# Complement of multiplicative set is a (prime) ideal.

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?

• @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 $x+y\notin S$? Jun 3, 2020 at 18:25
• Are you sure this is true? It's only true if the set is maximal Jun 3, 2020 at 18:27
• 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. Jun 3, 2020 at 18:30
• +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. Jun 3, 2020 at 18:31

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 $$r\in R$$.

• 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 :) Jun 4, 2020 at 11:05

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 $$q$$.

• 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. Jun 3, 2020 at 18:35
• Well, by Shivering Soldier's answer above, your claim seems to hold if $S$ is a maximal multiplicative subset of $R\setminus\{0\}$, no? Jan 22, 2021 at 19:49