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Suppose $R$ is a commutative ring with unity and that $I$ is an ideal of $R$. Then $I$ is a prime ideal iff $R-I$ is multiplicative (if $a,b\in R-I$, then $ab\in R-I$).

So far I have been able to prove the forward direction. If $ab\in I$, then $a\in I$ or $b\in I$. Assume that $a\in I$. Then $I$ contains elements of the form $ar=ra$, where $r\in R$. Thus, $ar\notin R-I$ for any $r\in R$. If $c,d\in R-I$, then $c\neq ar_1$ and $d\neq ar_2$ for some $r_1,r_2 \in R$. Thus, $cd\notin I$, implying that $cd\in R-I$, and hence, $R-I$ is multiplicative.

I'm having trouble with the other direction. I want to assume $R-I$ is multiplicative, so if $c,d \in R-I$, then $cd\in R-I$, so $cd\notin I$, but I don't see what this will do for me. Can someone help me out? Thank you in advance.

Tim
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    you have $c,d\notin I$ and you end up with the conclusion $cd\notin I$... does this imply (somehow) I being Prime ideal??? –  Sep 23 '13 at 05:57
  • What is a prime ideal by the way?? $I\subseteq R$ is prime ideal if $ab\in I$ implies $a\in I$ or $b\in I$... Can you form contra positive statement for this?? –  Sep 23 '13 at 05:59

4 Answers4

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In the following, I am going to assume that $I$ is a proper ideal of the ring $R$. This assumption allows me to bypass certain logical intracacies associated with the possibility of a vacuous multiplicatively closed set $R - I$, while still I hope capturing the algebraic essence of the situation.

These things being said:

Let $R$ be a commutative ring with unit, and let $I$ be a prime ideal in $R$. Then according to the definition (given in this wikipedia entry: http://en.m.wikipedia.org/wiki/Prime_ideal), $I \ne R$. Let $a, b \in R - I$. I claim $ab \in R - I$; for if $ab \notin R - I$, then clearly $ab \in I$, whence $a \in I$ or $b \in I$ since $I$ is prime. But this contradicts $a, b \in R - I$, whence we must have $R - I$ multiplicatively closed.

Conversely, let $R - I$ be multiplicatively closed, and suppose the proper ideal $I$ is not prime. Then there exist $a, b \in R$ with $ab \in I$ but $a \notin I$ and $b \notin I$. Then $a, b \in R - I$; but this forces $ab \in R - I$ contradicting the assumption that $ab \in I$. Thus $I$ must be prime.

I hope this helps clarify things somewhat. Cheers, and as always

Fiat Lux

Robert Lewis
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I don't know why you went so far in your proof of the forward direction: by contrapositive, $ab\not\in R-I \Leftrightarrow ab\in I\Rightarrow a\in I \mbox{ or } b\in I \Leftrightarrow a\not\in R-I \mbox{ or } b\not\in R-I$, and done.

For the second direction, assume $ab\in I$. Hence $ab\not\in R-I$. If both $a$ and $b$ were in $R-I$, then $ab\in R-I$ since you assume that $R-I$ is multiplicative. Hence one of them must lie in $I$, hence $I$ is prime.

zarathustra
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If $I$ is prime, then with $a$ and $b$ not in $I$, then $ab$ cannot be in $I$ (because using primeness of $I$ ,we would get $a\in I$ or $b\in I$.

If $R\backslash I$ is multiplicative, then let $ab\in I$. For sake of contradiction say $a\notin I$ and $b\notin I$, then $ab\in R\backslash I$ (by assumption), contradicting the fact that $ab\in I$. Hence at least one of $a$ and $b$ must be in $I$.

Daniel Montealegre
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The assertion is wrong. If $I$ is a prime ideal, then $R - I$ is a multiplicative set, however the converse does not hold. If $R -I$ is a multiplicative set, then you cannot even say that $I$ is an ideal. Of course, if $I$ is an ideal, then it's prime.

However it's equivalent to the Axiom of Choice that if $S$ is a multiplicative set, then there exists an ideal $I$ such that $I \subset R - S$.

user40276
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