Exercise 3, § 8 Algebra, X, p. 202 Let n be an integer ⩾1\geqslant 1⩾1. Now prove that dh(Mn(A))=dh(A)\mathrm{dh}(M_n(A)) = \mathrm{dh}(A)dh(Mn(A))=dh(A) (cf. VIII, § 4). Solution No solution has been written yet.