Exercise 6, § 8 Algebra, X, p. 202 Let III be the set of left ideals of AAA. Now prove that dh(A)=supa∈IdpA(A/a)\mathrm{dh}(A) = \sup_{a \in I} \mathrm{dp}_A(A/a)dh(A)=supa∈IdpA(A/a). Solution No solution has been written yet.