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
The rings are commutative, and \(R\subseteq B\subseteq A\). Write \(A=R[x_1,\dots,x_n]\) and \(A=\sum_{j=1}^m By_j\) with \(y_1=1\). Choose \(b_{ij},c_{jkl}\in B\) with \[x_i=\sum_{j=1}^m b_{ij}y_j,\qquad y_jy_k=\sum_{l=1}^m c_{jkl}y_l ,\] and let \(B_0=R[\{b_{ij}\}\cup\{c_{jkl}\}]\subseteq B\).
1. \(B_0\) is a Noetherian ring.
\(B_0\) is a quotient of a polynomial ring in finitely many variables over the Noetherian ring \(R\), so the Hilbert basis theorem applies.
2. \(A=\sum_{j=1}^m B_0y_j\); in particular \(A\) is a finitely generated \(B_0\)-module.
Put \(M=\sum_j B_0y_j\). Since \(y_1=1\), \(B_0\subseteq M\), and the relations \(y_jy_k=\sum_l c_{jkl}y_l\) with \(c_{jkl}\in B_0\) show that \(M\) is closed under multiplication. So \(M\) is an \(R\)-subalgebra of \(A\), and it contains each \(x_i=\sum_j b_{ij}y_j\). Hence \(A=R[x_1,\dots,x_n]\subseteq M\subseteq A\).
3. \(B\) is a finitely generated \(B_0\)-module.
By steps 1 and 2, \(A\) is a finitely generated module over the Noetherian ring \(B_0\), hence a Noetherian \(B_0\)-module. Its \(B_0\)-submodule \(B\) is therefore finitely generated.
4. Q.E.D.
If \(B=\sum_{r=1}^pB_0z_r\) (step <1>3), then \(B=R[\{b_{ij}\},\{c_{jkl}\},z_1,\dots,z_p]\) is a finitely generated \(R\)-algebra.