# Atiyah-Macdonald: Exercise 3.6

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:

$$"\impliedby"$$

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

$$"\implies"$$

$$S$$: 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?

Edit

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?

• 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 $A\setminus S$. Dec 26, 2020 at 5:03
• For a more "commutative algebra" approach, you can localize and work in $S^{-1}R$. Dec 26, 2020 at 20:30
• Can you please give a bit more hint on this? Dec 27, 2020 at 5:01
• Find a maximal ideal $\mathfrak{m}$ of $S^{-1}R$ and then show $\mathfrak{m}\cap R$ is the prime ideal you seek. Dec 27, 2020 at 7:20

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

$$"\implies"$$

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.

$$"\impliedby"$$

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 $$\Sigma.$$

• Your first proof is not complete. You cannot "contradict the maximality of $S$" without first showing $\mathfrak{p}$ is prime. Dec 26, 2020 at 22:35
• @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. Dec 27, 2020 at 4:58
• 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. Dec 27, 2020 at 7:18
• @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? Dec 31, 2020 at 13:16
• 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 $S$? 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 $$S'$$

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