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Let $R$ be a commutative ring with $1\neq0$. I'm trying to show that the complement $\mathfrak p$ of a multiplicative subset $S\subseteq R\setminus\{0\}$ is a (prime) ideal. In particular, I am having trouble showing that $\mathfrak p$ is additive in the first place.

I read the answers to this question, but all of those answers seem to pull out prime ideals from nowhere which just so happen to coincide with $\mathfrak p$. However, I am trying to find a more naive approach to show that $x+y\in\mathfrak p$ for any two $x,y\in\mathfrak p$ and $x\mathfrak p\subseteq\mathfrak p$ for every $x\in R$.

Any hints would be appreciated.


EDIT: Just to clarify what I am trying to achieve (for a voluntary homework exercise). I am given a commutative ring $R$ with $1\neq0$ and the set $\Sigma$ of all multiplicative subsets of $R\setminus\{0\}$. Using Zorn's lemma one easily shows that $\Sigma$ contains a maximal element. My exercise now is the following:

Show that $S\in\Sigma$ is maximal, if and only if $\mathfrak p:=R\setminus S$ is a minimal prime ideal.

Example 1 on page 38 of Introduction to Commutative Algebra by Atiyah reads

Let $\mathfrak p$ be a prime ideal of $R$. Then $S=R\setminus\mathfrak p$ is multiplicatively closed (in fact $R\setminus\mathfrak p$ is multiplicatively closed $\Leftrightarrow\mathfrak p$ is prime).

For my exercise I just need to apply the statement in the example, BUT I strongly suspect that the part in parentheses assumes a priori that $\mathfrak p$ is an ideal, which I don't yet know in the exercise.

Is the claim in the exercise correct?

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  • $\begingroup$ @M.Wang I don't understand. An ideal $I$ of a commutative ring $R$ is, by definition, a subgroup of $(R,+)$ such that $xI\subseteq I$ for every $x\in R$. I do not see why $\mathfrak p=R\setminus S$ satisfies $x+y\in\mathfrak p$ for every $x,y\in\mathfrak p$. If I take $x,y\in\mathfrak p=R\setminus S$, then the only thing I know so far is $x+y\in R$. But why do I have $x+y\notin S$? $\endgroup$
    – Cubi73
    Jun 3, 2020 at 18:25
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    $\begingroup$ Are you sure this is true? It's only true if the set is maximal $\endgroup$ Jun 3, 2020 at 18:27
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    $\begingroup$ I don't think this is true. For example, the set $\{1,x,x^2...\}$ in $\mathbb{Q}[X]$ is multiplicative but its complement isn't an ideal because both $\frac{1}{2}x+1$ and $\frac{1}{2}x-1$ are in the complement but their sum isn't. $\endgroup$
    – M. Wang
    Jun 3, 2020 at 18:30
  • $\begingroup$ +Don Thousand Not quite. I am supposed to show that if I have a maximal multiplicative subset $S$ of $R\setminus\{0\}$, then $\mathfrak p=R\setminus S$ is a minimal prime ideal, and I read that for any ideal $\mathfrak a$ the set $R\setminus\mathfrak a$ is multiplicative, if and only if $\mathfrak a$ is prime. However, that last statement doesn't help me much, because I don't yet know whether $\mathfrak p$ is actually an ideal. $\endgroup$
    – Cubi73
    Jun 3, 2020 at 18:31

2 Answers 2

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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 $r\in R$.

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    $\begingroup$ I was sooo close... I also noticed that $S_x:=\bigcup_{n\geq0}x^nS\supseteq S$ is multiplicatively closed and contains $x\notin S$, so we have $S\subsetneq S_x$, which by maximality of $S$ means $0\in S_x$ and therefore $0=sx^n$ for some $s\in S$ and $n\geq1$ (and analogously $0=ty^m$). But I didn't think of the binomial theorem to combine those identities... Thank you for your help :) $\endgroup$
    – Cubi73
    Jun 4, 2020 at 11:05
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You can't prove the complement of a multiplicative set is a prime ideal, as it would imply any ring of fractions is a local ring, which is false: an easy counterexample is the ring $\mathbf Z_{pq}=\mathbf Z\Bigl[\frac1{pq}\Bigr]$, and its prime ideals correspond bijectively to the primes of $\mathbf Z$ different from $p$ and $q$.

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  • $\begingroup$ Does my claim work, if I additionally assume $S$ to be a maximal multiplicative subset of $R\setminus\{0\}$? Because otherwise, one homework exercise of mine is nothing but humbug. $\endgroup$
    – Cubi73
    Jun 3, 2020 at 18:35
  • $\begingroup$ Well, by Shivering Soldier's answer above, your claim seems to hold if $S$ is a maximal multiplicative subset of $R\setminus\{0\}$, no? $\endgroup$ Jan 22, 2021 at 19:49

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