Complement of multiplicative set is a (prime) ideal.

Question

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.

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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?

Answer 1

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 $x+y\in\p$.

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Similarly, you can show that $rx\in\p$ for all $r\in R$.

Answer 2

I don't think complement of any given multiplicative set is actually an ideal because if it were true: Suppose $S$ is a multiplicative set of $A$. Then if $xy\in S$ and $x\notin S$ then $x\in A-S\implies xy\in A-S$ which is a contradiction. Thus it implies all multiplicative sets are saturated which is not true. (A saturated set $S$ is multiplicative and $xy\in S\iff x,y\in S$).

Answer 3

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 $q$.