... has the format(x)=xsn+ a1 (x) xs−1n+ ···+ as (x) ,where aj (x) ∈ k[x 0 , ,xn−1] and deg aj (x) a j or aj (x) =0 .Proof. We make a linear transformation x 0 = y 0 + λ 0 yn, ,xn−1 = ... + h0s 0 = h1i, the matrix Aα =( aij(α,x)) is again a homogeneous matrix withdeg(ai1(α,x)) + h 01 = ···=deg(ais 0 (α,x)) + h0s 0 = h1i.Therefore, the homomorphism φα:s1j=1Rα(−h1j) ... k(v )[x]andk(v)[y]whenineverypolynomialofk(v)[y ]the substitutionyi = n3j =0 vijxj,i =0 , 1, ,n,is carried out. The inverse transformationxi = n3j =0 wijyj,i =0 , 1, ,n,has its coefficients...