# Commutative Algebra/Radicals, strong Nakayama

## Contents

## 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]

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