Editing Ideal (ring theory) (section)

== Radical of a ring ==
{{main|Radical of a ring}}

Ideals appear naturally in the study of modules, especially in the form of a radical.

: ''For simplicity, we work with commutative rings but, with some changes, the results are also true for non-commutative rings.''

Let ''R'' be a commutative ring. By definition, a [[primitive ideal]] of ''R'' is the annihilator of a (nonzero) [[simple module|simple ''R''-module]]. The [[Jacobson radical]] <math>J = \operatorname{Jac}(R)</math> of ''R'' is the intersection of all primitive ideals. Equivalently,
: <math>J = \bigcap_{\mathfrak{m} \text{ maximal ideals}} \mathfrak{m}.</math>
Indeed, if <math>M</math> is a simple module and ''x'' is a nonzero element in ''M'', then <math>Rx = M</math> and <math>R/\operatorname{Ann}(M) = R/\operatorname{Ann}(x) \simeq M</math>, meaning <math>\operatorname{Ann}(M)</math> is a maximal ideal. Conversely, if <math>\mathfrak{m}</math> is a maximal ideal, then <math>\mathfrak{m}</math> is the annihilator of the simple ''R''-module {{tmath|1= R/\mathfrak{m} }}. There is also another characterization (the proof is not hard):
: <math>J = \{ x \in R \mid 1 - yx \, \text{ is a unit element for every } y \in R\}.</math>
For a not-necessarily-commutative ring, it is a general fact that <math>1 - yx</math> is a [[unit element]] if and only if <math>1 - xy</math> is (see the link) and so this last characterization shows that the radical can be defined both in terms of left and right primitive ideals.

The following simple but important fact ([[Nakayama's lemma]]) is built-in to the definition of a Jacobson radical: if ''M'' is a module such that {{tmath|1= JM = M }}, then ''M'' does not admit a [[maximal submodule]], since if there is a maximal submodule {{tmath|1= L \subsetneq M }}, <math>J \cdot (M/L) = 0</math> and so {{tmath|1= M = JM \subset L \subsetneq M }}, a contradiction. Since a nonzero [[finitely generated module]] admits a maximal submodule, in particular, one has:
: If <math>JM = M</math> and ''M'' is finitely generated, then {{tmath|1= M = 0 }}.

A maximal ideal is a prime ideal and so one has
: <math>\operatorname{nil}(R) =
\bigcap_{\mathfrak{p} \text { prime ideals }} \mathfrak{p} \subset \operatorname{Jac}(R)</math>
where the intersection on the left is called the [[Nilradical of a ring|nilradical]] of ''R''. As it turns out, <math>\operatorname{nil}(R)</math> is also the set of [[nilpotent element]]s of ''R''.

If ''R'' is an [[Artinian ring]], then <math>\operatorname{Jac}(R)</math> is nilpotent and {{tmath|1= \operatorname{nil}(R) = \operatorname{Jac}(R) }}. (Proof: first note the DCC implies <math>J^n = J^{n+1}</math> for some ''n''. If (DCC) <math>\mathfrak{a} \supsetneq \operatorname{Ann}(J^n)</math> is an ideal properly minimal over the latter, then <math>J \cdot (\mathfrak{a}/\operatorname{Ann}(J^n)) = 0</math>. That is, {{tmath|1= J^n \mathfrak{a} = J^{n+1} \mathfrak{a} = 0 }}, a contradiction.)