# Proof verification request: complement of multiplicative set is ideal iff the ideal is prime

I am attempting to complete the proof given here, and am attempting to prove the following:

The other direction is a little harder, and usually we use a lemma: every ideal maximal with respect to being disjoint from a multiplicative subset of R is a prime ideal of R.

I found that I can drop the maximal requirement if I already know that the complement of the multiplicative subset is an ideal. if I do not know that the complement of the multiplicative subset is an ideal, then I need to go for a more complicated proof as sketched here, as far as I can tell.

I am going to prove: Ideal $$I$$ is prime iff $$S \equiv R / I$$ is multiplicative.

### Part 1: Ideal $$I$$ is prime $$\implies S \equiv R \setminus I$$ is multiplicative

A prime ideal $$I$$ is an ideal such that if $$xy \in I$$, then $$x \in I \lor y \in I$$. Now, consider $$s_1, s_2 \in S$$ (that is, $$s_1, s_2 \not \in I$$). We wish to show that $$s_1 s_2 \in S$$ for $$S$$ to be multiplicative.

For contraditction, assume $$s_1 s_2 \not \in S$$. Translating to statements in $$I$$, this means that $$s_1 s_2 \in I$$. However, if $$s_1 s_2 \in I$$, since $$I$$ is prime, we must have $$s_1 \in I \lor s_2 \in I$$, leading to a contradiction.

Hence, $$S$$ is a multiplicative subset of $$R$$.

### Part 2: If $$I$$ is an ideal such that $$S \equiv R \setminus I$$ is multiplicative, then $$I$$ is prime.

Let $$xy \in I$$. For $$I$$ to be prime, we wish to show that $$x \in I$$ or $$y \in I$$.

For contradiction, assume $$x \not \in I \land y \not \in I$$. Hence, $$x, y \in S$$. Since $$S$$ is multiplicative, $$xy \in S$$. However, $$S$$ and $$I$$ are disjoint, hence $$xy \not \in I$$.

Is the proof correct? I think it is, but I am worried that I somehow missed a need for maximality.

am I correct in my understanding of where the maximality is necessary? I feel it would be needed in Part 2 if we did not know that $$I$$ was an ideal. If the statement had been:

If $$I$$ is an ideal such that $$S \equiv R \setminus I$$ is multiplicative, then $$I$$ is is a prime ideal.

this is not provable, but the statement:

If $$I$$ is a subset such that $$S \equiv R \setminus I$$ is a maximal multiplicative subset, then $$I$$ is a prime ideal.

Your proof is correct. Maybe we can reformulate a bit to ease your worries: let us say that a subset $$T\subset R$$ of a ring is "anti-multiplicative" (completely random terminology, don't use it elsewhere) if $$1\not\in T$$ and $$\forall x,y\in R,\, xy\in T \Longrightarrow (x\in T\lor y\in T).$$
What you proved is essentially the following easy result: $$T$$ is anti-multiplicative iff $$S=R\setminus T$$ is multiplicative (I assume that a multiplicative set must include the unit). This implies as you showed that if the complement if an ideal is a multiplicative set then this ideal is prime.
On the other hand, as you suspect, if you don't already know that $$T$$ is an ideal, then you can't deduce it. In other words, an anti-multiplicative set does not have to be an ideal (it might not be stable by sums or products by arbitrary elements). For instance, $$T=R\setminus \{1\}$$ is anti-multiplicative (since $$S=\{1\}$$ is multiplicative), but is obviously not an ideal.