# More About Ideals

Recall that we defined three operations on ideals: intersection, sum and product. We can take intersection and sum of *any collection* of ideals (even infinitely many of them), but we can only define the product of finitely many ideals.

As we mentioned earlier, ideals are a generalization of elements of the ring. In the case of with ideals , we have

Thus, philosophically, we have the following correspondences in mind

- product of ideals ↔ product,
- sum of ideals ↔ gcd,
- intersection of ideals ↔ lcm,
- reverse containment ↔ divides.

Armed with the above picture, the following properties become quite natural.

Proposition 1.Given any ideals and collection of ideals of the ring , we have:

- .
- .
- .

**Proof**

For the first claim, ⊆ follows from: if and , then

for some .

Then where each . Hence .

For the other containment ⊇, note that since we have and hence .

Second claim: if , then since it is a multiple of , and similarly it lies in since it is a multiple of . Thus and so must contain all finite sums with and .

Last claim: . ♦

# Coprime Ideals

Suppose are coprime. The following properties of are familiar to us.

- (Bezout’s theorem) There exist integers
*a*,*b*such that . - Any common multiple of
*m*and*n*is a multiple of*mn*. - (Chinese Remainder Theorem) For any
*c*mod*m*and*d*mod*n*, there is a unique*a*mod*mn*such that and . - Any powers , are also coprime.

Let us generalize this to the setting of ideals.

Definition.Let be a ring. We say that ideals are

coprimeif .

**Note**

It is clear that ideals are coprime if and only if there exist such that .

Also, if is a maximal ideal of , then it is coprime to any ideal *not* contained in it, because is an ideal which strictly contains , so by the maximality of , .

The following two results give us a recipe for producing more coprime pairs of ideals given existing ones. Philosophically the results make sense if we imagine coprime to mean “not sharing any factor”.

Proposition 2.If ideals are coprime, then so are any powers .

**Proof**

Pick such that . Taking this to the (*m*+*n*)-th power, the LHS is a sum of *m*+*n*+1 terms, each of the form where *C* is an integer, are integers such that . Since either or , this term is a multiple of or so it lies in or . Hence the whole sum lies in and . ♦

Proposition 3.If are ideals such that is coprime and is coprime, then is coprime.

**Proof**

Find such that . Also find such that . Then

and we are done. ♦

Corollary.If are ideals such that and are coprime for each , then and are coprime.

**Proof**

Repeatedly apply proposition 3: since and are coprime pairs, so is . Together with the fact that is coprime, we see that is coprime, etc. ♦

# Consequences

Finally we discuss the consequences given two coprime ideals of a ring *A*. Immediately we obtain Bezout’s theorem from the definition of coprimality: if are coprime, there exist such that . Next we have:

Proposition 4.If are coprime ideals, then .

**Proof**

By proposition 1, so equality holds throughout. ♦

Corollary.If are pairwise coprime ideals of A, then .

**Note**

In general, a collection of items is said to satisfy **pairwise** *X* if any two of them satisfy *X*.

**Proof**

When *n* = 1, clear. For , suppose it holds for *n* – 1 so that . By corollary to proposition 3, and are coprime. By proposition 4 we have

♦

Chinese Remainder Theorem (CRT).If are coprime ideals, then the natural map

,

is an isomorphism.

**Proof**

The kernel of the map is clearly which is by proposition 4. To prove that the map is surjective, pick such that . Now for any in the RHS, let . Since and we have

so maps to . ♦

As before, this immediately generalizes to the following.

Corollary.If are pairwise coprime ideals, then the natural map

is an isomorphism.

**Proof**

For *n* = 1, OK. For *n* > 1, by corollary to proposition 3, and are coprime so CRT gives us

via the natural map. Now proceed inductively. ♦

The following is an important special case.

Example.Let be distinct maximal ideals of the ring A. Then their powers are pairwise coprime for any . Hence we have: