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Let $R$ be a commutative ring and let $S \subset R$ is closed under multiplication. Let $a$ to be an ideal in $R$ such that $a \cap S = \emptyset$. We further assume that if $b$ is an ideal in $R$ such that $a \subset b$ and $a \ne b$, then $b\cap S \ne \emptyset$. Prove that a is a prime ideal in R.

So I am seriously lost on this question, I've been picking different elements to see if I could get some sort of direction but I simply don't know how to approach this problem. Any hints would be greatly help me.

pcon
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  • Your title suggests that you are trying to prove that an ideal is not prime but you then ask us to prove that the ideal is prime. – John Douma Mar 26 '19 at 03:32
  • I apologize for that. I was thinking of a contradictory approach and accidentally set the title as what I was thinking. – pcon Mar 26 '19 at 03:39
  • Always try the definitions first. Assume that $xy\in a$ and $x\not\in a$ and try to show that $y$ must be in $a$. I don't know if this is that straightforward but you must start somewhere. – John Douma Mar 26 '19 at 03:42
  • So following that approach $xy\notin S$, then because S is closed under multiplication, it must follow that $x,y\notin S$ as well? – pcon Mar 26 '19 at 03:55
  • It is impossible for $a\cap S\neq \emptyset$, and $a\subseteq b$ and $b\cap S=\emptyset$ all at the same time. I rather think you interchanged = and $\neq$. At that point, this question is a duplicate. – rschwieb Mar 26 '19 at 11:13
  • I think he means $a\cap S =\varnothing$ but $b\cap S\neq \varnothing$. – Ehsaan Mar 26 '19 at 20:54
  • See this answer – Bill Dubuque Mar 27 '19 at 01:11
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    I am very sorry about the inconsistency. I only realize that now that you pointed it out as I've been working on the "inconsistent" version for a long time. – pcon Mar 27 '19 at 03:07
  • @pcon Thanks for getting back to us and clearing that up, and also editing the question :) – rschwieb Mar 28 '19 at 15:06

1 Answers1

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This is actually a pretty tricky question, so don't worry too much that you haven't gotten it yet.

Basically your hypothesis is as follows: $a$ is an ideal which is maximal with respect to the property that $a\cap S=\varnothing$. Usually these maximality properties yield a prime ideal.

For example, I'll show you how to prove the following similar (but easier) statement:

Every maximal ideal is prime.

(The "best" proof is to take the quotient and show it's a field, but let's try the following naive proof just to illustrate the point.)

Suppose $m$ is a maximal ideal in a commutative ring $R$. We want to show that if $ab\in m$ then either $a\in m$ or $b\in m$. If neither $a$ nor $b$ lay in $m$, then look at the ideals $I=m+(a)$ and $J=m+(b)$ --- notice these each properly contain $m$. So by maximality of $m$, it must be that $I,J$ are not proper ideals of $R$, so $I=J=R$.

Now $m+(a)=R$ means that $x+ra=1$ for some $x\in m$ and $r\in R$. Similarly, $y+sb=1$ for some $y\in m$ and $s\in R$. This gives: $$1 = (x+ra)(y+sb) = xy + x(sb) + y(ra) + (rs)ab$$ which lays in $m$ because each term does --- it's important to note that here is where we use the assumption that $ab$ lays in $m$! This shows $1\in m$, so $m=R$ which contradicts the assumption that maximal ideals are proper ideals.

Often in commutative algebra, if you have a proper ideal $p$ which is maximal with respect to some useful property, then $p$ is actually prime ideal. And the proof usually goes along the lines I wrote above: suppose $ab\in p$ but $a\notin p$ and $b\notin p$, then consider the ideals $p+(a)$ and $p+(b)$ and use maximality here.

Good luck!

Ehsaan
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  • Remark - a version of the "best" proof can still be used: localize away from $S,$ and then the ideal becomes maximal in $S^{-1}R,$ by the correspondence of ideals of $S^{-1}R$ to ideals of $R$ not meeting $S.$ Then use the fact that this bijection of ideals restricts to a bijection on the subsets of prime ideals of $R$ not meeting $S$ and prime ideals of $S^{-1}R.$ – Stahl Mar 26 '19 at 08:00
  • First let me appreciate the work you have put into your answer. But then I need to ask if it can be re-written to correct what the poster has apparently written incorrectly. The users assumptions ($a\cap S\neq \emptyset$, $a\subseteq b$ and $b\cap S=\emptyset$) are inconsistent with each other. It seems that they inadvertently swapped the $\neq$ and $=$. I'd like to see your answer stand, but not with the user's original inconsistent assumptions. – rschwieb Mar 26 '19 at 13:22
  • @rschwieb: Ah, thank you. I hastily read OP's statement and interpreted it "correctly" (meaning the usual ring theory exercise). I'll think about whether OP's property implies prime. – Ehsaan Mar 26 '19 at 18:26
  • @Ehsaan The OP's set of hypotheses is logically inconsistent with itself. If I were in your place, I'd make a comment to that effect and answer the "corrected" version (which I think is probably the statement we are predicting.) – rschwieb Mar 26 '19 at 20:05
  • @rschwieb Thank you for the hint, I believe I have solved it although I'm not sure if it's correct. My approach is that I choose $st \in a$ and $s, t \notin a$. Then, $I = a + s, J = a + t$, now because $I, J \in b$ I can find elements $s_1, s_2 \in S$ such that $s_1\in I, s_2 \in J$. $s_1s_2$ would now belong to $a$ which is a contradiction to the original assumption. – pcon Mar 28 '19 at 14:37
  • @pcon You can compare it with the solution linked in the comments above. – rschwieb Mar 28 '19 at 15:07