Q: Let $A$ be a nonzero ring and let $\Sigma$ be the set of multiplicatively closed subsets of $A$ such that $0 \notin S.$ Show that $\Sigma$ has a maximal element, and that $S \in \Sigma$ is maximal iff $A \setminus S$ is a minimal prime ideal of $A$.

It can be shown using Zorn's Lemma $\Sigma$ has a maximal element, be showing every chain has an upper-bound. (for this $0 \notin S$ condition is not required.)

$\underline{Claim}$: $S$ maximal in $\Sigma$ iff $A \setminus S$ is a minimal prime ideal. I proceeded as follows:


As $A\setminus S$ is a prime ideal, $S \in \Sigma$.
Now how do I show that $S$ is maximal?


$S$: a maximal element in $\Sigma$.

  1. $0 \in A\setminus S$, and if $xy \in A\setminus S$ then either $(x \in A \setminus S)$ or $(y \in A \setminus S)$ as $S$ is multiplicatively closed.

  2. Also, $A \setminus S$ does not contain any prime ideal properly.
    If it does, let $\mathfrak{p}$ be one such prime ideal. Then $A \setminus \mathfrak{p} \in \Sigma$ containing S properly, a contradiction.

It remains to show is $A \setminus S$ is a subgroup of the additive group $A.$ How can I show this?


Using @SteveD's comment I have completed the above proof, which I have posted as an answer. But the proof is a bit set-theoretic. I would like to see a proof using tools from Commutative Algebra.

Is there any nice application of this result?

  • 2
    $\begingroup$ If you look at the ideals contained in $A\setminus S$, you can apply Zorn's lemma there to get a maximal such ideal, which you can prove is prime. Maximality of $S$ implies this prime ideal is the whole complement $A\setminus S$. $\endgroup$
    – Steve D
    Dec 26, 2020 at 5:03
  • $\begingroup$ For a more "commutative algebra" approach, you can localize and work in $S^{-1}R$. $\endgroup$
    – Steve D
    Dec 26, 2020 at 20:30
  • $\begingroup$ Can you please give a bit more hint on this? $\endgroup$
    – Saikat
    Dec 27, 2020 at 5:01
  • $\begingroup$ Find a maximal ideal $\mathfrak{m}$ of $S^{-1}R$ and then show $\mathfrak{m}\cap R$ is the prime ideal you seek. $\endgroup$
    – Steve D
    Dec 27, 2020 at 7:20

3 Answers 3


Using Steve D's suggestion I am trying to complete the above proof.


Let $\mathfrak{P}:=\{\text{set of all ideals contained in} \ A\setminus S\}$. $\mathfrak{P} \neq \phi$ as $\{0\} \in \mathfrak{P}$.
For any chain $\{\mathfrak{p}_i:i \in I\} \subset \mathfrak{P}, \cup \mathfrak{p_i}$ is an upperbound of that chain in $\mathfrak{P}$. i.e it contains a maximal element say $\mathfrak{p}$.
If $\left(\mathfrak{p} \subsetneq A\setminus S\right)$ then $\left(S \subsetneq A\setminus \mathfrak{p}(\in \Sigma) \right)$, contradicting the maximality of $S \implies \mathfrak{p=}A\setminus S$ $\left(as \ \mathfrak{p} \subset A\setminus S.\right)$ $\implies A\setminus S$ is an ideal.
From 1. and 2. in the question we can conclude:
$A\setminus S$ is a minimal prime ideal.


Let $\mathfrak{M:=}\{\text{Set of all multiplicative subset of A containing S}\}$. $\mathfrak{m} \subset \Sigma$.
From Zorn's Lemma, $\mathfrak{M}$ contains a maximal element say $\mathfrak{m}.$ Now $\mathfrak{m}$ also is maximal in $\Sigma$,(if not then it would contradict the maximality of $\mathfrak{m}$ in $\mathfrak{M}$.)
If $S \subsetneq \mathfrak{m}$ then $A\setminus \mathfrak{M} \subsetneq A\setminus S$. As $\mathfrak{m}$ is maximal in $\Sigma$ we have $A\setminus \mathfrak{m}$ is a (minimal) prime ideal, contradicting the minimality of $A\setminus S$. Hence $S=\mathfrak{m}$
i.e $S$ is maximal in $\Sigma.$

  • $\begingroup$ Your first proof is not complete. You cannot "contradict the maximality of $S$" without first showing $\mathfrak{p}$ is prime. $\endgroup$
    – Steve D
    Dec 26, 2020 at 22:35
  • $\begingroup$ @SteveD: I have shown that in the question, from which I draw the fact i.e "from 1. and 2." it follows that $A\setminus S$ is prime. $\endgroup$
    – Saikat
    Dec 27, 2020 at 4:58
  • $\begingroup$ No you have shown $A\setminus S$ is prime. That is not enough here. Please spend some time thinking about why this is not a complete proof. $\endgroup$
    – Steve D
    Dec 27, 2020 at 7:18
  • $\begingroup$ @SteveD: I still cannot figure out why this is not complete. I have shown that $\mathfrak{p}=A\setminus S$, which makes $A\setminus S$ an ideal, and I have separately shown that if $ab \in A\setminus S$ then either $a \in A\setminus S$ or $b \in A \setminus S$. Which makes $A \setminus S$ a prime ideal. Please tell where am I wrong? $\endgroup$
    – Saikat
    Dec 31, 2020 at 13:16
  • $\begingroup$ You only show that $\mathfrak{p}$ is an ideal, not a prime ideal. Thus you cannot conclude $A\setminus\mathfrak{p}$ is multiplicative. So how does that contradict the maximality of $S$? $\endgroup$
    – Steve D
    Dec 31, 2020 at 16:43

Here is a slightly different solution which is also less set-theoretic.

Arbitrary unions of elements of $\Sigma$ are again multiplicatively closed subsets which do not contain 0 and so Zorn's lemma tells us that $\Sigma$ has maximal elements.

Suppose that $S$ is a maximal element of $\Sigma,$ and consider the fact that the ring of fractions $S^{-1}A$ is in particular a ring. Therefore it has at least one maximal ideal, which corresponds to a prime ideal $\mathfrak{p} \subset A\backslash S$. But then $A \backslash \mathfrak{p}$ is a multiplicative system in $A$ and it contains $S$ which was assumed to be maximal so that $A \backslash \mathfrak{p} = S \implies \mathfrak{p} = A \backslash S$. Furthermore, $\mathfrak{p}$ cannot contain any smaller prime ideals for a similar reason and so $\mathfrak{p}$ is minimal.

Conversely, the complement of a minimal prime ideal $\mathfrak{p}$ is a multiplicative system, which must then lie inside a maximal multiplicative system. Then the complement of the maximal multiplicative system would be a prime ideal contained in $\mathfrak{p}$ which was assumed to be minimal and so $A \backslash \mathfrak{p}$ is a maximal multiplicative system.


This is essentially the same answer as that of the OP's, but written to my liking. I put it here for my own future reference. Kindly ignore.

Lemma 1: Let $S$ be a maximal multiplicative subset of $A$ not containing $0$, then $A\backslash S$ is an ideal.

Proof. Let $\mathcal{I}$ be the set of ideals of $A$ contained in $ A\backslash S$ and ordered by inclusion.

  1. $ 0 \in \mathcal{I}.$
  2. If $\{\mathfrak{a}_i\}$ is a chain of ideals of $\mathcal{I},$ then $ \cup \mathfrak{a}_i\in \mathcal{I}.$

By Zorn's lemma, $\mathcal{I}$ contains a maximal element, say $\mathfrak{m}.$ Assume for contradiction that $\mathfrak{m} \subsetneq A\backslash S,$ then there is $x \notin \mathfrak{m} $ and $x \in A\backslash S,$ i.e. $ x \notin S.$ Consider the ideal $\bar{\mathfrak{m}} =(\mathfrak{m}, x),$ I claim that $\bar{\mathfrak{m}}$ is contained in $A\backslash S.$

By maximality of $S,$ the monoid generated by $S \cup \{x\}$ contains $0,$ i.e. there exists $s \in S$ and $r \in \mathbb{N}$ such that $ x^rs =0.$ Now assume $ m + \alpha x \in \bar{\mathfrak{m}} \cap (A\backslash S)^c =\bar{\mathfrak{m}} \cap S,$ then $ m x^{r-1}s + \alpha x^rs = mx^{r-1}s \in S,$ but $mx^{r-1}s \in \mathfrak{m},$ thus we have a contradiction. Hence $ \mathfrak{m} = A \backslash S,$ as desired. $\square$

Lemma 2: Let $\mathfrak{p}$ be an ideal, then $$ \mathfrak{p} \text{ is prime }\iff A\backslash \mathfrak{p} \text{ is multiplicative.} $$

Proof. This the contrapositive of the definition of a prime ideal. $\square$

The claim now follows from the following:

If $S$ maximal, then for a prime ideal $ \mathfrak{p},$ $$ \mathfrak{p} \subsetneq A \backslash S \iff S \subsetneq A\backslash \mathfrak{p}.$$ If $\mathfrak{p}$ is minimal prime, then for a maximal multiplicative set $S'$

$$ A \backslash S' \subsetneq \mathfrak{p} \iff A \backslash \mathfrak{p} \subsetneq S'.$$


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