[proofplan] We prove the cycle $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$. The implication $(1) \Rightarrow (2)$ is immediate from definitions. For $(2) \Rightarrow (3)$, we use the monic relations satisfied by the generators to show that every monomial in the generators can be reduced to a bounded-degree monomial, producing a finite $R$-module generating set. For $(3) \Rightarrow (1)$, we show that any element of a finite $R$-algebra is $R$-integral by applying the [Faithful Module Criterion for Integrality](/theorems/2936) with $M = A$. [/proofplan] [step:Show $(1) \Rightarrow (2)$: a finitely generated integral $R$-algebra is generated by finitely many integral elements] By definition, condition (1) states that $A$ is generated as an $R$-algebra by finitely many elements $\alpha_1, \ldots, \alpha_m \in A$, and every element of $A$ is $R$-integral. In particular, each generator $\alpha_i$ is $R$-integral. This is precisely condition (2). [/step] [step:Show $(2) \Rightarrow (3)$: reduce monic relations to produce a finite $R$-module generating set] Suppose $A$ is generated as an $R$-algebra by $R$-integral elements $\alpha_1, \ldots, \alpha_m$. For each $i$, the integrality of $\alpha_i$ gives $n_i \geq 1$ and $r_{i,1}, \ldots, r_{i,n_i} \in R$ such that \begin{align*} \alpha_i^{n_i} + r_{i,1}\alpha_i^{n_i - 1} + \cdots + r_{i,n_i} = 0. \end{align*} Rearranging, $\alpha_i^{n_i} = -(r_{i,1}\alpha_i^{n_i - 1} + \cdots + r_{i,n_i})$, so $\alpha_i^{n_i} \in R \cdot 1_A + R \cdot \alpha_i + \cdots + R \cdot \alpha_i^{n_i - 1}$. By repeated substitution, every power $\alpha_i^k$ with $k \geq n_i$ can be expressed as an $R$-linear combination of $1_A, \alpha_i, \ldots, \alpha_i^{n_i - 1}$. Since $A = R[\alpha_1, \ldots, \alpha_m]$, every element of $A$ is an $R$-linear combination of monomials $\alpha_1^{e_1} \cdots \alpha_m^{e_m}$ with $e_i \geq 0$. Applying the reduction above to each factor, every such monomial lies in the $R$-span of \begin{align*} \mathcal{B} = \{\alpha_1^{f_1} \cdots \alpha_m^{f_m} : 0 \leq f_i \leq n_i - 1 \text{ for all } i\}. \end{align*} This is a finite set with $|\mathcal{B}| = n_1 \cdots n_m$ elements. Since every element of $A$ is an $R$-linear combination of elements of $\mathcal{B}$, the set $\mathcal{B}$ generates $A$ as an $R$-module. Therefore $A$ is a finitely generated $R$-module, i.e., $A$ is a finite $R$-algebra. [guided] The goal is to show that $A$ is finitely generated as an $R$-module (not just as an $R$-algebra). The key idea is that the monic polynomial satisfied by each generator $\alpha_i$ lets us "cap" the exponent of $\alpha_i$ at $n_i - 1$. Let $\alpha_1, \ldots, \alpha_m$ be $R$-integral generators of $A$ as an $R$-algebra. Each $\alpha_i$ satisfies a monic relation of degree $n_i$: \begin{align*} \alpha_i^{n_i} = -(r_{i,1}\alpha_i^{n_i - 1} + \cdots + r_{i,n_i}). \end{align*} This expresses $\alpha_i^{n_i}$ as an $R$-linear combination of $1_A, \alpha_i, \ldots, \alpha_i^{n_i - 1}$. By induction on $k \geq n_i$, every power $\alpha_i^k$ is an $R$-linear combination of $1_A, \alpha_i, \ldots, \alpha_i^{n_i - 1}$: the base case $k = n_i$ is the relation itself, and for the inductive step, $\alpha_i^{k+1} = \alpha_i \cdot \alpha_i^k$; by the inductive hypothesis, $\alpha_i^k$ is an $R$-linear combination of $1_A, \ldots, \alpha_i^{n_i - 1}$; multiplying by $\alpha_i$ gives an $R$-linear combination of $\alpha_i, \ldots, \alpha_i^{n_i}$; the term $\alpha_i^{n_i}$ is then replaced by the monic relation. Since $A = R[\alpha_1, \ldots, \alpha_m]$, every element of $A$ is a polynomial in $\alpha_1, \ldots, \alpha_m$ with $R$-coefficients, i.e., an $R$-linear combination of monomials $\alpha_1^{e_1} \cdots \alpha_m^{e_m}$. Applying the exponent reduction to each factor independently, every monomial $\alpha_1^{e_1} \cdots \alpha_m^{e_m}$ is an $R$-linear combination of monomials $\alpha_1^{f_1} \cdots \alpha_m^{f_m}$ with $0 \leq f_i \leq n_i - 1$. Therefore, the finite set \begin{align*} \mathcal{B} = \{\alpha_1^{f_1} \cdots \alpha_m^{f_m} : 0 \leq f_i \leq n_i - 1\} \end{align*} spans $A$ as an $R$-module. Since $|\mathcal{B}| = n_1 n_2 \cdots n_m < \infty$, the module $A$ is finitely generated over $R$. Why does this work? The monic polynomial relation for $\alpha_i$ effectively makes $\alpha_i$ "algebraically bounded" over $R$: every power of $\alpha_i$ is redundant beyond degree $n_i - 1$. When all generators are bounded in this way, the entire algebra collapses to a finite-dimensional $R$-module. [/guided] [/step] [step:Show $(3) \Rightarrow (1)$: every element of a finite $R$-algebra is $R$-integral] Suppose $A$ is a finite $R$-algebra, i.e., $A$ is finitely generated as an $R$-module. We need to show two things: (a) $A$ is a finitely generated $R$-algebra, and (b) every element of $A$ is $R$-integral. For (a): if $A$ is generated as an $R$-module by $a_1, \ldots, a_\ell$, then in particular $A$ is generated as an $R$-algebra by $a_1, \ldots, a_\ell$ (since every $R$-linear combination is an $R$-polynomial expression of degree at most one). For (b): let $\alpha \in A$ be arbitrary. Write $\rho: R \to A$ for the structure homomorphism of the $R$-algebra $A$. We apply the [Faithful Module Criterion for Integrality](/theorems/2936) with $M = A$ itself. We verify the two conditions. First, $A$ is finitely generated as an $R$-module by hypothesis. Second, $A$ is a faithful $R[\alpha]$-module: the $R[\alpha]$-module structure on $A$ is given by multiplication in $A$, and $\alpha A \subseteq A$ since $A$ is a ring. For faithfulness, let $0 \neq p \in R[\alpha] \subseteq A$. Then $p \cdot 1_A = p \neq 0$ (since $1_A$ is the identity of $A$ and $p \neq 0$ in $A$). So $1_A \in A$ witnesses that $p$ does not annihilate $A$. By the [Faithful Module Criterion for Integrality](/theorems/2936), $\alpha$ is $R$-integral. Since $\alpha$ was arbitrary, every element of $A$ is $R$-integral. Combined with (a), this gives condition (1). [guided] Suppose $A$ is finitely generated as an $R$-module. We first observe that $A$ is also finitely generated as an $R$-algebra: if $a_1, \ldots, a_\ell$ generate $A$ as an $R$-module, they certainly generate $A$ as an $R$-algebra (an $R$-algebra generated by $a_1, \ldots, a_\ell$ contains all polynomial expressions in $a_1, \ldots, a_\ell$ with $R$-coefficients, which includes all $R$-linear combinations). The main content is showing every element $\alpha \in A$ is $R$-integral. We apply the [Faithful Module Criterion for Integrality](/theorems/2936), which states: $\alpha$ is $R$-integral if and only if there exists a faithful, finitely generated $R$-submodule $M$ of $A$ with $\alpha M \subseteq M$. We take $M = A$. We verify the conditions: - **Finitely generated over $R$**: this is the hypothesis of condition (3). - **$\alpha M \subseteq M$**: since $M = A$ and $A$ is a ring containing $\alpha$, we have $\alpha A \subseteq A$. - **Faithful as an $R[\alpha]$-module**: let $0 \neq p \in R[\alpha]$. We need to find $m \in A$ with $pm \neq 0$. Take $m = 1_A$: then $p \cdot 1_A = p$. Since $R[\alpha] \subseteq A$ and $p \neq 0$ in $R[\alpha]$, we have $p \neq 0$ in $A$ (because $R[\alpha]$ is a subring of $A$). Therefore $p \cdot 1_A = p \neq 0$. All hypotheses of the Faithful Module Criterion are verified. The criterion yields that $\alpha$ is $R$-integral. Since $\alpha \in A$ was arbitrary, every element of $A$ is integral over $R$. Together with the fact that $A$ is a finitely generated $R$-algebra, this establishes condition (1). [/guided] [/step]