THE JACOBSON RADICAL OF A RING

E. L. Lady

THEOREM. Let R be a non-trivial ring and r ∈ R. The following conditions are equivalent and the set of r ∈ R satisying these conditions forms a (two-sided) ideal.

(1) rM = 0 for all simple left R-modules M.

(2) Nr = 0 for all simple right R-modules N.

(3) φ(r) = 0 for all homomorphisms φ from R into a simple left R-module.

(4) ψ(r) = 0 for all homomorphisms ψ from R into a simple right R-module.

(5) r belongs to all maximal left ideals in R.

(6) r belongs to all maximal right ideals in R.

(7) For all x ∈ R, 1 − xr is invertible.

(8) For all x ∈ R, 1 − rx is invertible. proof: Let J denote the set of elements satisfying condition (1). It is easy to see that J is a two-sided ideal.

(1) ⇒ (3): Let φ: R → M, where M is a simple left R-module. If rM = 0 then φ(r) = rφ(1) = 0.

(3) ⇒ (1): Let M be a simple R-module and m ∈ M. There exists an R-linear map φ R → M given by φ(x) = xm. Then rm = φ(r) so if φ(r) = 0 for all φ: R → M then rm = 0 for all m ∈ M, i. e. rM = 0.

(3) ⇔ (5): If M is a simple left R-module and φ: R → M is non-trivial then Ker φ must be a maximal left ideal L (why?). On the other hand, if L is a maximal left ideal and φ: R → R/L is the quotient map then φ is a map from R into a simple left R-module and Ker φ = L. Thus φ(r) = 0 for all maps φ from R into simple left R-modules if and only if r belongs to all maximal left ideals.

(2) ⇔ (4) ⇔ (6): Analogous.

(5) ⇒ (7): If r belongs to every maximal left ideal then so does xr for any x ∈ R. It follows easily that 1 − xr does not belong to any maximal left ideal and therefore 1 − xr must be left invertible. Let s ∈ R be such that s(1 − xr) = 1. Then 1 − s = −sxr ∈ J so by the preceding s = 1 − (1 − s) is left invertible, say ts = 1. But then t = t1 = ts(1 − xr) = 1(1 − xr) = 1 − xr, so s(1 − xr) = 1 = ts = (1 − xr)s and 1 − xr is in fact invertible.

(7) ⇒ (5): Let L be a maximal left ideal. If 1 − xr is invertible for all x ∈ R then 1 − xr ∉ L. Thus 1 ∉ Rr + L, so L ⊆ Rr + L ⊊ R. Since L is a maximal left ideal, this forces L = Rr + L, and therefore r ∈ L.

(6) ⇔ (8): Analogous.

(7) ⇒ (1): [Alternate proof] Suppose that 1 − xr is invertible for all x ∈ R and suppose bwoc that rM ≠ 0 for some simple left R-module M. Choose m ∈ M with rm ≠ 0. Then Rrm is a non-trivial submodule of M so M = Rrm since M is simple. Thus m ∈ Rrm so m = xrm for some x ∈ R. Thus (1 − xr)m = 0. But by assumption there exists u ∈ R such that u(1 − xr) = 1. Therefore m = u(1 − xr)m = 0, a contradiction. [This is a special case of Nakayama’s Lemma.]

(7) ⇒ (8): Since we have seen that (1) is equivalent to (7), the set of elements satisfying (7) is the two-sided ideal J. Thus if r satisfies (7) then so does rx for any x, so by (7) 1 − rx is invertible for all x.

(8) ⇒ (7): Analogous. ✓