Answer to Question #17656 in Abstract Algebra for Tsit Lam
Using the definition of rad R as the intersection of the maximal left ideals, show directly that rad R is an ideal.
For y ∈rad R, r ∈ R, and many maximal left ideal, we must show that yr ∈m. Assume otherwise; then Rr + m = R. Consider the left R-modulehomomorphism ϕ: R → R/m definedby ϕ(x) = xr (mod m).Since Rr + m = R, ϕis onto. This implies thatker(ϕ) is a maximal left ideal.Therefore, y ∈ker(ϕ), so we have 0 = ϕ(y) = yr (mod m), acontradiction.