Corollary

The degrees $d_i$ are determined up to order by S.

Indeed, the inverse of $P_S(T)$ is the polynomial $N(T) = \prod_{i=1}^m (1 - T^{d_i})$, which is thus uniquely determined. If q is an integer $\geq 1$ and if $\zeta \in \mathbf{C}$ is a primitive qth root of unity, the multiplicity of the root $\zeta$ of $N(T)$ is equal to the number of $d_i$ that are multiples of $q$. This number is zero for $q$ sufficiently large. The number of $d_i$ equal to $q$ is thus determined uniquely by descending induction.

The integers $d_i$ are called the characteristic degrees of S. The number of them is equal to the transcendence degree of S over K when K is a field; we shall also call it the transcendence degree of S over K in the general case. It is the multiplicity of the root 1 of the polynomial $N(T)$.

Let $S = \bigoplus_{n \geq 0} S_n$ be a commutative graded K-algebra, and $R = \bigoplus_{n \geq 0} R_n$ a graded subalgebra of S. Assume that each $R_n$ is free and of finite type, and that the R-module S admits a finite basis consisting of homogeneous elements $z_1, z_2, \ldots, z_N$ of degrees $f_1, f_2, \ldots, f_N$. Then, if M denotes the graded K-module $\sum_{j=1}^N Kz_j$, the graded K-module S is isomorphic to $R \otimes_K M$, so each $S_n$ is free and of finite type and