What if ?
, and is non invertible (or ) and does not divide .
So there is a maximal ideal in such that (Krull's theorem). Thus isn't an element of , because it doesn't belong to .
Let R be an integral domain with quotient field , and let I and J be ideals of R. Let X denote the set of maximal ideals of R.
Let . Prove that if , then
pf.
Now, , so
To be honest, I'm quite lost after this, how should I go on with this problem? thanks.
unfortunately the conclusion is not correct! you can have both and for example i will show that we actually have : define
obviously is an ideal of if then and hence if then there exists a maximal ideal now since there exist such that
thus hence which is impossible since this proves that also note that for all maximal ideals because the map
defined by is injective. thus hence:
Thanks for your massive help!!!
My last question here is: Prove that if and only if , for each maximal ideal M of R.
Proof so far.
Suppose that I=J, then , since I and J are ideals.
Conversely, suppose that , pick an element in , then it has the form .
Now, I=J implies that there exists an element such that , thus , so we have , and the reverse is the same.
Is this correct? My solution look a little bit too easy compare to the other problems...
Thanks!!!
no! your proof is not correct: it's not true that if is an integral domain, then we can only say that but we may have
for example if then anyway, here's how you should solve the problem:
it's obvious that implies that because by definition:
conversely suppose for all maximal ideals of we'll show that : let and define see that is an
ideal of if then and thus otherwise, for some maximal ideal now since we have for some
but then and thus which is impossible because this completes the proof of an exactly the
same argument shows that and therefore