Jacobson Radical Annihilates Simple Modules

Every element of the Jacobson radical acts as 0 on any simple module.
Jacobson Radical Annihilates Simple Modules

Statement

Let RR be a commutative ring and MM a RR-module.

Theorem.
The satisfies

J(R)AnnR(M), J(R)\subseteq \mathrm{Ann}_R(M),

so in particular

J(R)M=0. J(R)\,M=0.

Reason (standard): AnnR(M)\mathrm{Ann}_R(M) is a , and by we have J(R)J(R) contained in every maximal ideal, hence in AnnR(M)\mathrm{Ann}_R(M). (See also .)

This fact is one of the key inputs behind .

Examples

  1. Local ring: the maximal ideal kills the residue field.
    If RR is with maximal ideal m\mathfrak m, then J(R)=mJ(R)=\mathfrak m.
    The simple RR-module is R/mR/\mathfrak m (the ), and indeed m(R/m)=0\mathfrak m\cdot (R/\mathfrak m)=0.

    Example: R=k[x]/(x2)R=k[x]/(x^2), m=(xˉ)\mathfrak m=(\bar x). Then xˉ\bar x acts by 00 on kR/(xˉ)k\cong R/(\bar x).

  2. Integers.
    J(Z)=(0)J(\mathbb Z)=(0), so J(Z)M=0J(\mathbb Z)M=0 for every simple Z\mathbb Z-module.
    The simple Z\mathbb Z-modules are Z/pZ\mathbb Z/p\mathbb Z, and the zero ideal acts as 00 as expected.

  3. A finite quotient of Z\mathbb Z.
    Let R=Z/12ZR=\mathbb Z/12\mathbb Z. Then J(R)=(6)J(R)=(6) (intersection of the maximal ideals (2)(2) and (3)(3)).
    The simple RR-modules are R/(2)Z/2ZR/(2)\cong \mathbb Z/2\mathbb Z and R/(3)Z/3ZR/(3)\cong \mathbb Z/3\mathbb Z.
    In both cases, the class of 66 acts as 00, so J(R)J(R) annihilates each simple module.