[proofplan] To show $\mathcal{O}_L$ is closed under addition, subtraction, and multiplication, we combine the two preceding module-theoretic results. Given $\alpha, \beta \in \mathcal{O}_L$, both are integral over $\mathbb{Z}$, so by [Integral Implies Finitely Generated](/theorems/1565) the ring $\mathbb{Z}[\alpha, \beta]$ is finitely generated as a $\mathbb{Z}$-module. By [Finitely Generated Implies Integral](/theorems/1566), every element of $\mathbb{Z}[\alpha, \beta]$ is integral over $\mathbb{Z}$. Since $\alpha + \beta$, $\alpha - \beta$, and $\alpha \beta$ all lie in $\mathbb{Z}[\alpha, \beta]$, they are algebraic integers, hence elements of $\mathcal{O}_L$. [/proofplan]

[step:Verify the additive identity $0$ and multiplicative identity $1$ lie in $\mathcal{O}_L$] The integers $0$ and $1$ satisfy the monic polynomials $x$ and $x - 1$ respectively, both in $\mathbb{Z}[x]$. Since $0, 1 \in \mathbb{Q} \subseteq L$, they are algebraic integers in $L$, i.e., $0, 1 \in \mathcal{O}_L$. This shows $\mathcal{O}_L$ contains both a zero element and a multiplicative identity. [/step]

[step:Form the subring $\mathbb{Z}[\alpha, \beta]\subseteq L$ and verify it is finitely generated over $\mathbb{Z}$] Fix $\alpha, \beta \in \mathcal{O}_L$. By definition of algebraic integer, $\alpha$ and $\beta$ are both integral over $\mathbb{Z}$: there exist monic polynomials $f_\alpha, f_\beta \in \mathbb{Z}[x]$ with $f_\alpha(\alpha) = 0$ and $f_\beta(\beta) = 0$. Define $R = \mathbb{Z}[\alpha, \beta]$, the smallest subring of $L$ containing $\mathbb{Z}$, $\alpha$, and $\beta$. We apply [Integral Implies Finitely Generated](/theorems/1565), part (2), to the ring extension $\mathbb{Z} \subseteq \mathbb{Z}[\alpha, \beta]$. The hypotheses of that theorem require: 1. **$\mathbb{Z}[\alpha, \beta]$ is of the form $\mathbb{Z}[s_1, \ldots, s_n]$ with each $s_i$ integral over $\mathbb{Z}$.** Here $n = 2$, $s_1 = \alpha$, $s_2 = \beta$; both are integral over $\mathbb{Z}$ by hypothesis. The theorem then concludes that $\mathbb{Z}[\alpha, \beta]$ is finitely generated as a $\mathbb{Z}$-module.[/step]

[guided]Fix $\alpha, \beta \in \mathcal{O}_L$. By definition of algebraic integer, $\alpha$ and $\beta$ are each integral over $\mathbb{Z}$, witnessed by monic polynomials $f_\alpha, f_\beta \in \mathbb{Z}[x]$. We form the subring \begin{align*} R = \mathbb{Z}[\alpha, \beta] \subseteq L, \end{align*} the smallest subring of $L$ containing $\mathbb{Z}$, $\alpha$, and $\beta$; explicitly, it consists of $\mathbb{Z}$-polynomial expressions in $\alpha$ and $\beta$. Our first step is to show $R$ is **finitely generated as a $\mathbb{Z}$-module**. This is where we invoke [Integral Implies Finitely Generated](/theorems/1565), part (2). The theorem says: if $S = R[s_1, \ldots, s_n]$ with each $s_i$ integral over $R$, then $S$ is finitely generated over $R$. We verify the hypotheses in our setting: - **Ring extension**: $\mathbb{Z} \subseteq \mathbb{Z}[\alpha, \beta] \subseteq L$, with $L$ a commutative ring (actually a field), so both intermediate rings are commutative. - **Generators are integral over the base**: $s_1 = \alpha$ and $s_2 = \beta$ are integral over $\mathbb{Z}$ by the hypothesis $\alpha, \beta \in \mathcal{O}_L$. All hypotheses hold. The theorem concludes that $\mathbb{Z}[\alpha, \beta]$ is finitely generated as a $\mathbb{Z}$-module. Note that this is a statement about $R$ as a module; $R$ is, of course, also finitely generated as a ring by $\{\alpha, \beta\}$, but the module statement is strictly stronger and is what we need.[/guided]

[step:Conclude that every element of $\mathbb{Z}[\alpha, \beta]$ is integral over $\mathbb{Z}$] We apply [Finitely Generated Implies Integral](/theorems/1566) to the ring extension $\mathbb{Z} \subseteq \mathbb{Z}[\alpha, \beta]$. The hypothesis of that theorem requires: 1. **$\mathbb{Z}[\alpha, \beta]$ is finitely generated as a $\mathbb{Z}$-module**, which we verified in the previous step. The theorem then concludes that $\mathbb{Z}[\alpha, \beta]$ is integral over $\mathbb{Z}$, i.e., every $\gamma \in \mathbb{Z}[\alpha, \beta]$ is integral over $\mathbb{Z}$.[/step]

[guided]We now apply [Finitely Generated Implies Integral](/theorems/1566): if $S$ is finitely generated as an $R$-module, then $S$ is integral over $R$. We check the hypothesis: $\mathbb{Z}[\alpha, \beta]$ is finitely generated as a $\mathbb{Z}$-module by the previous step. The theorem yields: every element of $\mathbb{Z}[\alpha, \beta]$ is integral over $\mathbb{Z}$. Why does this two-step dance work? Integrality is not self-evidently closed under ring operations — if $\alpha$ satisfies a monic polynomial $f_\alpha$ and $\beta$ satisfies $f_\beta$, there is no formula producing a monic integer polynomial satisfied by $\alpha + \beta$ from $f_\alpha$ and $f_\beta$ directly (there is a formula using resultants, but it is intricate and produces a polynomial of potentially very high degree). The module-theoretic approach finesses this: instead of manipulating polynomials, we package $\alpha$ and $\beta$ into a **finite-rank $\mathbb{Z}$-module** $\mathbb{Z}[\alpha, \beta]$, and the determinant trick (the heart of [Finitely Generated Implies Integral](/theorems/1566)) produces monic polynomials for **every** element of this module at once.[/guided]

[step:Verify $\alpha \pm \beta$ and $\alpha \beta$ lie in $\mathbb{Z}[\alpha, \beta]$] The elements $\alpha + \beta$, $\alpha - \beta$, and $\alpha\beta$ are obtained from $\alpha$ and $\beta$ by ring operations (addition, subtraction, multiplication). Since $\mathbb{Z}[\alpha, \beta]$ is by definition a ring containing $\alpha$ and $\beta$, it is closed under these operations; hence \begin{align*} \alpha + \beta, \; \alpha - \beta, \; \alpha \beta \in \mathbb{Z}[\alpha, \beta]. \end{align*} [/step]

[step:Conclude $\mathcal{O}_L$ is a ring]By the previous two steps, $\alpha + \beta$, $\alpha - \beta$, and $\alpha\beta$ all lie in $\mathbb{Z}[\alpha, \beta]$ and every element of $\mathbb{Z}[\alpha, \beta]$ is integral over $\mathbb{Z}$. Moreover, each of these elements lies in $L$ (since $\mathbb{Z}[\alpha, \beta] \subseteq L$), so they are integral elements of $L$ over $\mathbb{Z}$, which is the definition of being in $\mathcal{O}_L$: \begin{align*} \alpha + \beta, \; \alpha - \beta, \; \alpha \beta \in \mathcal{O}_L. \end{align*} Combined with the first step ($0, 1 \in \mathcal{O}_L$), this shows $\mathcal{O}_L$ is closed under addition, subtraction, and multiplication, and contains $0$ and $1$. Since $\mathcal{O}_L \subseteq L$ and $L$ is a commutative ring (in fact a field), the inherited operations on $\mathcal{O}_L$ satisfy associativity, commutativity, and distributivity. Therefore $\mathcal{O}_L$ is a subring of $L$, in particular a commutative ring.[/step]

[guided]We collect the pieces. Let $\alpha, \beta \in \mathcal{O}_L$. The previous steps show: - $\alpha + \beta$, $\alpha - \beta$, $\alpha \beta \in \mathbb{Z}[\alpha, \beta]$ (closure of a ring under its operations). - Every element of $\mathbb{Z}[\alpha, \beta]$ is integral over $\mathbb{Z}$ (by [Finitely Generated Implies Integral](/theorems/1566) applied to the finitely-generated module $\mathbb{Z}[\alpha, \beta]$). - $\mathbb{Z}[\alpha, \beta] \subseteq L$ (the ambient field). An element of $L$ that is integral over $\mathbb{Z}$ is by definition in $\mathcal{O}_L$. So \begin{align*} \alpha + \beta, \; \alpha - \beta, \; \alpha \beta \in \mathcal{O}_L. \end{align*} Together with the earlier verification that $0, 1 \in \mathcal{O}_L$, this proves $\mathcal{O}_L$ is closed under $+$, $-$, and $\cdot$ and contains $0, 1$. Since $\mathcal{O}_L \subseteq L$ inherits its operations from the field $L$, associativity, commutativity, and distributivity come for free. Therefore $\mathcal{O}_L$ is a (commutative) subring of $L$. A final remark on the logical flow: the proof is a two-theorem sandwich. We pass from the "integral" world (elements) to the "finitely generated module" world via [Integral Implies Finitely Generated](/theorems/1565), do closure inside the module world (where closure under ring operations is automatic), then pass back via [Finitely Generated Implies Integral](/theorems/1566). This is a standard pattern in commutative algebra: ring-theoretic closure properties of an integrally closed subring are most cleanly proved by detouring through finitely generated modules.[/guided]