[proofplan] We must show that $\mathcal{O}$ is closed under addition, multiplication, and contains the image of $R$. For sums and products, the key idea is that if $x$ and $y$ are each $R$-integral, then the $R$-subalgebra $R[x, y]$ they generate is a finite $R$-algebra by the [Equivalence of Finite and Integral Extensions](/theorems/2937). Since $x + y$ and $xy$ both lie in this finite $R$-algebra, they are themselves $R$-integral by the same equivalence. For the image of $R$, each element $\rho(r)$ satisfies a monic polynomial of degree one. [/proofplan] [step:Show that the image of $R$ under the structure map lies in $\mathcal{O}$] Let $\rho: R \to A$ denote the structure homomorphism. For any $r \in R$, the element $\rho(r) \in A$ satisfies the monic polynomial $T - \rho(r) \in \rho(R)[T]$, since substituting $T = \rho(r)$ gives $\rho(r) - \rho(r) = 0$. This is a monic polynomial of degree one with coefficients in $\rho(R)$, so $\rho(r)$ is $R$-integral. Hence $\rho(R) \subset \mathcal{O}$. [/step] [step:Show that $\mathcal{O}$ is closed under addition and multiplication using finiteness] Let $x, y \in \mathcal{O}$. Since $x$ is $R$-integral, there exist $r_1, \ldots, r_n \in R$ such that $x^n + \rho(r_1) x^{n-1} + \cdots + \rho(r_n) = 0$. Similarly, since $y$ is $R$-integral, there exist $s_1, \ldots, s_m \in R$ such that $y^m + \rho(s_1) y^{m-1} + \cdots + \rho(s_m) = 0$. Consider the $R$-subalgebra $R[x, y] := \rho(R)[x, y] \subset A$ generated by $x$ and $y$. Both $x$ and $y$ are $R$-integral elements that generate $R[x, y]$ as an $R$-algebra. By the [Equivalence of Finite and Integral Extensions](/theorems/2937), implication $(2) \implies (3)$, the $R$-algebra $R[x, y]$ is a finite $R$-algebra (i.e., finitely generated as an $R$-module). The elements $x + y$ and $xy$ both lie in $R[x, y]$. Since $R[x, y]$ is a finite $R$-algebra, it is in particular a finitely generated integral $R$-algebra. By the [Equivalence of Finite and Integral Extensions](/theorems/2937), implication $(3) \implies (1)$, every element of $R[x, y]$ is $R$-integral. In particular, $x + y \in \mathcal{O}$ and $xy \in \mathcal{O}$. [guided] We want to show that sums and products of $R$-integral elements remain $R$-integral. The direct approach --- trying to construct explicit monic polynomials for $x + y$ and $xy$ from the monic polynomials for $x$ and $y$ --- is possible but combinatorially involved. Instead, we use the structural characterisation of integrality. Since $x$ and $y$ are $R$-integral, they each satisfy monic polynomial equations over $\rho(R)$. The $R$-subalgebra $R[x, y] := \rho(R)[x, y]$ is generated as an $R$-algebra by the two $R$-integral elements $x$ and $y$. By the [Equivalence of Finite and Integral Extensions](/theorems/2937), the implication $(2) \implies (3)$ tells us that an $R$-algebra generated by finitely many $R$-integral elements is finite (i.e., finitely generated as an $R$-module). To verify the hypothesis: $R[x, y]$ is generated as an $R$-algebra by the finite set $\{x, y\}$, and both $x$ and $y$ are $R$-integral by assumption. Therefore $R[x, y]$ is a finite $R$-algebra. Now $x + y$ and $xy$ are both elements of $R[x, y]$. Since $R[x, y]$ is a finite $R$-algebra, the implication $(3) \implies (1)$ of the [Equivalence of Finite and Integral Extensions](/theorems/2937) tells us that every element of $R[x, y]$ is $R$-integral. In particular, $x + y$ and $xy$ are $R$-integral, so $x + y \in \mathcal{O}$ and $xy \in \mathcal{O}$. [/guided] [/step] [step:Conclude that $\mathcal{O}$ is an $R$-subalgebra of $A$] We have shown that $\rho(R) \subset \mathcal{O}$, and that $\mathcal{O}$ is closed under addition and multiplication. These are precisely the conditions for $\mathcal{O}$ to be an $R$-subalgebra of $A$. [/step]