Question from the proof of the thread dealing with showing an ideal maximal in the set of ideals not intersecting multiplicative sets is prime.

Question

I have a question from the following thread: A maximal ideal among those avoiding a multiplicative set is prime

I didn't ask in the thread, because it looks like user38268's account is deleted. In particular, I am just asking about problem $1$ that Spook asked. That is: "Let $S$ be a multiplicatively closed subset of a ring $R$ (which I will assume is nonzero, commutative with identity) and let $I$ be an ideal of $R$ which is maximal among ideals disjoint from $S$. Show that $I$ is prime."

Now, in user38268's answer, they talk about using Zorns to show there is an ideal, $P$, with $I\subset P$ maximal to the condition that $P\cap S=\emptyset$. $\textbf{why did they do this? Aren't we already given that $I$ is precisely that from the statement of the problem?}$

Next, in the comments on the original question, Spook answers this question by saying "If $x,y$ are outside $I$, then by maximality, $(x,I)$, $(y,I)$ both intersect $S$. So, for some $r,s\in R$, $i,j\in I$, $xr+i, ys+j \in S$. So $(xr+i)(ys+j)\in S$, and on expanding one sees $(xy,I)$ intersects $S$, so $xy$ is outside $I$. Hence $I$ is prime."

Is this a complete argument? I'm assuming, "by expanding", just means multiplying $(xr+i)(ys+j)$ together. My confusion is coming from the answer from user38268 in which they give the hint that "if $f,g\notin P$, then what can you say about the ideals $P+(f)$ and $P+(g)$"? So, I take it as this...

Suppose $f, g\notin P$, then let $x\in P+(f)$ and $y\in P+(g)$. Then, $x=fr+p_1$ and $y=gs+p_2$ for $r,s\in S$. Then, we multiply $xy$ together to get something that is not in $P$, hence $P$ is prime... but that feels off. I am wondering if what they have is correct, or if my thought process is correct, but maybe I am not executing it as I should. Any help is greatly appreciated! Thank you.

Answer 1

You are absolutely correct, that Zorn is not needed here, as you are already given that $I$ is maximal subject to the constraint. The point being made is that even if you were not given this, you could always find some $P\supseteq I$ which was maximal subject to the constraint, and $P$ is prime. This is the more common application of the argument, so the answer covered this general case, even though the question only deals with the case where $I$ is already maximal subject to the constraint.

Now consider:

$$(xr+i)(ys+j)=xyrs+iys+jxr+ij$$

The first term lies in $(xy)$ and the other three lie in $I$ so the whole things lies in $(xy)+I$ and in $S$. Thus $xy\notin I$ which is what you are trying to prove.

Regarding your last paragraph, $r,s\in R$, not necessarily in $S$ as you wrote. The letters $f$ and $g$ here are playing the role of the letters $x$ and $y$ in the comment. Also $x$ and $y$ are used here to mean something different to the comment.

Essentially the hint and the comment are exactly the same argument, but with letters swapped around. It is unnecessarily confusing to think about both at the same time! I suggest you concentrate on the comment, as it is a more complete argument.

Does it make sense now, after expanding and looking at the individual terms?