[proofplan] For part (1), we express $C$ as a finitely generated $B$-module and $B$ as a finitely generated $A$-module, then show the set of all pairwise products of the two generating sets spans $C$ over $A$. For part (2), we take an arbitrary $c \in C$, use its integral equation over $B$ to identify a finitely generated $A$-subalgebra $A' \subset B$ containing the coefficients, show $A'[c]$ is finite over $A'$ and $A'$ is finite over $A$, then apply part (1) to conclude $A'[c]$ is finite over $A$, which gives integrality of $c$ over $A$ via the [Equivalence of Finite and Integral Extensions](/theorems/2937). [/proofplan] [step:Prove part (1): show $C$ is spanned over $A$ by the products $\gamma_i \beta_j$] Since $C$ is finite over $B$, there exist $\gamma_1, \ldots, \gamma_n \in C$ such that $C = \sum_{i=1}^n B \gamma_i$ (i.e., every element of $C$ is a $B$-linear combination of $\gamma_1, \ldots, \gamma_n$). Since $B$ is finite over $A$, there exist $\beta_1, \ldots, \beta_\ell \in B$ such that $B = \sum_{j=1}^\ell A \beta_j$. Let $c \in C$. Then $c = \sum_{i=1}^n b_i \gamma_i$ for some $b_1, \ldots, b_n \in B$. Each $b_i$ can be written as $b_i = \sum_{j=1}^\ell a_{ij} \beta_j$ for some $a_{ij} \in A$. Substituting: \begin{align*} c = \sum_{i=1}^n \left( \sum_{j=1}^\ell a_{ij} \beta_j \right) \gamma_i = \sum_{i=1}^n \sum_{j=1}^\ell a_{ij} (\gamma_i \beta_j). \end{align*} This shows $C = \sum_{i,j} A (\gamma_i \beta_j)$, so $C$ is generated as an $A$-module by the finite set $\{\gamma_i \beta_j : 1 \leq i \leq n, \, 1 \leq j \leq \ell\}$. Hence $C$ is finite over $A$. [guided] We want to show that $C$ is finitely generated as an $A$-module, given that $C$ is finitely generated as a $B$-module and $B$ is finitely generated as an $A$-module. Write $C = \sum_{i=1}^n B \gamma_i$ and $B = \sum_{j=1}^\ell A \beta_j$. Take an arbitrary $c \in C$. Since $\gamma_1, \ldots, \gamma_n$ generate $C$ over $B$, we can write $c = \sum_{i=1}^n b_i \gamma_i$ with $b_i \in B$. Each coefficient $b_i$ lives in $B$, which is spanned over $A$ by $\beta_1, \ldots, \beta_\ell$, so $b_i = \sum_{j=1}^\ell a_{ij} \beta_j$ with $a_{ij} \in A$. Substituting: \begin{align*} c = \sum_{i=1}^n \left( \sum_{j=1}^\ell a_{ij} \beta_j \right) \gamma_i = \sum_{i=1}^n \sum_{j=1}^\ell a_{ij} (\gamma_i \beta_j). \end{align*} Since this holds for arbitrary $c \in C$, the finite set $\{\gamma_i \beta_j : 1 \leq i \leq n, \, 1 \leq j \leq \ell\}$ generates $C$ as an $A$-module. This set has at most $n\ell$ elements, so $C$ is finite over $A$. [/guided] [/step] [step:Prove part (2): reduce integrality of $c$ over $A$ to a finiteness statement] Let $c \in C$. Since $C$ is integral over $B$, there exist $b_1, \ldots, b_n \in B$ such that \begin{align*} c^n + b_1 c^{n-1} + \cdots + b_n = 0. \end{align*} Define $A' := A[b_1, \ldots, b_n]$, the $A$-subalgebra of $B$ generated by $b_1, \ldots, b_n$. Since $B$ is integral over $A$, each $b_i$ is integral over $A$. By the [Equivalence of Finite and Integral Extensions](/theorems/2937), implication $(2) \implies (3)$, the $A$-algebra $A'$ is finite over $A$: the hypothesis is satisfied because $A'$ is generated as an $A$-algebra by the finite set $\{b_1, \ldots, b_n\}$ of $A$-integral elements. The equation $c^n + b_1 c^{n-1} + \cdots + b_n = 0$ shows that $c$ is integral over $A'$ (the coefficients $b_1, \ldots, b_n$ all lie in $A'$). By the [Equivalence of Finite and Integral Extensions](/theorems/2937), implication $(1) \implies (3)$ applied to the $A'$-algebra $A'[c]$, the ring $A'[c]$ is finite over $A'$. [guided] Take an arbitrary $c \in C$. We need to show $c$ is integral over $A$. Since $c$ is integral over $B$, there is a monic polynomial relation $c^n + b_1 c^{n-1} + \cdots + b_n = 0$ with coefficients $b_i \in B$. The coefficients $b_1, \ldots, b_n$ live in $B$, not in $A$. To bridge the gap, we introduce the intermediate ring $A' := A[b_1, \ldots, b_n] \subset B$. Since $B$ is integral over $A$, each $b_i$ is integral over $A$. The [Equivalence of Finite and Integral Extensions](/theorems/2937) requires, for $(2) \implies (3)$, that $A'$ be generated as an $A$-algebra by finitely many $A$-integral elements. This is exactly our setup: $A'$ is generated by $\{b_1, \ldots, b_n\}$, all of which are $A$-integral. The conclusion is that $A'$ is a finite $A$-algebra. Now the monic equation $c^n + b_1 c^{n-1} + \cdots + b_n = 0$ has all its coefficients in $A'$, so $c$ is integral over $A'$. The $A'$-algebra $A'[c]$ is therefore generated by the single $A'$-integral element $c$. Applying the [Equivalence of Finite and Integral Extensions](/theorems/2937) again ($(1) \implies (3)$ for the $A'$-algebra $A'[c]$), we conclude that $A'[c]$ is finite over $A'$. [/guided] [/step] [step:Apply transitivity of finiteness to conclude $c$ is integral over $A$] We have established: - $A'$ is finite over $A$ (from the previous step), - $A'[c]$ is finite over $A'$ (from the previous step). The chain $A \subset A' \subset A'[c]$ satisfies the hypotheses of part (1). Applying part (1) to this chain, we conclude that $A'[c]$ is finite over $A$. Since $c \in A'[c]$ and $A'[c]$ is a finite $A$-algebra, the [Equivalence of Finite and Integral Extensions](/theorems/2937), implication $(3) \implies (1)$, shows that every element of $A'[c]$ is $A$-integral. In particular, $c$ is integral over $A$. Since $c \in C$ was arbitrary, $C$ is integral over $A$. [guided] We now have $A \subset A' \subset A'[c]$ with $A'[c]$ finite over $A'$ and $A'$ finite over $A$. This is exactly the setup of part (1), which gives that $A'[c]$ is finite over $A$. To conclude that $c$ is integral over $A$, we invoke the [Equivalence of Finite and Integral Extensions](/theorems/2937) one final time. The implication $(3) \implies (1)$ says: if an $A$-algebra is finite over $A$, then it is an integral $A$-algebra. Since $A'[c]$ is finite over $A$ and $c \in A'[c]$, the element $c$ is $A$-integral. Since $c \in C$ was arbitrary, we have shown that every element of $C$ is integral over $A$, i.e., $C$ is integral over $A$. [/guided] [/step]