a chain is just an increasing “nested” set of ideals.

Given such a chain, I claim that its union is an upper bound. If the union is in fact within the set; i.e., if the union is itself an ideal that’s disjoint from $S$, then the union is obviously an upper bound of the chain under set inclusion.

But it’s easy to see that the union is such an ideal. Since each component of the union is disjoint from $S$, the whole union also must be disjoint from $S$. If the chain is $\mathscr I=\{ I_\alpha \mid \alpha \lt \kappa \}$ (where $\alpha \lt \beta \Rightarrow I_{\alpha} \subseteq I_{\beta}$) and $x, y \in I= \bigcup \mathscr I$, then $x \in I_{\alpha}, y \in I_{\beta}$, where without loss of generality $\alpha \leq \beta$. Then $x \in I_{\beta}$, and since $I_{\beta}$ is an ideal containing $x$ and $y$, $x+y \in I_{\beta} \subseteq I$ and $I$ is closed under addition. Finally, $\forall r \in R~(ry \in I_{\beta} \subseteq I$), so $I$ is in fact an ideal.

Zorn’s Lemma now applies to tell us there’s a maximal element of this partially ordered set. That maximal element is an ideal that’s maximal with respect to the property of being disjoint from $S$.


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