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?
Zorn’s Lemma and Prime Ideals
abstract-algebracommutative-algebramaximal-and-prime-idealsring-theory
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If you are familiar with localizations, they provide a very nice alternate way to frame this argument. By considering the ring $S=R/B$, it suffices to show the following: if $x\in S$ is in every prime ideal of $S$, then $x^n=0$ for some $n\in\mathbb{N}$. To prove this, consider the localization $S[x^{-1}]$ (i.e., the localization with respect to the multiplicatively closed subset $\{1,x,x^2,x^3,\dots,\}$). By the explicit description of localizations as fractions under an equivalence relation, we see that $1/1=0/1$ in the ring $S[x^{-1}]$ iff $x^n=0$ for some $n$. Thus if $x^n\neq 0$ for all $n$, the ring $S[x^{-1}]$ is not the zero ring. There thus exists a maximal ideal $M\subset S[x^{-1}]$. If $\varphi:S\to S[x^{-1}]$ is the canonical map, $\varphi^{-1}(M)$ is then a prime ideal (because the inverse image of a prime ideal is always prime) such that $x\not\in\varphi^{-1}(M)$ (since $\varphi(x)$ is a unit).
Let $S$ be a maximal multiplicative subset of $R\setminus\{0\}$ and $\mathfrak{p}:=R\setminus{S}.$ As you mentioned above, it's enough to prove that $\mathfrak p$ is an ideal.
Clearly, $0\in\newcommand{\p}{\mathfrak{p}}\p.$ Let $x,y\in \p$. If we can show that $s(x+y)=0$ for some $s\in S$, then $x+y\in \p$(because $s(x+y)=0\notin S$ implies that $s\notin S$ or $x+y\notin S$ and the only possibility is $x+y\notin S$). With that in mind, consider the smallest multiplicatively closed set containing $S$ and $x$; it is the set $\tilde S=\{sx^n\mid s\in S, n\geq0\}.$ Since $S$ is a maximal multiplicative subset of $R\setminus\{0\}$ and $\tilde S$ properly contains $S$, we have $sx^n=0$ for some $s$ and $n$. Similarly, we get $ty^m=0$ for some $t\in S$ and $m$. Thus, for a large enough number, say $N$, we have $st(x+y)^N=0$(Ok, this is not what we wanted, but we are close). Since $st\in S$, we see that $(x+y)^N\in\p$. Write $(x+y)^N=(x+y)(x+y)^{N-1}$. If $x+y\in\p$, then we are done. Otherwise, $x+y\in S$ and by the above argument, $(x+y)^{N-1}\in\p$. So after a finite number of steps, we'll see that $x+y\in\p$.
Similarly, you can show that $rx\in\p$ for all $r\in R$.
Best Answer
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 $S$.