Prove that a maximal ideal is also a prime ideal

Question

If $S$ is a multiplication closed subset of a ring $R$, suppose $P$ is the maximal element of the set of ideals and is disjoint from S, then $P$ is the prime ideal. I'm a little bit stuck in step 2: (1) suppose $ab \in P$, if $a \notin P$ and $b \notin P$, then the ideal generated by a $(P, a) \gt P$

(2) then we have some S$_1$=P$_1$+r$_1$a（ S$_1$ $\in$ S, r$_1$$\in$ R, P$_1$ $\in$ P）and some S$_2$=P$_2$+r$_2$b（ S$_2$ $\in$ S, r$_2$$\in$ R, P$_2$ $\in$ P）, here Richard says S$_1$S$_2$ $\in$ P, and hence contradictory to our assumption that P is disjoint from S.

I do not quite follow here. By computing, I have $S_1S_2 = P_1P_2 + P_1r_2b + P_2r_a + r_1r_2ab$, I don't see why it is in $P$. Can anyone help with this?

The video is at https://www.youtube.com/watch?v=cXLiOgHQHP4. The proof part ends around 24:43.

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Jun 23 '24 at 09:24
The first 3 terms have a term from P in them explicitly. The last term has ab, which by assumption is in P. Ideals are closed under addition and multiplication by arbitrary elements of R, so the sum of the 4 terms is an element of P. — Ragib Zaman, Jun 23 '24 at 09:27
I can understand the first and fourth, but the second and third also containing a and b, which are not in P, so why they are also in P? — Yi Shen, Jun 23 '24 at 09:29
Multiplying $P_1$ and $P_2$ by arbitrary elements of the ring $R$ produces elements inside the ideals, and $r_2b, r_1a$ are just some elements in the ring $R$. — TravorLZH, Jun 23 '24 at 09:40
Your title is extremely misleading, and you are not reproducing faithfully the hypothesis of Richard's theorem. $S$ is a multiplicatively closed subset of $R$, and $P$ is *a* maximal element of the set of those ideals or $R$ *which are* disjoint from $S$. He uses this hypothesis for the existence of your $S_1\in S\cap(P,a)$ and $S_2\in S\cap (P,b)$. — Anne Bauval, Jun 23 '24 at 10:06

Prove that a maximal ideal is also a prime ideal

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