- Let \(R\) be a Noetherian ring and \(A,B\) algebras over \(R\). Suppose \(A\) is finite type over \(R\) and finite over B. Then \(B\) is finite type over \(R\).
Solution
Assume \(B\subseteq A\) as \(R\)-subalgebras. Choose elements \[ x_1,\dots,x_n\in A \] generating \(A\) as an \(R\)-algebra, and elements \[ y_1,\dots,y_m\in A \] generating \(A\) as a \(B\)-module. Enlarge the module generating set if necessary so that \(1=y_1\).
1. There are coefficients \(a_{ij},b_{ijk}\in B\) such that\[ x_i=\sum_j a_{ij}y_j, \qquad y_i y_j=\sum_k b_{ijk}y_k. \]
Both assertions follow because the \(y_j\) generate \(A\) as a \(B\)-module.
2. Let \(B_0\subseteq B\) be the \(R\)-subalgebra generated by all \(a_{ij}\) and \(b_{ijk}\). Then \(B_0\) is Noetherian.
It is finitely generated as an \(R\)-algebra, and \(R\) is Noetherian. Hilbert’s basis theorem implies that \(B_0\) is Noetherian.
3. The \(B_0\)-submodule\[ M=B_0y_1+\cdots+B_0y_m \]is a subring of \(A\) containing \(R\) and all \(x_i\).
It contains \(1=y_1\). The multiplication formulas in 1 show that products of the generators \(y_i\) remain in \(M\), hence \(M\) is closed under multiplication. The first formulas in 1 show that every \(x_i\in M\).
4. Therefore \(A=M\), so \(A\) is finite as a \(B_0\)-module.
Since \(A=R[x_1,\dots,x_n]\) and \(M\) is a subring containing \(R\) and every \(x_i\), one has \(A\subseteq M\). The reverse inclusion is tautological.
5. The subalgebra \(B\) is finite as a \(B_0\)-module.
Since \(B_0\subseteq B\subseteq A\), the ring \(B\) is a \(B_0\)-submodule of the finite \(B_0\)-module \(A\). By 2, \(B_0\) is Noetherian, so every submodule of a finite module is finite.
6. Hence \(B\) is of finite type over \(R\).
5. A finite module algebra over a finite-type algebra is finite type over the base.