# Abstract Nonsense

## Prime Ideals (Pt. III)

Point of Post: This is a continuation of this post.

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So, why is this lemma so important? It clearly isn’t a ‘practical’ method to show that certain ideals are prime. I mean, theoretically it shows that $2\mathbb{Z}$ and $(x)$ are prime in $\mathbb{Z}$ and $R[x]$ (for an integral domain $x$) but it is much harder to show that they are maximal (in terms of containment) ideals disjoint from the multiplicative sets $O$ and $\mathbb{Z}^\times$. Indeed, in both these cases the primality of the ideals follows from the fact that the multiplicative set isn’t in their complement–it is their complement. Moreover, it’s clear that even if we had some multiplicative set $S\subseteq R$ it doesn’t seem likely that we would actually (in most cases) find the maximal (in terms of containment) ideal of $R-S$. But, if we are able to show that $R-S$ always has a maximal (in terms of containment) ideal, then we may conclude that $R$ has ‘a lot’ of prime ideal–at least one, in certain cases as we shall see below.

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Theorem: Let $R$ be a commutative unital ring and $S$ a multiplicative subset of $R$. Then the set $\mathfrak{Z}$ of all ideals disjoint from $S$, ordered by containment, is either empty or has a maximal element.

Proof: Let $\mathcal{C}$ be a chain in $\mathfrak{Z}$. We claim then that $\displaystyle \mathfrak{a}=\bigcup_{\mathfrak{c}\in C}\mathfrak{c}$ is an element of $\mathfrak{Z}$. In particular, we claim that $\mathfrak{a}$ is an ideal and it is disjoint from $S$. To see that $\mathfrak{a}$ really is an ideal we merely note that if $x,y\in\mathfrak{a}$ then by definition we have that $x\in\mathfrak{c}_1$ and $y\in\mathfrak{c}_2$ for $\mathfrak{c}_1,\mathfrak{c}_2\in\mathcal{C}$, but since $\mathcal{C}$ is a chain we may assume without loss of generality that $\mathfrak{c}_1\subseteq\mathfrak{c}_2$ and so $x,y\in\mathfrak{c}_2$ and thus $x-y\in\mathfrak{c}_2\subseteq\mathfrak{a}$. Next, if $r\in R$ and $x\in\mathfrak{a}$ then $x\in\mathfrak{c}$ for some $\mathfrak{c}\in\mathcal{C}$ and so $rx\in\mathfrak{c}\subseteq\mathfrak{a}$. Thus, $\mathfrak{a}$ is an ideal as claimed. The fact that $\mathfrak{a}$ is basic set theory since to assume that $s\in\mathfrak{a}\cap S$ is to assume that $s\in\mathfrak{c}\cap S$ for some $\mathfrak{c}\in\mathcal{C}$ contrary to assumption.

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Since $\mathcal{C}$ was arbitrary we may conclude from Zorn’s Lemma. $\blacksquare$

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Remark: To see why $\mathfrak{Z}$ may be empty consider the multiplicative set $\{0\}$ sitting inside $\mathbb{Z}$.

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From this and the previous theorem we may conclude the following:

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Theorem: Let $R$ be a commutative unital ring and $S$ a multiplicative subset of $R$. If there exists an ideal disjoint from $S$ then there exists a prime ideal disjoint from $S$.

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Or, in the more user-friendly version:

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Theorem: Let $R$ be a commutative unital ring and $S$ a multiplicative subset of $R$. If $\mathfrak{a}$ is an ideal disjoint from $S$ then there exists a prime ideal $\mathfrak{p}\supseteq\mathfrak{a}$ disjoint from $S$.

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This existence theorem is only the tip of the iceberg concerning the consequences of the innocuous lemma. The next consequence we’d like to discuss shall seem somewhat contrived at first, but hopefully the usefulness will become readily apparent soon after. Namely, it’s not hard to see from the above discussion that the focus of our recent discussion has been more on multiplicative properties of rings $R$ and less on the additive properties. What I have in mind is that we defined a prime ideal $\mathfrak{p}$ to be a subgroup of $(R,+)$ with the property that $ab\in\mathfrak{p}$ if and only if $a\in\mathfrak{p}$ or $b\in\mathfrak{p}$. In particular, $\mathfrak{p}$ has the absorption property of an ideal and has a multiplicative complement. Well, what if we forewent the seemingly cumbersome subgroup structure and just wondered about sets with the second property. Namely, we want to see if we can somehow characterize subsets $Q$ of a commutative unital ring $R$ with the property that $ab\in Q$ if and only if $a\in Q$ or $p\in Q$. Let’s start with $\mathbb{Z}$ (as always) and work our way up. The place to begin (once again, as always) is with the primes and inevitably$1$. Namely, if $Q\subseteq\mathbb{Z}$ is of the type described, where does $Q$ ‘put the primes and $1$‘? Well, the first thing we note is that just like in the case where $Q$ is actually a subgroup, so that it would be an ideal, one has that $Q=\mathbb{Z}$ if and only if $1\in Q$. Thus, where $1$ goes becomes an almost afterthought and we are left with the real meat–the primes. To see where they ‘go’ we begin by noticing that if a prime $p\in Q$ then $p^n\in Q$ for every $n\in\mathbb{N}$. Moreover, it’s clear that if $p^n\in Q$ for some $n\in\mathbb{N}$ and $p$ prime then $p\in Q$ since, inductively we see that $p^n\in Q$ implies $p\in Q$ or $p^{n-1}\in Q$, etc. Thus, a dichotomy emerges for primes to powers. Namely we have that $\{p,p^2,p^3,\cdots\}\subseteq Q$ or $\{p,p^2,p^3,\cdots,\}\subseteq\mathbb{Z}-Q$. But, from the absorption property we see that this is equivalent to $(p)\subseteq Q$ or $(p)\subseteq \mathbb{Z}-Q$. Therefore, we may conclude that for the case when $1\notin Q$ we have that $Q$ is precisely the union of $(p)$ where $(p)\cap Q\ne\varnothing$. Seeing that sets of this form are, in fact, the type of sets we wish to study we may conclude that:

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Theorem: A set $Q\subseteq\mathbb{Z}$ has the property that $ab\in Q$ if and only if $a\in Q$ or $b\in Q$ if and only $Q=\mathbb{Z}$ or if it is of the form

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$\displaystyle \bigcup_{p\in\mathbb{P}}(p)$

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where $\mathbb{P}$ is a set of (some) primes.

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Recalling our earlier characterization of $\text{Spec}(\mathbb{Z})$ we may phrase this in language more amenable to our current goals:

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Theorem: For a set $Q\subseteq\mathbb{Z}$ the following are equivalent

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\begin{aligned}&\mathbf{(1)}\quad ab\in Q\text{ if and only if }a\in Q\text{ or }b\in Q\\ &\mathbf{(2)}\quad Q=\mathbb{Z}\text{ or }Q\text{ is a union of prime ideals}\end{aligned}

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This gives us the conviction to state and prove the following:

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Theorem: Let $R$ be a commutative unital ring, then the following are equivalent:

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\begin{aligned}&\mathbf{(1)}\quad ab\in Q\text{ if and only if }a\in Q\text{ or }b\in Q\\ &\mathbf{(2)}\quad Q=R\text{ or }Q\text{ is a union of prime ideals}\end{aligned}

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Proof: Clearly the case $Q=R$ is simple, so we assume that $Q\ne R$. Evidently if $Q$ is a union of prime ideals then it satisfies property $\mathbf{(1)}$. Conversely, suppose that $Q$ satisfies property $\mathbf{(1)}$ and let $x\in Q$ be arbitrary. Since $R-Q$ is multiplicative and $(x)=Rx$ is disjoint from $R-Q$ we have that there exists some prime ideal $\mathfrak{p}_x$ with the property that $(x)\subseteq\mathfrak{p}_x\subseteq Q$. Thus, evidently we have that

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$\displaystyle Q=\bigcup_{x\in Q}\mathfrak{p}_x$

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from where the conclusion follows. $\blacksquare$

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References:

1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

3.  Reid, Miles. Undergraduate Commutative Algebra. Cambridge: Cambridge UP, 1995. Print.