# Answer on Abstract Algebra Question for Hym@n B@ss

Question #23870

Show that any quasi-Frobenius ring is left (and right) Kasch.

Expert's answer

For such a ring, we have always thefollowing Double Annihilator Property: if

Assuming this, consider now anysimple left

*I*is any left ideal, then ann*l*(ann*r*(*I*))=*I.*Assuming this, consider now anysimple left

*R*-module*V*, say*V*=*R/*m, where m is amaximal left ideal. By the property quoted above, we see that ann*r*(m)*<>*0. Fix a nonzero element*a**∈**R*such thatm*a*= 0, and define*ϕ*:*R → R*by*ϕ*(*r*) =*ra*. Then*ϕ*is a nonzero endomorphism of*,with m*_{R}R*⊆**ker(**ϕ*). Since m is a maximal left ideal,we have m = ker(*ϕ*). Therefore*ϕ*(*R*)*∼**R/*m =*V.*So ϕ(R) is the minimal left ideal wesought*.*Need a fast expert's response?

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