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

Solution 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?