[proofplan] The direction ($\Leftarrow$) is immediate: if $p_\alpha \in \mathbb{Z}[x]$ is monic and vanishes at $\alpha$, the definition of $\mathcal{O}_L$ gives $\alpha \in \mathcal{O}_L$. For ($\Rightarrow$), assume $\alpha$ satisfies a monic $h \in \mathbb{Z}[x]$. Then $p_\alpha \mid h$ in $\mathbb{Q}[x]$, so in a splitting field $M$ of $h$ over $\mathbb{Q}$, every root of $p_\alpha$ is a root of $h$ and hence an algebraic integer. The coefficients of $p_\alpha$ are $\pm$ elementary symmetric functions of its roots, hence lie in the ring $\mathcal{O}_M$ (closed under sums and products); they also lie in $\mathbb{Q}$ by construction. Intersecting, they lie in $\mathcal{O}_M \cap \mathbb{Q} = \mathcal{O}_\mathbb{Q} = \mathbb{Z}$, where the final equality is the classical "algebraic integers in $\mathbb{Q}$ are rational integers" theorem. [/proofplan]

[step:Prove the easy direction ($\Leftarrow$) directly from the definition of $\mathcal{O}_L$]Assume $p_\alpha \in \mathbb{Z}[x]$. By definition of the minimal polynomial, $p_\alpha$ is monic (its leading coefficient is $1$) and satisfies $p_\alpha(\alpha) = 0$. Hence $p_\alpha \in \mathbb{Z}[x]$ is a monic polynomial in $\mathbb{Z}[x]$ annihilating $\alpha$. 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 $\alpha$ to be in $\mathcal{O}_L$.[/step]

[guided]This direction is a direct application of the definition. Recall: - An element $\beta$ of a number field $L$ is called an **algebraic integer** (written $\beta \in \mathcal{O}_L$) if there exists a monic polynomial $g \in \mathbb{Z}[x]$ with $g(\beta) = 0$. - The minimal polynomial $p_\alpha \in \mathbb{Q}[x]$ of $\alpha$ over $\mathbb{Q}$ is monic with $p_\alpha(\alpha) = 0$. If we are given $p_\alpha \in \mathbb{Z}[x]$ (as a polynomial with integer coefficients, not merely rational ones), then $p_\alpha$ itself plays the role of the witness monic integer polynomial: it is monic, it has coefficients in $\mathbb{Z}$, and it annihilates $\alpha$. Hence $\alpha \in \mathcal{O}_L$. No further argument is needed — we read off the conclusion from the definition.[/guided]

[step:For the hard direction ($\Rightarrow$), extract a monic $h \in \mathbb{Z}[x]$ with $h(\alpha) = 0$ and invoke divisibility by $p_\alpha$] Assume $\alpha \in \mathcal{O}_L$. By the definition of $\mathcal{O}_L$, there exists a monic polynomial \begin{align*} h(x) = x^m + c_{m-1} x^{m-1} + \cdots + c_1 x + c_0 \in \mathbb{Z}[x], \qquad c_i \in \mathbb{Z}, \end{align*} with $h(\alpha) = 0$. Every integer is rational, so $h \in \mathbb{Q}[x]$ as well, and $h$ remains monic when regarded as an element of $\mathbb{Q}[x]$. Since $h \in \mathbb{Q}[x]$ satisfies $h(\alpha) = 0$, we apply the [Minimal Polynomial Divides Any Annihilating Polynomial](/theorems/1569) theorem with $K = \mathbb{Q}$ and $f = h$. The hypotheses are: - The minimal polynomial $p_\alpha$ of $\alpha$ over $\mathbb{Q}$ exists (it does, by definition of a minimal polynomial of an algebraic element over a field). - $h \in \mathbb{Q}[x]$ and $h(\alpha) = 0$: verified. The theorem concludes $p_\alpha \mid h$ in $\mathbb{Q}[x]$.[/step]

[guided]We assume $\alpha \in \mathcal{O}_L$. Unpacking the definition: \begin{align*} \alpha \in \mathcal{O}_L \iff \exists\, h \in \mathbb{Z}[x] \text{ monic with } h(\alpha) = 0. \end{align*} Fix such a polynomial: \begin{align*} h(x) = x^m + c_{m-1} x^{m-1} + \cdots + c_1 x + c_0, \qquad c_i \in \mathbb{Z}, \quad h(\alpha) = 0. \end{align*} Our goal is to prove $p_\alpha \in \mathbb{Z}[x]$. We do **not** have $h = p_\alpha$ in general — the integer-coefficient polynomial $h$ may well have higher degree than $p_\alpha$, and $p_\alpha$ might a priori have rational (non-integer) coefficients. So we cannot just declare "$p_\alpha = h$". What we need is a relation between $p_\alpha$ and $h$ that constrains the coefficients of $p_\alpha$ to lie in $\mathbb{Z}$. The relation is divisibility. Regard $h$ as an element of the larger ring $\mathbb{Q}[x]$ (via the inclusion $\mathbb{Z}[x] \hookrightarrow \mathbb{Q}[x]$): $h$ is still monic and still vanishes at $\alpha$. By the [Minimal Polynomial Divides Any Annihilating Polynomial](/theorems/1569) theorem, the minimal polynomial of $\alpha$ over $\mathbb{Q}$ divides every element of $\mathbb{Q}[x]$ that annihilates $\alpha$. We verify the hypotheses: - The theorem requires $K$ a field (applied with $K = \mathbb{Q}$): $\mathbb{Q}$ is a field, so $\mathbb{Q}[x]$ is a polynomial ring over a field. - The theorem requires $f \in K[x]$ with $f(\alpha) = 0$ (here $f = h$): we verified $h \in \mathbb{Q}[x]$ and $h(\alpha) = 0$. The theorem concludes $p_\alpha \mid h$ in $\mathbb{Q}[x]$. Divisibility in $\mathbb{Q}[x]$ means: there exists $q \in \mathbb{Q}[x]$ with $h = p_\alpha q$. We will next use this divisibility to show that the roots of $p_\alpha$ (in a suitable extension field) are all algebraic integers.[/guided]

[step:Pass to a splitting field $M$ of $h$ over $\mathbb{Q}$ containing all roots of $p_\alpha$]Let $M \supseteq \mathbb{Q}$ be a [splitting field](/pages/???) of $h$ over $\mathbb{Q}$: a finite extension of $\mathbb{Q}$ over which $h$ factors into linear factors and that is generated by the roots of $h$. Such a field exists and is unique up to isomorphism. Since $p_\alpha \mid h$ in $\mathbb{Q}[x]$, and $\mathbb{Q}[x] \subseteq M[x]$, the divisibility persists in $M[x]$: $p_\alpha \mid h$ in $M[x]$. In $M$, the polynomial $h$ splits as a product of monic linear factors by construction: \begin{align*} h(x) = \prod_{j=1}^m (x - \rho_j), \qquad \rho_j \in M. \end{align*} Since $M[x]$ is a [UFD](/pages/???) (in fact a PID) and $p_\alpha$ is a divisor of $h$ of degree $r := \deg p_\alpha$, any irreducible factor of $p_\alpha$ in $M[x]$ must equal some $x - \rho_j$ up to a unit; combined with monicity of $p_\alpha$, we obtain a factorisation \begin{align*} p_\alpha(x) = \prod_{i=1}^r (x - \alpha_i), \qquad \alpha_i \in \{\rho_1, \ldots, \rho_m\} \subseteq M. \end{align*} In particular, each $\alpha_i$ is a root of $h$ in $M$.[/step]

[guided]Divisibility gives us $h = p_\alpha q$ in $\mathbb{Q}[x]$. To exploit this algebraically we pass to a field where the factors can be seen — a [splitting field](/pages/???) $M$ of $h$ over $\mathbb{Q}$. By general field theory, such an $M$ exists as a finite extension of $\mathbb{Q}$: $M$ is constructed by iteratively adjoining the roots of $h$ to $\mathbb{Q}$, and by standard theory $M$ is a finite extension unique up to $\mathbb{Q}$-isomorphism. In $M[x]$, by the defining property of a splitting field, $h$ factors as a product of linear polynomials: \begin{align*} h(x) = \prod_{j=1}^m (x - \rho_j), \qquad \rho_j \in M. \end{align*} (Each $\rho_j$ is a root of $h$; since $h$ is monic of degree $m$, there are exactly $m$ roots counted with multiplicity.) The divisibility $p_\alpha \mid h$ in $\mathbb{Q}[x]$ lifts to $M[x]$: the inclusion $\mathbb{Q}[x] \hookrightarrow M[x]$ is a ring homomorphism (in fact, an inclusion of integral domains), and if $h = p_\alpha q$ in $\mathbb{Q}[x]$ then the same identity holds in $M[x]$. Now, $M[x]$ is a [UFD](/pages/???) (polynomial rings over fields are PIDs, hence UFDs). Any monic divisor $p_\alpha$ of $h = \prod_j (x - \rho_j)$ in $M[x]$ must be a product of a subset of the linear factors $\{x - \rho_j\}$, because these are the irreducible factors of $h$ in $M[x]$ and the factorisation is unique up to units and ordering. Since $p_\alpha$ is monic and divides $\prod_j (x - \rho_j)$, we obtain: \begin{align*} p_\alpha(x) = \prod_{i=1}^r (x - \alpha_i), \qquad r = \deg p_\alpha, \quad \alpha_i \in \{\rho_1, \ldots, \rho_m\} \subseteq M. \end{align*} So the roots $\alpha_1, \ldots, \alpha_r$ of $p_\alpha$ (in the field $M$) are a subset of the roots of $h$. In particular $h(\alpha_i) = 0$ for every $i$. Why do we need to work in the splitting field? Because the coefficients of $p_\alpha$ are symmetric functions of its roots, and we need the roots to exist somewhere in order to reason about them. The roots $\alpha_i$ need not lie in $L$ or $\mathbb{Q}$; we pass to $M$ to ensure they exist.[/guided]

[step:Show that each root $\alpha_i$ of $p_\alpha$ is an algebraic integer, i.e., lies in $\mathcal{O}_M$]For each $i \in \{1, \ldots, r\}$, the element $\alpha_i \in M$ satisfies the monic integer polynomial $h \in \mathbb{Z}[x]$: \begin{align*} h(\alpha_i) = 0, \end{align*} by the previous step. By the definition of the ring of integers $\mathcal{O}_M := \{\beta \in M : \beta \text{ is integral over } \mathbb{Z}\}$, this means $\alpha_i \in \mathcal{O}_M$.[/step]

[guided]Having established that each root $\alpha_i$ of $p_\alpha$ is also a root of $h$, we invoke the definition of the ring of integers in $M$. The definition: $\mathcal{O}_M = \{\beta \in M : \text{there exists a monic } g \in \mathbb{Z}[x] \text{ with } g(\beta) = 0\}$. For each $i$: - $\alpha_i \in M$ (by construction — the roots live in the splitting field). - The polynomial $h$ is a monic element of $\mathbb{Z}[x]$ with $h(\alpha_i) = 0$. Hence $\alpha_i \in \mathcal{O}_M$. Every root of $p_\alpha$ is an algebraic integer in $M$. This is the crucial mechanism: by extending to a field where all roots of $p_\alpha$ exist, we pick up a uniform witness — the single polynomial $h$ — that demonstrates the integrality of all of them simultaneously.[/guided]

[step:Express the coefficients of $p_\alpha$ as elementary symmetric polynomials in the $\alpha_i$, hence as elements of $\mathcal{O}_M$]Expand the factorisation $p_\alpha(x) = \prod_{i=1}^r (x - \alpha_i)$ using the [Vieta formulas](/pages/???) (equivalently, Newton's identities for elementary symmetric polynomials). For $k = 0, 1, \ldots, r$, define the $k$-th elementary symmetric polynomial \begin{align*} e_k(\alpha_1, \ldots, \alpha_r) &= \sum_{\substack{S \subseteq \{1, \ldots, r\} \\ |S| = k}} \prod_{i \in S} \alpha_i \in M. \end{align*} Expanding the product gives \begin{align*} p_\alpha(x) = \prod_{i=1}^r (x - \alpha_i) = \sum_{k=0}^r (-1)^k e_k(\alpha_1, \ldots, \alpha_r)\, x^{r-k}. \end{align*} Reading off coefficients: the coefficient of $x^{r-k}$ in $p_\alpha$ is $(-1)^k e_k(\alpha_1, \ldots, \alpha_r)$. Now each $\alpha_i \in \mathcal{O}_M$ by the previous step. By the [Ring of Integers is a Ring](/theorems/1567) theorem, $\mathcal{O}_M$ is a subring of $M$, hence closed under addition, subtraction, and multiplication. The elementary symmetric polynomial $e_k(\alpha_1, \ldots, \alpha_r)$ is a sum of products of $k$ of the $\alpha_i$; such an expression is a sum of products of elements of the ring $\mathcal{O}_M$, hence lies in $\mathcal{O}_M$. Multiplying by the sign $(-1)^k \in \mathbb{Z} \subseteq \mathcal{O}_M$ preserves this. Therefore every coefficient of $p_\alpha$ lies in $\mathcal{O}_M$.[/step]

[guided]We factored $p_\alpha$ as $\prod_{i=1}^r (x - \alpha_i)$. To access the coefficients of $p_\alpha$ from this factorisation, expand the product using the [Vieta formulas](/pages/???). The elementary symmetric polynomials in $r$ variables are defined by \begin{align*} e_0 &= 1, \\ e_1(y_1, \ldots, y_r) &= y_1 + y_2 + \cdots + y_r, \\ e_2(y_1, \ldots, y_r) &= \sum_{i < j} y_i y_j, \\ &\vdots \\ e_r(y_1, \ldots, y_r) &= y_1 y_2 \cdots y_r. \end{align*} Expanding the product $\prod_{i=1}^r (x - \alpha_i)$ by distributivity: each term corresponds to a choice, for each factor $(x - \alpha_i)$, of either $x$ or $-\alpha_i$. If we choose $-\alpha_i$ for a subset $S \subseteq \{1, \ldots, r\}$ of size $k$, and $x$ for the rest, the resulting summand is $\left(\prod_{i \in S} -\alpha_i\right) \cdot x^{r-k} = (-1)^k \left(\prod_{i \in S} \alpha_i\right) x^{r-k}$. Summing over all subsets $S$ of size $k$ and grouping by $k$: \begin{align*} \prod_{i=1}^r (x - \alpha_i) = \sum_{k=0}^r (-1)^k e_k(\alpha_1, \ldots, \alpha_r) \, x^{r-k}. \end{align*} So the coefficient of $x^{r-k}$ in $p_\alpha$ is $(-1)^k e_k(\alpha_1, \ldots, \alpha_r)$. Now apply the [Ring of Integers is a Ring](/theorems/1567) theorem: $\mathcal{O}_M$ is a ring (specifically, a subring of $M$), hence closed under addition, subtraction, and multiplication. We verify the hypotheses of the theorem: $M$ is a number field (a finite extension of $\mathbb{Q}$, which it is by construction — $M$ is a splitting field of a polynomial over $\mathbb{Q}$, hence a finite algebraic extension of $\mathbb{Q}$). The conclusion is that $\mathcal{O}_M$ is a subring of $M$. Each $\alpha_i \in \mathcal{O}_M$. The expression \begin{align*} e_k(\alpha_1, \ldots, \alpha_r) = \sum_{\substack{S \subseteq \{1, \ldots, r\} \\ |S| = k}} \prod_{i \in S} \alpha_i \end{align*} is built from the elements $\alpha_i \in \mathcal{O}_M$ by finitely many additions and multiplications. Since $\mathcal{O}_M$ is closed under these operations, $e_k(\alpha_1, \ldots, \alpha_r) \in \mathcal{O}_M$. Multiplying by $(-1)^k \in \mathbb{Z} \subseteq \mathcal{O}_M$ (the inclusion is because $\mathcal{O}_\mathbb{Q} = \mathbb{Z}$, and $\mathbb{Z} \subseteq M$ is obvious; or more directly, any integer is a root of a monic integer polynomial, e.g., $x - n$), the product stays in $\mathcal{O}_M$. Therefore every coefficient of $p_\alpha$ lies in $\mathcal{O}_M$. The leading coefficient, which is $(-1)^0 e_0 = 1$, also lies in $\mathcal{O}_M$ since $1 \in \mathbb{Z} \subseteq \mathcal{O}_M$.[/guided]

[step:Conclude that the coefficients of $p_\alpha$ lie in $\mathcal{O}_M \cap \mathbb{Q} = \mathbb{Z}$]The coefficients of $p_\alpha$ lie in $\mathcal{O}_M$ by the previous step. They also lie in $\mathbb{Q}$ by the very definition of the minimal polynomial: $p_\alpha \in \mathbb{Q}[x]$. Therefore each coefficient of $p_\alpha$ lies in the intersection $\mathcal{O}_M \cap \mathbb{Q}$. By the [Algebraic Integers in $\mathbb{Q}$](/theorems/1564) theorem, $\mathcal{O}_\mathbb{Q} = \mathbb{Z}$, i.e., the rational numbers that are integral over $\mathbb{Z}$ are exactly the rational integers. We verify the hypotheses: the theorem is an equality of two subsets of $\mathbb{Q}$, and we apply it as the chain \begin{align*} \mathcal{O}_M \cap \mathbb{Q} = \{\beta \in \mathbb{Q} : \beta \text{ is integral over } \mathbb{Z}\} = \mathcal{O}_\mathbb{Q} = \mathbb{Z}. \end{align*} The first equality uses that an element $\beta \in \mathbb{Q} \subseteq M$ lies in $\mathcal{O}_M$ precisely when it is integral over $\mathbb{Z}$ (the definition is independent of the ambient field $M$, provided $\beta \in M$). The second equality is by definition of $\mathcal{O}_\mathbb{Q}$. The third equality is [Algebraic Integers in $\mathbb{Q}$](/theorems/1564). Hence every coefficient of $p_\alpha$ is in $\mathbb{Z}$, i.e., $p_\alpha \in \mathbb{Z}[x]$. This completes the proof of the ($\Rightarrow$) direction.[/step]

[guided]We have the coefficients of $p_\alpha$ satisfying two simultaneous constraints: - They lie in $\mathcal{O}_M$ (from the symmetric-function argument of the previous step). - They lie in $\mathbb{Q}$ (because, by definition, $p_\alpha \in \mathbb{Q}[x]$: the minimal polynomial of $\alpha$ over $\mathbb{Q}$ has rational coefficients). So each coefficient lives in $\mathcal{O}_M \cap \mathbb{Q}$. We claim this intersection is exactly $\mathbb{Z}$. First, by definition, an element $\beta \in M$ is in $\mathcal{O}_M$ iff it is integral over $\mathbb{Z}$. This integrality condition is intrinsic to $\beta$ — it does not depend on which ambient number field we consider $\beta$ to live in. So if $\beta \in \mathbb{Q} \subseteq M$, then $\beta \in \mathcal{O}_M$ iff $\beta$ is integral over $\mathbb{Z}$ iff $\beta \in \mathcal{O}_\mathbb{Q}$. Therefore: \begin{align*} \mathcal{O}_M \cap \mathbb{Q} = \mathcal{O}_\mathbb{Q}. \end{align*} Second, the [Algebraic Integers in $\mathbb{Q}$](/theorems/1564) theorem asserts $\mathcal{O}_\mathbb{Q} = \mathbb{Z}$. We verify the hypotheses: the theorem is a specific identity about $\mathbb{Q}$ — no further hypotheses needed. Combining: \begin{align*} \mathcal{O}_M \cap \mathbb{Q} = \mathcal{O}_\mathbb{Q} = \mathbb{Z}. \end{align*} So every coefficient of $p_\alpha$ lies in $\mathbb{Z}$. That is exactly the statement $p_\alpha \in \mathbb{Z}[x]$, which is what we needed. Combining this with the easy ($\Leftarrow$) direction established in the first step, we have the full if-and-only-if: \begin{align*} \alpha \in \mathcal{O}_L \iff p_\alpha \in \mathbb{Z}[x]. \end{align*} This completes the proof. A structural observation: the proof transfers information about $\alpha$ (an element of a number field $L$) to information about all the roots $\alpha_i$ of its minimal polynomial (elements of a splitting field $M$, possibly larger than $L$). The [Galois](/pages/???) philosophy is that the algebraic properties of $\alpha$ are shared by all its conjugates — here, specifically, integrality is a "Galois-invariant" property: if one root of an irreducible polynomial over $\mathbb{Q}$ is an algebraic integer, all roots are. The coefficients of $p_\alpha$, being symmetric in the roots, then inherit integrality automatically.[/guided]