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.
8.
From this formula, we have .
This fact is mentioned on pg.7 of Atiyah-Macdonald.
9. .
From this formula, we have .