Proposition 8

Let $A[(X_i)_{i \in I}]$ be a polynomial algebra, $J$ a subset of $I$ and $K$ the complement of $J$ in $I$. Writing $A' = A[(X_j)_{j \in J}]$ and denoting by $X'_k$ ($k \in K$) the indeterminates in the polynomial algebra Libasc_{A'}(K) = A'[(X'_k)_{k \in K}], there exists a unique ring isomorphism of A'[(X'_k)_{k \in K}] onto A[(X_i)_{i \in I}] which coincides with the identity on A' and maps X'_k to X_k for all k \in K.

Clearly A[(X_i)_{i \in I}] is an A'-algebra generated by the X_k for k \in K. On the other hand, as a polynomial relator between the X_k (k \in K) with coefficients in A' can be written uniquely as $\sum_v h_v((X_j)_{j \in J}) X^v$ where v runs through a finite subset of $\mathbf{N}^{(K)}$ and where the h_v are elements of A[(X_j)_{j \in J}], the h_v must be polynomial relators between the X_j with coefficients in A and hence are all zero, which proves the proposition.

The isomorphism described in Proposition 8 is often used to identify the elements of A[(X_i)_{i \in I}] with polynomials with coefficients in A' = A[(X_j)_{j \in J}]. If u is an element $\neq 0$ of A[(X_i)_{i \in I}], its total degree considered as an element of A'[(X_k)_{k \in K}] is also called its degree with respect to the X_i of index i \in K.