Corollary 1

Let B be a regular ring, graded in positive degrees. Suppose that every finitely generated projective $B_0$-module is free. a) The ring $B_0$ is an integral domain and regular, and the $B_0$-algebra $B$ is a finitely generated graded polynomial $B_0$-algebra. b) Let A be a graded subring of B such that $A_0 = B_0$ and that B is a finitely generated A-module. The following conditions are equivalent: (i) the A-module B is graded free; (ii) the A-module B is flat; (iii) A is a finitely generated graded polynomial $A_0$-algebra.

By theorem 2, the ring $B_0$ is regular, hence a product of regular integral rings ($§ 4$, No. 2, example 2). For every idempotent element $e$ of $B_0$, the $B_0$-module $B_0e$ is projective, hence free, which implies that $e = 0$ or $1$; thus $B_0$ is integral. Assertion a) then follows from theorem 2 and from the fact that a graded finitely generated projective $B_0$-module is graded free.

Let us prove b).

(i) $\Leftrightarrow$ (ii): flat finitely generated graded $A_0$-modules are graded free (II, $§ 5$, No. 2, cor. 2 to theorem 1). The equivalence of (i) and (ii) therefore follows from A, X, p. 144, prop. 8.

(ii) $\Leftrightarrow$ (iii): since $B$ is integral over $A$ and is a finitely generated $A_0$-algebra, $A$ is a finitely generated $A_0$-algebra (V, $§ 1$, No. 9, lemma 5), hence a noetherian ring. The equivalence of (ii) and (iii) then follows from the corollary to proposition 8 of $§ 4$, No. 5 and from a) applied to $A$.