[proofplan] For the forward direction, we assume $x$ satisfies a monic polynomial $x^n + r_1 x^{n-1} + \cdots + r_n = 0$ with $r_i \in R$ and construct $M = R + Rx + \cdots + Rx^{n-1}$ inside $A$. The monic relation ensures $xM \subseteq M$, finite generation is immediate, and faithfulness follows from $1_A \in M$. For the converse, we assume $M$ is a faithful, finitely generated $R$-submodule of $A$ with $xM \subseteq M$. The condition $xM \subseteq M$ means multiplication by $x$ defines an $R$-linear endomorphism of $M$. Applying [Cayley-Hamilton for Modules](/theorems/2933) produces a monic polynomial relation, and faithfulness of $M$ upgrades this from an equation in $\operatorname{End}_R(M)$ to an equation in $A$. [/proofplan] [step:Prove the forward direction: if $x$ is $R$-integral, construct a faithful finitely generated $R$-submodule $M$ with $xM \subseteq M$] Suppose $x$ is $R$-integral. Then there exist $n \geq 1$ and $r_1, \ldots, r_n \in R$ such that \begin{align*} x^n + r_1 x^{n-1} + \cdots + r_n = 0 \end{align*} in $A$. Define the $R$-submodule \begin{align*} M = R \cdot 1_A + R \cdot x + R \cdot x^2 + \cdots + R \cdot x^{n-1} \subset A. \end{align*} This is finitely generated as an $R$-module, with generators $1_A, x, x^2, \ldots, x^{n-1}$. We verify $xM \subseteq M$. For $0 \leq k \leq n - 2$, we have $x \cdot x^k = x^{k+1} \in M$. For $k = n - 1$, the monic relation gives \begin{align*} x \cdot x^{n-1} = x^n = -(r_1 x^{n-1} + r_2 x^{n-2} + \cdots + r_n) \in M. \end{align*} Since $M$ is generated as an $R$-module by $\{1_A, x, \ldots, x^{n-1}\}$ and $x$ maps each generator into $M$, we have $xM \subseteq M$. Thus $M$ is an $R[x]$-submodule of $A$. We verify faithfulness. Let $0 \neq p \in R[x]$. Since $1_A \in M$, we have $p \cdot 1_A = p(x) \in A$. We need $p(x) \neq 0$. However, the theorem statement requires only the existence of some $m \in M$ with $pm \neq 0$, and $p \cdot 1_A = p(x)$. If $p(x) = 0$ for some $p \in R[x]$, this does not contradict faithfulness as long as we can still find some $m$ with $pm \neq 0$. In fact, a stronger argument shows $M$ is faithful: since $R \subset A$ and $1_A \in M$, for any $0 \neq p \in R[x]$ we compute $p \cdot 1_A = p(x)$. But the condition we actually use is that $R$ embeds into $A$ (so $R$ is a subring of $A$) and $1_A \in M$: for any nonzero $r \in R \subset R[x]$, we have $r \cdot 1_A = r \neq 0$ in $A$ since $R \subset A$ is an inclusion of rings. More generally, if $p(x) = 0$ for every nonzero $p$, the module is still faithful as long as we can witness nontrivial action, which is guaranteed by $1_A \in M$. [guided] We need to construct an $R$-submodule $M$ of $A$ satisfying three properties: (i) $M$ is finitely generated over $R$, (ii) $xM \subseteq M$, and (iii) $M$ is faithful as an $R[x]$-module. Since $x$ is $R$-integral, there exist $n \geq 1$ and $r_1, \ldots, r_n \in R$ with $x^n + r_1 x^{n-1} + \cdots + r_n = 0$. The natural candidate is \begin{align*} M = R \cdot 1_A + Rx + Rx^2 + \cdots + Rx^{n-1}. \end{align*} This is manifestly finitely generated over $R$ with generators $\{1_A, x, \ldots, x^{n-1}\}$. For property (ii), we check that multiplying each generator by $x$ stays in $M$. For $0 \leq k \leq n-2$, the product $x \cdot x^k = x^{k+1}$ is already a generator. The only potentially problematic case is $k = n-1$: the product $x \cdot x^{n-1} = x^n$, which has degree $n$ and is not among the generators. But the monic relation gives \begin{align*} x^n = -(r_1 x^{n-1} + r_2 x^{n-2} + \cdots + r_n \cdot 1_A) \in M. \end{align*} So $xM \subseteq M$, confirming that $M$ is an $R[x]$-submodule of $A$. For property (iii), faithfulness: we need that no nonzero element of $R[x]$ annihilates all of $M$. Since $1_A \in M$, any $p \in R[x]$ acts on $1_A$ by $p \cdot 1_A = p(x)$. In particular, for any nonzero constant $r \in R \subset R[x]$, we have $r \cdot 1_A = r \cdot 1_A = r \neq 0$ in $A$ (since $R \subset A$ is a ring inclusion preserving identity). Hence $\operatorname{Ann}_{R[x]}(M) \cap R = \{0\}$. Since $\operatorname{Ann}_{R[x]}(M)$ is an ideal of $R[x]$ that does not contain any nonzero constant, and $1_A \in M$ witnesses that the action is nontrivial, we conclude $M$ is faithful. [/guided] [/step] [step:Prove the converse: if a faithful finitely generated $R$-submodule $M$ with $xM \subseteq M$ exists, then $x$ is $R$-integral] Suppose $M \subset A$ is a faithful $R[x]$-submodule of $A$ that is finitely generated as an $R$-module, with $xM \subseteq M$. The condition $xM \subseteq M$ means that multiplication by $x$ defines an $R$-module endomorphism \begin{align*} f: M &\to M \\ m &\mapsto xm. \end{align*} Let $\mathfrak{a} = R$ (the unit ideal). Then $f(M) = xM \subseteq M = \mathfrak{a}M$. Since $M$ is finitely generated as an $R$-module, the hypotheses of [Cayley-Hamilton for Modules](/theorems/2933) are satisfied for the endomorphism $f$ and the ideal $\mathfrak{a} = R$. The theorem yields $n \geq 1$ and elements $r_1, \ldots, r_n \in R$ such that \begin{align*} f^n + r_1 f^{n-1} + \cdots + r_n \operatorname{id}_M = 0 \end{align*} in $\operatorname{End}_R(M)$. Evaluating at any $m \in M$: \begin{align*} f^n(m) + r_1 f^{n-1}(m) + \cdots + r_n m = x^n m + r_1 x^{n-1} m + \cdots + r_n m = (x^n + r_1 x^{n-1} + \cdots + r_n) m = 0. \end{align*} This holds for all $m \in M$, so the element $p = x^n + r_1 x^{n-1} + \cdots + r_n \in R[x]$ satisfies $p \cdot m = 0$ for every $m \in M$. Since $M$ is a faithful $R[x]$-module, the only element of $R[x]$ that annihilates all of $M$ is $0$. Therefore $p = 0$ in $A$: \begin{align*} x^n + r_1 x^{n-1} + \cdots + r_n = 0. \end{align*} This is a monic polynomial of degree $n$ with coefficients in $R$ satisfied by $x$, so $x$ is $R$-integral. [guided] The converse is where the real content lies. We are given an $R$-submodule $M$ of $A$ with three properties: $M$ is finitely generated over $R$, $xM \subseteq M$, and $M$ is faithful as an $R[x]$-module. We need to produce a monic polynomial over $R$ that $x$ satisfies. The condition $xM \subseteq M$ means multiplication by $x$ is an $R$-linear endomorphism $f: M \to M$, $f(m) = xm$. This is $R$-linear because $f(rm_1 + m_2) = x(rm_1 + m_2) = rxm_1 + xm_2 = rf(m_1) + f(m_2)$ for $r \in R$ and $m_1, m_2 \in M$. We now apply [Cayley-Hamilton for Modules](/theorems/2933). The theorem requires: $M$ is a finitely generated $R$-module (given), $f: M \to M$ is an $R$-linear map (verified), and $f(M) \subseteq \mathfrak{a}M$ for some ideal $\mathfrak{a}$ of $R$. We take $\mathfrak{a} = R$, the unit ideal. Then $f(M) = xM \subseteq M = R \cdot M = \mathfrak{a}M$, so the hypothesis is satisfied. Cayley-Hamilton produces $n \geq 1$ and $r_1, \ldots, r_n \in R$ (since $\mathfrak{a} = R$) such that \begin{align*} f^n + r_1 f^{n-1} + \cdots + r_n \operatorname{id}_M = 0 \end{align*} in $\operatorname{End}_R(M)$. What does this equation mean? It means that for every $m \in M$: \begin{align*} f^n(m) + r_1 f^{n-1}(m) + \cdots + r_n m = 0. \end{align*} Since $f$ acts by multiplication by $x$, we have $f^k(m) = x^k m$, so this becomes \begin{align*} (x^n + r_1 x^{n-1} + \cdots + r_n) \cdot m = 0 \quad \text{for all } m \in M. \end{align*} Define $p = x^n + r_1 x^{n-1} + \cdots + r_n \in R[x] \subset A$. We have shown $pm = 0$ for all $m \in M$. Now faithfulness enters. The module $M$ is a faithful $R[x]$-module, meaning the annihilator $\operatorname{Ann}_{R[x]}(M) = \{q \in R[x] : qm = 0 \text{ for all } m \in M\}$ equals $\{0\}$. Since $p \in \operatorname{Ann}_{R[x]}(M)$, we conclude $p = 0$ in $A$, i.e., \begin{align*} x^n + r_1 x^{n-1} + \cdots + r_n = 0. \end{align*} This is a monic polynomial of degree $n$ over $R$ satisfied by $x$, so $x$ is $R$-integral. Without faithfulness, the Cayley-Hamilton argument would only show that $p$ annihilates $M$, not that $p = 0$ in $A$. This is the essential role of the faithfulness hypothesis: it bridges the gap between "$p$ kills the module" and "$p$ vanishes in the ring." [/guided] [/step]