Proposition 3

Let $k$ be a ring, $A$ a $k$-algebra and $B$ an $A$-algebra. If $A$ is essentially of finite type over $k$ and $B$ essentially of finite type over $A$, then $B$ is essentially of finite type over $k$.

Let us denote by $\rho : A \to B$ the canonical mapping. Let $x = (x_i)_{i \in I}$ be a finite essentially generating family of the $k$-algebra $A$, and let $A'$ be the subalgebra generated by it; let $y = (y_j)_{j \in J}$ be a finite essentially generating family of the $A$-algebra $B$. Let us denote by $B'$ the sub-$k$-algebra of $B$ generated by the $\rho(x_i)$ and the $y_j$. Let $b \in B$; by hypothesis there exist polynomials $P$ and $Q$ in $A[(Y_j)_{j \in J}]$ such that $Q(y)$ is invertible in $B$ and such that one has $Q(y) b = P(y)$. The non-zero coefficients of $P$ and $Q$ are finite in number; there exists a polynomial $R \in k[(X_i)_{i \in I}]$ such that $R(x)$ is invertible in $A$ and that $R(x) P$ and $R(x) Q$ belong to $A'[(Y_j)_{j \in J}]$. Then $\rho(R(x)) Q(y)$ is invertible in $B$, and one has

Thus the elements $\rho(x_i)$ for $i \in I$ and $y_j$ for $j \in J$ form a finite essentially generating family of the $k$-algebra $B$.