Commutative Algebra/Radicals, strong Nakayama

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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.