[proofplan] The proof reduces immediately to the [Integral Characterisation via Finite Generation](/theorems/???). Since $s_1$ and $s_2$ are both integral over $R$, the ring $R[s_1, s_2]$ is a finitely generated $R$-module: $R[s_1]$ is finitely generated over $R$ (by integrality of $s_1$), and $R[s_1, s_2]$ is finitely generated over $R[s_1]$ (by integrality of $s_2$ over $R[s_1]$), so transitivity of finite generation gives the claim. The elements $s_1 + s_2$ and $s_1 s_2$ both belong to $R[s_1, s_2]$, and the $(\Leftarrow)$ direction of the characterisation theorem implies they are integral over $R$. Closure under additive inverses follows because $-1 \in R \subseteq \tilde{R}$ and $\tilde{R}$ is closed under multiplication; containment of $R$ follows because every $r \in R$ satisfies $t - r \in R[t]$. These four axioms establish that $\tilde{R}$ is a subring. [/proofplan] [step:Show $R[s_1, s_2]$ is a finitely generated $R$-module] Since $s_1$ is integral over $R$, the [Integral Characterisation via Finite Generation](/theorems/???) (($\Rightarrow$) direction, applied with $n = 1$) gives that $R[s_1]$ is a finitely generated $R$-module. Since $s_2$ is integral over $R$, it is also integral over the larger ring $R[s_1] \supseteq R$ (every monic polynomial over $R$ is in particular a monic polynomial over $R[s_1]$). The same theorem applied to $s_2$ over $R[s_1]$ gives that $R[s_1][s_2] = R[s_1, s_2]$ is a finitely generated $R[s_1]$-module. A finitely generated module over a finitely generated $R$-algebra is finitely generated over $R$: if $R[s_1] = \sum_{i=1}^{a} R e_i$ and $R[s_1, s_2] = \sum_{j=1}^{b} R[s_1] f_j$, then $R[s_1, s_2] = \sum_{i,j} R \, e_i f_j$, which is generated by $ab$ elements over $R$. [guided] The argument is an instance of the transitivity of finite generation. Let us trace through it carefully. Since $s_1$ is integral over $R$, there exists a monic polynomial $f \in R[x]$ of degree $d_1$ with $f(s_1) = 0$. By the [Integral Characterisation via Finite Generation](/theorems/???) (the ($\Rightarrow$) direction), $R[s_1]$ is generated by $\{1, s_1, \ldots, s_1^{d_1 - 1}\}$ as an $R$-module. Similarly, since $s_2$ is integral over $R$, there exists a monic polynomial $g \in R[x]$ of degree $d_2$ with $g(s_2) = 0$. In particular $g \in R[s_1][x]$, so $s_2$ is integral over $R[s_1]$. The same theorem gives that $R[s_1][s_2] = R[s_1, s_2]$ is generated by $\{1, s_2, \ldots, s_2^{d_2 - 1}\}$ as an $R[s_1]$-module. Combining: every element of $R[s_1, s_2]$ has the form $\sum_{j=0}^{d_2 - 1} p_j(s_1) s_2^j$ with $p_j \in R[s_1]$, and each $p_j(s_1) = \sum_{i=0}^{d_1-1} r_{ij} s_1^i$ with $r_{ij} \in R$. So $R[s_1, s_2]$ is generated over $R$ by the $d_1 d_2$ elements $\{s_1^i s_2^j : 0 \leq i < d_1,\ 0 \leq j < d_2\}$. [/guided] [/step] [step:Conclude that $s_1 + s_2$ and $s_1 s_2$ are integral over $R$] Since $s_1 + s_2$ and $s_1 s_2$ are both elements of $R[s_1, s_2]$, and $R[s_1, s_2]$ is a finitely generated $R$-module, the [Integral Characterisation via Finite Generation](/theorems/???) (($\Leftarrow$) direction) implies that $s_1 + s_2$ and $s_1 s_2$ are integral over $R$. [guided] We need to show that $s_1 + s_2$ and $s_1 s_2$ are each integral over $R$, i.e., each satisfies a monic polynomial with coefficients in $R$. The key is that both elements belong to $R[s_1, s_2]$, and we have just shown $R[s_1, s_2]$ is a finitely generated $R$-module, generated by the $d_1 d_2$ elements $\{s_1^i s_2^j : 0 \leq i < d_1,\ 0 \leq j < d_2\}$. By the $(\Leftarrow)$ direction of the [Integral Characterisation via Finite Generation](/theorems/???): if $M$ is a finitely generated $R$-module with $M \supseteq R \cdot 1$ and $\alpha \cdot M \subseteq M$ (i.e., $M$ is stable under multiplication by $\alpha$), then $\alpha$ is integral over $R$. Apply this with $M = R[s_1, s_2]$ and $\alpha = s_1 + s_2$. Multiplication by $s_1 + s_2$ maps $R[s_1, s_2]$ into itself (since $R[s_1, s_2]$ is a ring), so $M$ is stable under multiplication by $\alpha$. Therefore $s_1 + s_2$ is integral over $R$. The identical argument applies to $s_1 s_2$. [/guided] [/step] [step:Verify the integral closure $\tilde{R}$ is a subring of $S$] Define $\tilde{R} = \{s \in S : s \text{ is integral over } R\}$. We verify the subring axioms. Recall that a subset of a ring $S$ is a subring if it contains $1$, is closed under addition and multiplication, and is closed under additive inverses. **Contains $R$:** Every $r \in R$ satisfies the monic polynomial $x - r \in R[x]$, so $r$ is integral over $R$ and $R \subseteq \tilde{R}$. **Closed under addition and multiplication:** If $s_1, s_2 \in \tilde{R}$, then $s_1 + s_2$ and $s_1 s_2$ are integral over $R$ by the result just proved, so $s_1 + s_2, s_1 s_2 \in \tilde{R}$. **Closed under additive inverses:** If $s \in \tilde{R}$, then $-s = (-1) \cdot s$. Since $-1 \in R$ (as $R$ is a ring) and $R \subseteq \tilde{R}$ (established above), we have $-1 \in \tilde{R}$. By closure under multiplication (established above), the product $(-1) \cdot s = -s$ lies in $\tilde{R}$. Therefore $\tilde{R}$ is a subring of $S$, which is the definition of the integral closure being a ring. Note that the argument for closure under additive inverses uses closure under multiplication together with the fact that $-1 \in R$. If $R$ were only a semiring (lacking additive inverses), this step would fail. The subring axiom for $1_S$ is also immediate: $1_S$ satisfies $x - 1 \in R[x]$, so $1_S$ is always integral over $R$. [/step]