Commutative Algebra/Radicals, strong Nakayama

Radicals of ideals edit

Definition 13.1:

Let   be a ring,   an ideal. The radical of  , denoted  , is

 .

A radical ideal is an ideal   such that  .

Theorem 13.2:

Let   be a ring,   an ideal. Then

 .

Proof:

For  , note that  . For  , assume for no    . Form the quotient ring  . By theorem 12.3, pick a prime ideal   disjoint from the multiplicatively closed set  . Form the ideal  .   is a prime ideal which contains   and does not intersect  . Hence   is not in the right hand side. 

Corollary 13.3:

The radical of an ideal is an ideal.

Proof:

Intersection of ideals is an ideal. 

Definition 13.4:

Let   be a ring,  . The Jacobson radical of   is defined as thus:

 .

Theorem 13.5:

Let   be a ring,  .   is a radical ideal.

Proof:

Clearly,  . Further   from theorem 13.2; the last equality from  . 

The radicals of the zero ideal edit

Definition 13.6:

Let   be a ring.   is an ideal. The nilradical of  , written  , is defined as

 .

Note that by definition

 ,

the set of nilpotent elements.

Theorem 13.7:

 .

Proof:

Theorem 13.2. 

Definition 13.8:

Let   be a ring.   is an ideal. The Jacobson radical of  , written  , is defined as

 .

We have  .

If   is a ring,   is the set of units of  .

Theorem 13.9:

Let   be a ring,   its Jacobson radical. Then

 .

Proof:

 : Let  ,  . Assume  . Form the ideal  ; by theorem 12.8 there exists   maximal with  , hence  . If  , then  , contradiction.

 : Assume   for all   and  . Then there is a maximal ideal   not containing  . Hence   and   for a   and an  . Hence   is not a unit. 

Radicals and localisation edit

Definition 13.10:

If   is a ring,   is a multiplicative subset,   an ideal, set

 ,

the localisation of the ideal   with respect to  .

Theorem 13.11:

Let   be a ring,   an ideal,   a multiplicatively closed subset. Then

 .

Proof:

Let  , that is,  . Then  ,  ,  . There exists   such that  . Thus  , whence   and  . Thus,  .

Let  . We may assume  . Choose   such that  . Then  , whence  . 

Strong Nakayama lemma edit

Exercises edit

  1. Prove that whenever   is a reduced ring, then the canonical homomorphism   is injective.