Proof. By assumption, T is noetherian; so, Spec(T ) is a finite union of integral irreducible components. Let Spec(T0) ⊂ Spec(T ) be an irreducible component of maximal dimension; so, dimT = dimT0. Thus we have the surjective algebra homomorphism T T0. Composing this with B[[x1, . . . , xm]] T , we have a surjective algebra homomorphism: B[[x1, . . . , xm]] T0. Applying the above lemma to this morphism, we get B[[x1, . . . , xm]] ∼= T0, and hence B[[x1, . . . , xm]] ∼= T .