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where g1, g2, . . . , gn

∈ a .

Let g ∈ a be dened by g = r j =0 g j f r − j . Now g−f r = g−f r h + f r (h −1).

Also

g −f r h =r

j =0

g j f r− j (1 −f j Y j ) ∈ (1 −fY ),

since the polynomial 1 −f j Y j is divisible by the polynomial 1 −fY for allpositive integers j . It follows that g−f r ∈ (1−fY ). But the polynomial g−f ris a polynomial in the indeterminates X 1, X 2, . . . , X n , and, if non-zero, wouldbe of degree zero when considered as a polynomial in the indeterminate Y

with coefficients in the ring R. Also any non-zero element of the ideal (1 −fY ) of S is divisible by the polynomial 1 −fY , and is therefore of strictlypositive degree when considered as a polynomial in the indeterminate Y with coefficients in R. We conclude, therefore that g −f r = 0. But g ∈ a .Therefore f r ∈ a , as required.