[proofplan] Given $\alpha \in L$, let $g \in \mathbb{Q}[x]$ be its monic minimal polynomial over $\mathbb{Q}$. Clear denominators: let $n$ be the least common multiple of the denominators of the rational coefficients of $g$. Define $h(x) = n^{d} g(x/n)$ where $d = \deg g$. The rescaling $x \mapsto x/n$ composed with multiplication by $n^d$ sends a monic polynomial of degree $d$ to a monic polynomial of degree $d$ with coefficients $n^{d-i} \cdot (\text{coefficient of } x^i \text{ in } g)$. The factor $n^{d-i}$ kills the denominators by the choice of $n$, giving $h \in \mathbb{Z}[x]$. Finally $h(n\alpha) = n^d g(\alpha) = 0$ realises $n\alpha$ as a root of a monic integer polynomial, hence $n\alpha \in \mathcal{O}_L$. This exhibits $\alpha = (n\alpha)/n \in \operatorname{Frac}(\mathcal{O}_L)$; combined with the obvious inclusion $\operatorname{Frac}(\mathcal{O}_L) \subseteq L$, we obtain $\operatorname{Frac}(\mathcal{O}_L) = L$. [/proofplan] [step:Fix $\alpha \in L$ and extract the monic minimal polynomial $g \in \mathbb{Q}[x]$] Let $\alpha \in L$. Since $L$ is a number field — a finite extension of $\mathbb{Q}$ — every element $\alpha \in L$ is algebraic over $\mathbb{Q}$. Indeed, the powers $1, \alpha, \alpha^2, \ldots$ lie in the finite-dimensional $\mathbb{Q}$-vector space $L$, so they are linearly dependent over $\mathbb{Q}$; scaling the minimal dependence relation produces a nonzero polynomial $f \in \mathbb{Q}[x]$ with $f(\alpha) = 0$. Let \begin{align*} g(x) = x^d + q_{d-1} x^{d-1} + \cdots + q_1 x + q_0 \in \mathbb{Q}[x], \qquad q_i \in \mathbb{Q}, \end{align*} be the [minimal polynomial](/pages/???) of $\alpha$ over $\mathbb{Q}$: the unique monic polynomial of least degree in $\mathbb{Q}[x]$ annihilating $\alpha$. Here $d = \deg g \geq 1$. [guided] Fix an arbitrary $\alpha \in L$. We want to show that some nonzero integer multiple of $\alpha$ lies in $\mathcal{O}_L$; equivalently, $\alpha$ is a fraction of algebraic integers. Since $L/\mathbb{Q}$ is a number field, it is by definition a finite extension of $\mathbb{Q}$, i.e., finite-dimensional as a $\mathbb{Q}$-vector space. Let $[L:\mathbb{Q}] = N < \infty$. Then the $N + 1$ elements $1, \alpha, \alpha^2, \ldots, \alpha^N$ of $L$ are linearly dependent over $\mathbb{Q}$ (any $N+1$ vectors in an $N$-dimensional space are dependent). This produces a nonzero polynomial $f \in \mathbb{Q}[x]$ with $f(\alpha) = 0$ and $\deg f \leq N$. So $\alpha$ is algebraic over $\mathbb{Q}$. For algebraic elements, there is a distinguished annihilating polynomial — the [minimal polynomial](/pages/???) over $\mathbb{Q}$: the unique monic polynomial $g \in \mathbb{Q}[x]$ of least degree with $g(\alpha) = 0$. Write \begin{align*} g(x) = x^d + q_{d-1} x^{d-1} + \cdots + q_1 x + q_0, \qquad q_i \in \mathbb{Q}, \end{align*} with $d = \deg g \geq 1$ (since $g$ is nonzero and $\alpha$ is not automatically $0$; even if $\alpha = 0$, the minimal polynomial would be $x$, of degree $1$). The issue to overcome: the coefficients $q_i$ are rational, not necessarily integers, so $g$ itself does not witness integrality of $\alpha$. We need to modify the polynomial — specifically, by a change of variable that clears denominators. [/guided] [/step] [step:Choose $n \in \mathbb{Z}_{\geq 1}$ clearing the denominators of the coefficients of $g$] Write each rational coefficient in lowest terms as $q_i = a_i / b_i$ with $a_i \in \mathbb{Z}$, $b_i \in \mathbb{Z}_{\geq 1}$, $\gcd(a_i, b_i) = 1$ (and by convention $q_i = 0 \implies a_i = 0, b_i = 1$). Let \begin{align*} n := \operatorname{lcm}(b_0, b_1, \ldots, b_{d-1}) \in \mathbb{Z}_{\geq 1}. \end{align*} This is a positive integer with $b_i \mid n$ for every $i \in \{0, 1, \ldots, d-1\}$. By definition of $n$, for each $i$ the product $n \cdot q_i = n \cdot a_i / b_i = (n / b_i) \cdot a_i \in \mathbb{Z}$, since $n / b_i \in \mathbb{Z}$. More generally, for any integer $k \geq 1$: \begin{align*} n^k q_i = (n / b_i) \cdot a_i \cdot n^{k-1} \in \mathbb{Z}. \end{align*} [guided] To clear denominators uniformly, we need a single integer $n$ that contains every $b_i$ as a factor. The natural choice is the least common multiple: \begin{align*} n := \operatorname{lcm}(b_0, b_1, \ldots, b_{d-1}). \end{align*} This is a positive integer (an [lcm](/pages/???) of positive integers), and by construction $b_i \mid n$ for every $i$. In particular $n / b_i \in \mathbb{Z}_{\geq 1}$. Consequence: multiplying $q_i = a_i / b_i$ by $n$ gives \begin{align*} n q_i = \frac{n \cdot a_i}{b_i} = \frac{n}{b_i} \cdot a_i \in \mathbb{Z}, \end{align*} since both $n/b_i$ and $a_i$ are integers. More generally, for $k \geq 1$: \begin{align*} n^k q_i = n^{k-1} \cdot (n q_i) \in \mathbb{Z}, \end{align*} because $\mathbb{Z}$ is closed under multiplication. Why the lcm specifically, and not just the product $\prod_i b_i$? The product also works, but the lcm is minimal; we could use any common multiple. The argument only requires that each $b_i$ divides $n$. [/guided] [/step] [step:Define $h(x) := n^d g(x/n) \in \mathbb{Q}[x]$ and verify $h \in \mathbb{Z}[x]$] Define the polynomial \begin{align*} h: \mathbb{Q} &\to \mathbb{Q} \\ h(x) &:= n^d \cdot g\!\left(\frac{x}{n}\right) \in \mathbb{Q}[x]. \end{align*} Expanding using the explicit form of $g$: \begin{align*} h(x) = n^d \left( \left(\frac{x}{n}\right)^d + q_{d-1}\left(\frac{x}{n}\right)^{d-1} + \cdots + q_1 \left(\frac{x}{n}\right) + q_0 \right). \end{align*} Distributing the $n^d$: \begin{align*} h(x) = \sum_{i=0}^{d} n^d \cdot q_i \cdot \frac{x^i}{n^i} = \sum_{i=0}^{d} n^{d-i} q_i \cdot x^i, \end{align*} where we set $q_d := 1$ (the leading coefficient of $g$). Reading off the coefficient of $x^i$ in $h$: \begin{align*} [x^i] h(x) = n^{d-i} q_i, \qquad i = 0, 1, \ldots, d. \end{align*} We verify $h \in \mathbb{Z}[x]$ coefficient by coefficient: - **$i = d$ (leading coefficient):** $n^{d-d} q_d = n^0 \cdot 1 = 1 \in \mathbb{Z}$. Moreover this shows $h$ is monic. - **$0 \leq i \leq d - 1$:** $n^{d-i} q_i$ with $d - i \geq 1$. By the previous step, $n^k q_i \in \mathbb{Z}$ for every $k \geq 1$, so $n^{d-i} q_i \in \mathbb{Z}$. Hence every coefficient of $h$ lies in $\mathbb{Z}$: $h \in \mathbb{Z}[x]$. Furthermore, $h$ is monic of degree $d$. [guided] Define the new polynomial \begin{align*} h(x) := n^d \cdot g\!\left(\frac{x}{n}\right). \end{align*} This is a polynomial identity in $x$: we substitute $x/n$ into $g$ and multiply through by $n^d$. The substitution $x \mapsto x/n$ is a polynomial operation (replace every occurrence of the variable with $x/n$), and multiplying by the constant $n^d$ preserves polynomiality. Expanding $g(x/n)$: \begin{align*} g(x/n) = \left(\frac{x}{n}\right)^d + q_{d-1}\left(\frac{x}{n}\right)^{d-1} + \cdots + q_1 \left(\frac{x}{n}\right) + q_0 = \sum_{i=0}^d q_i \frac{x^i}{n^i}, \end{align*} where we set $q_d := 1$. Multiplying by $n^d$: \begin{align*} h(x) = n^d \sum_{i=0}^d q_i \frac{x^i}{n^i} = \sum_{i=0}^d q_i \cdot \frac{n^d}{n^i} \cdot x^i = \sum_{i=0}^d n^{d-i} q_i \cdot x^i. \end{align*} So the coefficient of $x^i$ in $h$ is $n^{d-i} q_i$. We check each coefficient is an integer: - The **leading coefficient** ($i = d$) is $n^0 \cdot q_d = 1 \cdot 1 = 1$. So $h$ is monic of degree $d$. Why is monicity important? Because integrality is defined via *monic* polynomials in $\mathbb{Z}[x]$; a polynomial with non-unit leading coefficient does not witness integrality. - For the other coefficients ($0 \leq i \leq d - 1$), the power $n^{d-i}$ has exponent $d - i \geq 1$. By the previous step, $n^k q_i \in \mathbb{Z}$ for any $k \geq 1$. So $n^{d-i} q_i \in \mathbb{Z}$. Thus $h \in \mathbb{Z}[x]$ is monic of degree $d$. Remark on the structure: this construction is a special case of the general principle that if $g \in K[x]$ is monic and $n \in K^\times$, then $h(x) = n^d g(x/n)$ is the polynomial whose roots are exactly $n$ times the roots of $g$. Geometrically, we scale the variable by $n$, and we compensate by $n^d$ to preserve monicity. [/guided] [/step] [step:Verify $h(n\alpha) = 0$ and conclude $n\alpha \in \mathcal{O}_L$] Evaluate $h$ at $n\alpha \in L$. By the definition of $h$: \begin{align*} h(n\alpha) = n^d \cdot g\!\left(\frac{n\alpha}{n}\right) = n^d \cdot g(\alpha). \end{align*} By the defining property of the minimal polynomial, $g(\alpha) = 0$, so \begin{align*} h(n\alpha) = n^d \cdot 0 = 0. \end{align*} Hence $h \in \mathbb{Z}[x]$ is a monic polynomial with $h(n\alpha) = 0$. By the definition of the [ring of integers](/pages/???), $\mathcal{O}_L = \{\beta \in L : \beta \text{ is integral over } \mathbb{Z}\}$, this is exactly what it means for $n\alpha \in L$ to lie in $\mathcal{O}_L$. (Note $n\alpha \in L$ because $n \in \mathbb{Z} \subseteq \mathbb{Q} \subseteq L$ and $\alpha \in L$, with $L$ closed under multiplication.) So \begin{align*} n\alpha \in \mathcal{O}_L. \end{align*} [guided] Recall the defining identity of $h$: $h(x) = n^d g(x/n)$. Plug in $x = n\alpha$: \begin{align*} h(n\alpha) = n^d \cdot g\!\left(\frac{n\alpha}{n}\right) = n^d \cdot g(\alpha). \end{align*} By the defining property of the minimal polynomial $g$, we have $g(\alpha) = 0$. Therefore \begin{align*} h(n\alpha) = n^d \cdot 0 = 0. \end{align*} This was the whole point of the substitution: transform the (possibly non-integer-coefficient) polynomial $g$ vanishing at $\alpha$ into an integer-coefficient polynomial $h$ vanishing at $n\alpha$. Now apply the definition of the ring of integers. An element $\beta$ of a number field lies in $\mathcal{O}_L$ iff there exists a monic $p \in \mathbb{Z}[x]$ with $p(\beta) = 0$. We take $\beta := n\alpha$ and $p := h$, and verify: - $n\alpha \in L$: since $n \in \mathbb{Z} \subseteq \mathbb{Q} \subseteq L$ and $\alpha \in L$, the product $n\alpha$ lies in $L$ by closure of $L$ under multiplication. - $h \in \mathbb{Z}[x]$ monic: verified in the previous step. - $h(n\alpha) = 0$: just computed. All conditions met, so $n\alpha \in \mathcal{O}_L$. [/guided] [/step] [step:Conclude $\operatorname{Frac}(\mathcal{O}_L) = L$ from the existence of such an $n$ for every $\alpha$] We have shown: for every $\alpha \in L$, there exists $n \in \mathbb{Z}_{\geq 1}$ (in particular $n \neq 0$) with $n\alpha \in \mathcal{O}_L$. This is precisely the more-precise statement in the theorem. It remains to deduce $\operatorname{Frac}(\mathcal{O}_L) = L$. Recall that $\mathcal{O}_L$ is an integral domain (a subring of the field $L$, and every subring of a field is an integral domain), so the [fraction field](/pages/???) $\operatorname{Frac}(\mathcal{O}_L)$ is well-defined as the localisation of $\mathcal{O}_L$ at $\mathcal{O}_L \setminus \{0\}$, embedded canonically as the smallest field containing $\mathcal{O}_L$. We verify the two inclusions: - **$\operatorname{Frac}(\mathcal{O}_L) \subseteq L$**: Since $\mathcal{O}_L \subseteq L$ and $L$ is a field containing $\mathcal{O}_L$, by the universal property of the fraction field there is a unique ring homomorphism \begin{align*} \operatorname{Frac}(\mathcal{O}_L) &\to L \\ \beta/\gamma &\mapsto \beta \gamma^{-1} \end{align*} extending the inclusion $\mathcal{O}_L \hookrightarrow L$. This is injective (a field homomorphism between fields is always injective, being a ring homomorphism with trivial kernel since fields have no nontrivial ideals). Its image consists of elements of $L$ of the form $\beta \gamma^{-1}$ with $\beta, \gamma \in \mathcal{O}_L$ and $\gamma \neq 0$. Identifying $\operatorname{Frac}(\mathcal{O}_L)$ with its image, $\operatorname{Frac}(\mathcal{O}_L) \subseteq L$. - **$L \subseteq \operatorname{Frac}(\mathcal{O}_L)$**: For $\alpha \in L$, if $\alpha = 0$ then $\alpha = 0/1 \in \operatorname{Frac}(\mathcal{O}_L)$. If $\alpha \neq 0$, the previous steps produced $n \in \mathbb{Z}_{\geq 1}$ with $\beta := n\alpha \in \mathcal{O}_L$. Since $\mathbb{Z} \subseteq \mathbb{Z}_{\geq 1} \cup \{0\} \subseteq \mathbb{Z} \subseteq \mathcal{O}_\mathbb{Q} = \mathcal{O}_L \cap \mathbb{Q} \subseteq \mathcal{O}_L$ (by [Algebraic Integers in $\mathbb{Q}$](/theorems/1564)), we have $n \in \mathcal{O}_L$ as well, and $n \neq 0$. Therefore \begin{align*} \alpha = \frac{n\alpha}{n} = \frac{\beta}{n} \in \operatorname{Frac}(\mathcal{O}_L). \end{align*} Combining the two inclusions: $\operatorname{Frac}(\mathcal{O}_L) = L$. This completes the proof. [guided] The work of the previous four steps gives us: for every $\alpha \in L$, there exists a nonzero integer $n$ with $n\alpha \in \mathcal{O}_L$. This is the "more precise" statement in the theorem — it is a quantitative version of "$\alpha$ is a fraction of algebraic integers". To derive the field-theoretic statement $\operatorname{Frac}(\mathcal{O}_L) = L$, we unpack the [fraction field](/pages/???) construction. Recall: - $\mathcal{O}_L$ is an integral domain. This is because $\mathcal{O}_L$ is a subring of $L$, and subrings of a field are integral domains (they inherit commutativity and the absence of zero divisors, and they contain $1$). - $\operatorname{Frac}(\mathcal{O}_L)$ is the localisation of $\mathcal{O}_L$ at the multiplicative set $\mathcal{O}_L \setminus \{0\}$: formal fractions $\beta/\gamma$ with $\gamma \neq 0$, modulo the equivalence $\beta_1/\gamma_1 \sim \beta_2/\gamma_2 \iff \beta_1 \gamma_2 = \beta_2 \gamma_1$. - The universal property: $\operatorname{Frac}(\mathcal{O}_L)$ is the smallest field in which $\mathcal{O}_L$ embeds. More precisely, any injective ring homomorphism from $\mathcal{O}_L$ into a field $F$ extends uniquely to a field homomorphism $\operatorname{Frac}(\mathcal{O}_L) \to F$. We verify both inclusions. **($\operatorname{Frac}(\mathcal{O}_L) \subseteq L$):** We have the inclusion $\mathcal{O}_L \hookrightarrow L$ (an injective ring homomorphism, since $\mathcal{O}_L$ is literally a subset of the field $L$). By the universal property of the fraction field, this extends uniquely to a field homomorphism $\operatorname{Frac}(\mathcal{O}_L) \to L$ sending $\beta/\gamma$ to $\beta\gamma^{-1}$ (the inverse exists in $L$ because $\gamma \neq 0$ and $L$ is a field). This homomorphism is injective because field homomorphisms are always injective: the kernel is an ideal of a field, hence $\{0\}$ or the whole field, and the latter is excluded because the homomorphism sends $1 \mapsto 1 \neq 0$. Identifying $\operatorname{Frac}(\mathcal{O}_L)$ with its image in $L$, we obtain $\operatorname{Frac}(\mathcal{O}_L) \subseteq L$. **($L \subseteq \operatorname{Frac}(\mathcal{O}_L)$):** Take any $\alpha \in L$. If $\alpha = 0$, then $\alpha = 0/1$ is an element of $\operatorname{Frac}(\mathcal{O}_L)$ (both $0$ and $1$ lie in $\mathcal{O}_L$, and $1 \neq 0$). Otherwise, by the previous four steps, there exists $n \in \mathbb{Z}_{\geq 1}$ with $n\alpha \in \mathcal{O}_L$. Now, $n$ itself lies in $\mathcal{O}_L$: we have $\mathbb{Z} \subseteq \mathcal{O}_L$ because every integer $k$ is a root of the monic integer polynomial $x - k$, hence integral over $\mathbb{Z}$, hence in $\mathcal{O}_L$. (Alternatively, [Algebraic Integers in $\mathbb{Q}$](/theorems/1564) gives $\mathbb{Z} = \mathcal{O}_\mathbb{Q} \subseteq \mathcal{O}_L$.) And $n \neq 0$ by choice. So we can form the fraction \begin{align*} \frac{n\alpha}{n} \in \operatorname{Frac}(\mathcal{O}_L), \end{align*} and in $L$ it equals $\alpha$. Identifying $\operatorname{Frac}(\mathcal{O}_L)$ with its image in $L$ from the first inclusion, $\alpha \in \operatorname{Frac}(\mathcal{O}_L)$. Combining both inclusions, $\operatorname{Frac}(\mathcal{O}_L) = L$, as claimed. A philosophical remark: this result is the algebraic number theory analogue of $\operatorname{Frac}(\mathbb{Z}) = \mathbb{Q}$. Just as $\mathbb{Q}$ is built from $\mathbb{Z}$ by inverting nonzero integers, $L$ is built from $\mathcal{O}_L$ by inverting nonzero algebraic integers. This justifies the slogan: "$\mathcal{O}_L$ stands to $L$ as $\mathbb{Z}$ stands to $\mathbb{Q}$." [/guided] [/step]