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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 $xy \not \in I$.

This contradicts our assumption.

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.

Siddharth Bhat
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1 Answers1

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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 $$\forall x,y\in R,\, xy\in T \Longrightarrow (x\in T\lor y\in T).$$

Then a prime ideal is exactly an ideal which is also anti-multiplicative.

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.

Then the stronger result would be that minimal anti-multiplicative sets are actually ideals, so they are prime ideals.

Captain Lama
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  • Thank you, this perspective of ideals as "anti-multiplicative" is helpful for organising my knowledge. I'll try and re-examine some of my ideas from this lens – Siddharth Bhat Mar 05 '20 at 13:07