Today we shall discuss some interesting properties of elements of a ring.
1. If is not a unit, then it is present in some maximal ideal of the ring . Self-explanatory.
2. If is present in every maximal ideal, then is a unit for all . Proof: Let not be a unit. Then it is present in some maximal ideal (from 1). Let , where is an element from the maximal ideal is a part of. Then . Hence, is also a member of the maximal ideal, which is absurd.
Let’s break down this theorem into elementary steps, and present a better proof (than given on pg.6 of “Commutative Algebra” by Atiyah-Macdonald). If for some maximal ideal , then for all . Similarly, If for some maximal ideal , then for all . This argument can then be extended to the fact that if all maximal ideals, then any maximal ideal for all . An element not there in any maximal ideal is a unit. Hence, is a unit.
3. If is a unit for all , then is part of every maximal ideal in . Proof: Let is assume is not part of some maximal ideal. Then there exists some that maximal ideal such that . This implies that , which is impossible as is a unit. The same argument can be used for .
On pg.6 of Atiyah-Macdonald, it is mentioned that if are ideals, then if and . It is not elaborated in the book, and the flippant reader may be confused. I’d like to elaborate on this concept.
consists of those elements in which are also there in . Now elements of wrt those elements of both and that are in ] [ wrt those elements of that are in and those elements of that are not] [ wrt those elements of that are not in and those elements of that are] [ wrt those elements of both and that are not in ]
The second and the third terms are null sets, as can be easily seen.
The fourth term is NOT necessarily empty. However, it becomes an empty set if . It may also become an empty set under other conditions which ensure that if both and are not in , then .
In summation, is definitely true when . However, it is also true under other conditions which ensure that the fourth term is an empty set.
I have extended a small paragraph in Atiyah-Macdonald to a full-fledged exposition: ( and are ideals in commutative ring )
1. – Both sides contain all elements of and . Remember that . However, the right hand side also contains elements of the form and , which the left hand side does not contain.
2. – This has already been explained above.
3. – Both are exactly the same.
4. – There might be such that . However, any element in will definitely be present in .
5. – LHS contains elements of the form , which the RHS doesn’t. In fact, LHS is an ideal while the RHS isn’t. You might wonder how LHS is an ideal. I have just extended the algorithm used to make an ideal when and are both ideals to situations in which is an ideal and is any subset of .
6. – The RHS may contain elements of the form for .
7. – Easy enough to see.
From this formula, we have .
This fact is mentioned on pg.7 of Atiyah-Macdonald.
From this formula, we have .