3

I am currently studying a course on commutative algebra and came across this statement:

An Ideal $I$ in a ring $R$ is prime if and only if $R\setminus I$ is a multiplicative set.

I have proved it the following way and was wondering if anyone would be kind enough to go through the proof to check for any errors.

So assume $R\setminus I$ is multiplicative, then $f\in R\setminus I$, $g \in R\setminus I\implies fg \in R\setminus I$. This is the same as saying $f,g$ not in $I$ implies $f.g$ not in $I$. So we just take the contrapositive of this to get our implication that $f.g$ $\in$ $I$ $\implies$ $f$ $\in$ $I$ or $g$ $\in$ $I$. Hence I is prime.

Now suppose $I$ is prime. Then,

$f.g$ $\in$ $I$ $\implies$ $f$ $\in$ $I$ or $g$ $\in$ $I$. Again taking the contrapositive of this, we can deduce the relation from above which would show that $R\setminus I$ is indeed multiplicative.

Is this a valid enough proof or would I need to be more "rigourous"?

user 1
  • 7,447
user1314
  • 1,498

2 Answers2

2

I correct one direction. can you correct the other?

I use Transposition: $P \implies Q \equiv (\lnot Q\implies \lnot P)$. (and De Morgan's Theorem)

Assume $R\setminus I$ is multiplicative set. So:
$$(f \in R\setminus I) \land (g\in R\setminus I)\implies f g \in R\setminus I.$$ This is the same as saying
$$f g \notin R\setminus I\implies (f\notin R\setminus I) \lor (g\notin R\setminus I)$$ This is the same as saying $$fg\in I\implies (f\in I) \lor (g\in I).$$ So $I$ is is prime.

user 1
  • 7,447
0

Multiplicativity of $\bar P\,$ is just the contrapositive of primality of $P$

$\begin{align}\ \ P\,\ {\rm prime} \iff&\ ab\in P\Rightarrow a\in P\,\ {\rm or}\,\ b\in P\\ \iff&\ a,b\in \bar P\Rightarrow\,ab\in \bar P\quad \rm [contrapositive\ of\ prior]\\ \iff&\ \bar P\,\ {\rm is\ closed\ under\ multiplication}\end{align}$

Note $\bar P$ closed under multiplication is, in $R/P,\,$ $\,a,b\neq 0\,\Rightarrow\,ab\neq 0,\,$ i.e. the domain property.

Remark $\ $ More generally: $S$ is multiplicative and saturated (closed under divisors) $\iff$ its complement is a union of prime ideals. This becomes clearer when one studies localizations

Bill Dubuque
  • 272,048
  • Sorry just to comment on your remark, is S the complement of the prime ideal? Also, I am studying localisations of rings now but I haven't come across any such result about union of prime ideals and I can't seem to link your statements together. Could you please explain a bit more? – user1314 Apr 11 '15 at 13:34
  • @user1314 In the remark $S$ is any subset of the ring. It can be viewed as a generalization which includes a converse of the result you mention. – Bill Dubuque Apr 11 '15 at 13:38
  • Ok, but I dont see how that would be related to localisations? Sorry if I sound like an idiot – user1314 Apr 11 '15 at 13:41
  • @user1314 Think about what happens in $,S^{-1} R.,$ By the correspondence theorem its maximal ideals correspond to prime ideals in $,R,$ that are maximal wrt to being disjoint from $,S.\ \ $ – Bill Dubuque Apr 11 '15 at 13:49
  • I dont think I have done the correspondence theorem yet and havent actually gone too much in depth into localisation so maybe that's why I don't understand you. I will read further and try to make some sense! thanks for the help – user1314 Apr 11 '15 at 13:56
  • 1
    @user1314 Then you can safely ignore the remakr till when you reach that point. – Bill Dubuque Apr 11 '15 at 13:57