[proofplan] The equivalence (1) $\Leftrightarrow$ (2) is the [Integrality Criterion via Minimal Polynomial](/theorems/1570). It remains to prove (2) $\Leftrightarrow$ (3), which links the minimal polynomial $p_\alpha$ of $\alpha$ over $\mathbb{Q}$ to the characteristic polynomial $\chi_\alpha(x) := \det(xI - m_\alpha)$ of the $\mathbb{Q}$-linear multiplication map $m_\alpha: L \to L$. The [Characteristic Polynomial via Minimal Polynomial](/theorems/1573) identifies these by $\chi_\alpha = p_\alpha^{[L:\mathbb{Q}(\alpha)]}$. From this factorisation both implications are transparent: $p_\alpha \in \mathbb{Z}[x]$ forces $\chi_\alpha \in \mathbb{Z}[x]$ (products of integer-coefficient polynomials stay integer-coefficient), and conversely if $\chi_\alpha \in \mathbb{Z}[x]$ then $\alpha$ satisfies a monic integer polynomial, so $\alpha \in \mathcal{O}_L$, whence (1) and (2) follow. The integrality of norm and trace is then immediate because both are, up to sign, coefficients of $\chi_\alpha \in \mathbb{Z}[x]$. [/proofplan] [step:Recall the standing objects and the prior equivalence of (1) and (2)] Let $L/\mathbb{Q}$ be a number field of degree $n := [L:\mathbb{Q}]$, let $\alpha \in L$, and let $p_\alpha \in \mathbb{Q}[x]$ denote the monic minimal polynomial of $\alpha$ over $\mathbb{Q}$. The multiplication-by-$\alpha$ map is the $\mathbb{Q}$-linear endomorphism \begin{align*} m_\alpha: L &\to L, \\ \beta &\mapsto \alpha \beta, \end{align*} and its characteristic polynomial is \begin{align*} \chi_\alpha(x) := \det(xI - m_\alpha) \in \mathbb{Q}[x], \end{align*} a monic polynomial of degree $n$. By the [Integrality Criterion via Minimal Polynomial](/theorems/1570), we have \begin{align*} \alpha \in \mathcal{O}_L \iff p_\alpha \in \mathbb{Z}[x], \end{align*} so (1) $\Leftrightarrow$ (2) is already established. It remains to show (2) $\Leftrightarrow$ (3), i.e. $p_\alpha \in \mathbb{Z}[x] \iff \chi_\alpha \in \mathbb{Z}[x]$. [guided] We set the stage. $L/\mathbb{Q}$ is a number field of degree $n := [L:\mathbb{Q}]$, $\alpha \in L$, and $p_\alpha \in \mathbb{Q}[x]$ is the monic minimal polynomial of $\alpha$ over $\mathbb{Q}$ — the unique monic polynomial of least positive degree in $\mathbb{Q}[x]$ with $p_\alpha(\alpha) = 0$. The **multiplication-by-$\alpha$** map \begin{align*} m_\alpha: L &\to L, \\ \beta &\mapsto \alpha \beta, \end{align*} is $\mathbb{Q}$-linear: it respects addition and $\mathbb{Q}$-scaling because $L$ is a $\mathbb{Q}$-algebra. Since $L$ is a finite-dimensional $\mathbb{Q}$-vector space of dimension $n$, the endomorphism $m_\alpha$ has a well-defined characteristic polynomial \begin{align*} \chi_\alpha(x) := \det(xI - m_\alpha) \in \mathbb{Q}[x], \end{align*} which is monic of degree $n$ (the determinant of $xI$ minus an $n \times n$ matrix contains the leading $x^n$ with coefficient $1$). Among the three conditions in the theorem, the equivalence (1) $\Leftrightarrow$ (2) — "$\alpha$ is an algebraic integer iff $p_\alpha$ has integer coefficients" — was established in the [Integrality Criterion via Minimal Polynomial](/theorems/1570). The task of this proof is therefore the equivalence (2) $\Leftrightarrow$ (3). Once that is proved, by transitivity we obtain the full three-way equivalence. [/guided] [/step] [step:Invoke [Characteristic Polynomial via Minimal Polynomial](/theorems/1573) to relate $\chi_\alpha$ and $p_\alpha$] Let $r := [L:\mathbb{Q}(\alpha)]$, so that $n = r \cdot \deg p_\alpha$ by the [tower law](/theorems/???). We apply [Characteristic Polynomial via Minimal Polynomial](/theorems/1573) with base field $K = \mathbb{Q}$, extension $L/\mathbb{Q}$, and element $\alpha \in L$. The hypotheses required are: 1. **$L/K$ is a finite field extension**: $L/\mathbb{Q}$ is a number field, hence finite over $\mathbb{Q}$ by definition. 2. **$\alpha \in L$ and $p_\alpha$ is its minimal polynomial over $K = \mathbb{Q}$**: these are our standing assumptions. Both hold. The theorem concludes \begin{align*} \chi_\alpha(x) = \det(xI - m_\alpha) = p_\alpha(x)^r. \tag{$\ast$} \end{align*} [guided] We need a precise relation between $\chi_\alpha$ and $p_\alpha$ if we are to transfer integer-coefficient information between them. The tool is [Characteristic Polynomial via Minimal Polynomial](/theorems/1573), which states: for a finite extension $L/K$ and $\alpha \in L$ with minimal polynomial $p_\alpha \in K[x]$, the characteristic polynomial of the $K$-linear map $m_\alpha: L \to L$ equals $p_\alpha^{[L:K(\alpha)]}$. We apply this with $K = \mathbb{Q}$. The hypotheses are: - **$L/\mathbb{Q}$ is a finite extension**: yes, $[L:\mathbb{Q}] = n < \infty$ because $L$ is a number field (finite-degree extension of $\mathbb{Q}$). - **$p_\alpha \in \mathbb{Q}[x]$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$**: this is our standing notation. Setting $r := [L:\mathbb{Q}(\alpha)]$, the theorem concludes \begin{align*} \chi_\alpha(x) = p_\alpha(x)^r. \end{align*} Note for the sanity check: the [tower law](/theorems/???) gives $n = [L:\mathbb{Q}] = [L:\mathbb{Q}(\alpha)] \cdot [\mathbb{Q}(\alpha):\mathbb{Q}] = r \cdot \deg p_\alpha$, and indeed $\deg \chi_\alpha = n = r \cdot \deg p_\alpha = \deg(p_\alpha^r)$, consistent with $(\ast)$. This identity is the whole content: knowledge of $p_\alpha$ determines $\chi_\alpha$ and conversely (as we shall see below, the factorisation can be inverted in $\mathbb{Z}[x]$). [/guided] [/step] [step:Deduce $p_\alpha \in \mathbb{Z}[x] \implies \chi_\alpha \in \mathbb{Z}[x]$] Suppose $p_\alpha \in \mathbb{Z}[x]$. The ring $\mathbb{Z}[x]$ is closed under multiplication (it is a commutative ring). Hence the $r$-fold product \begin{align*} \chi_\alpha = p_\alpha^r = \underbrace{p_\alpha \cdot p_\alpha \cdots p_\alpha}_{r \text{ factors}} \end{align*} also lies in $\mathbb{Z}[x]$, i.e. $\chi_\alpha \in \mathbb{Z}[x]$. [guided] This direction is the easy one. Suppose $p_\alpha \in \mathbb{Z}[x]$. The polynomial ring $\mathbb{Z}[x]$ is a commutative ring (more specifically, a commutative $\mathbb{Z}$-algebra), so it is closed under finite products. The identity $(\ast)$ expresses $\chi_\alpha$ as a product of $r$ copies of $p_\alpha$: \begin{align*} \chi_\alpha = p_\alpha^r = \underbrace{p_\alpha \cdot p_\alpha \cdots p_\alpha}_{r \text{ factors}}. \end{align*} Each factor has integer coefficients, so the product has integer coefficients: $\chi_\alpha \in \mathbb{Z}[x]$. [/guided] [/step] [step:Deduce $\chi_\alpha \in \mathbb{Z}[x] \implies \alpha \in \mathcal{O}_L$, hence $p_\alpha \in \mathbb{Z}[x]$] Suppose $\chi_\alpha \in \mathbb{Z}[x]$. We show $\alpha \in \mathcal{O}_L$; combined with the [Integrality Criterion via Minimal Polynomial](/theorems/1570), this will give $p_\alpha \in \mathbb{Z}[x]$. By the [Cayley-Hamilton theorem](/theorems/???) applied to the $\mathbb{Q}$-linear endomorphism $m_\alpha: L \to L$ of the finite-dimensional $\mathbb{Q}$-vector space $L$, \begin{align*} \chi_\alpha(m_\alpha) = 0 \quad \text{as an endomorphism of } L. \end{align*} Evaluating this zero endomorphism on the identity element $1 \in L$ and using $m_\alpha^k(1) = \alpha^k$ (by induction on $k$: $m_\alpha^0(1) = 1 = \alpha^0$ and $m_\alpha^{k+1}(1) = m_\alpha(\alpha^k) = \alpha \cdot \alpha^k = \alpha^{k+1}$), we obtain \begin{align*} \chi_\alpha(\alpha) = \chi_\alpha(m_\alpha)(1) = 0. \end{align*} Since $\chi_\alpha \in \mathbb{Z}[x]$ is monic of degree $n$ with $\chi_\alpha(\alpha) = 0$, the element $\alpha$ satisfies a monic integer polynomial, so by definition $\alpha \in \mathcal{O}_L$. Now applying the [Integrality Criterion via Minimal Polynomial](/theorems/1570) in the direction $\alpha \in \mathcal{O}_L \implies p_\alpha \in \mathbb{Z}[x]$ completes this direction. [guided] Now the converse: assume $\chi_\alpha \in \mathbb{Z}[x]$ and show $p_\alpha \in \mathbb{Z}[x]$. The strategy is indirect: we first use $\chi_\alpha \in \mathbb{Z}[x]$ to show $\alpha \in \mathcal{O}_L$ (via the definition of algebraic integer), and then invoke the [Integrality Criterion via Minimal Polynomial](/theorems/1570) to deduce $p_\alpha \in \mathbb{Z}[x]$. **Step A: $\chi_\alpha(\alpha) = 0$.** This is the Cayley-Hamilton theorem applied to the multiplication map. The [Cayley-Hamilton theorem](/theorems/???) asserts: for any endomorphism $T: V \to V$ of a finite-dimensional vector space $V$ over a field $K$, the characteristic polynomial $\chi_T(x) = \det(xI - T) \in K[x]$ satisfies $\chi_T(T) = 0$ (the zero endomorphism of $V$). We verify its hypotheses for $T = m_\alpha$: $L$ is a finite-dimensional $\mathbb{Q}$-vector space (dimension $n$) and $m_\alpha$ is a $\mathbb{Q}$-linear endomorphism. Hence \begin{align*} \chi_\alpha(m_\alpha) = 0 \quad \text{as an endomorphism of } L. \end{align*} Evaluating both sides on the element $1 \in L$ (which exists because $L$ is a field): the right side gives $0 \in L$; the left side gives the polynomial $\chi_\alpha$ evaluated at $m_\alpha$, then applied to $1$. We compute: \begin{align*} \chi_\alpha(m_\alpha)(1) = \sum_{k=0}^n c_k\, m_\alpha^k(1), \qquad \text{where } \chi_\alpha(x) = \sum_{k=0}^n c_k x^k. \end{align*} A routine induction on $k$ shows $m_\alpha^k(1) = \alpha^k$: the base case is $m_\alpha^0(1) = 1 = \alpha^0$, and the inductive step is \begin{align*} m_\alpha^{k+1}(1) = m_\alpha(m_\alpha^k(1)) = m_\alpha(\alpha^k) = \alpha \cdot \alpha^k = \alpha^{k+1}. \end{align*} Substituting: \begin{align*} 0 = \chi_\alpha(m_\alpha)(1) = \sum_{k=0}^n c_k \alpha^k = \chi_\alpha(\alpha) \quad \text{in } L. \end{align*} **Step B: $\alpha \in \mathcal{O}_L$.** The polynomial $\chi_\alpha \in \mathbb{Z}[x]$ is monic of degree $n$ with $\chi_\alpha(\alpha) = 0$. By definition, an element of $L$ is in $\mathcal{O}_L$ precisely when it satisfies a monic polynomial with integer coefficients. We have exhibited such a polynomial, so $\alpha \in \mathcal{O}_L$. **Step C: $p_\alpha \in \mathbb{Z}[x]$.** Now apply the [Integrality Criterion via Minimal Polynomial](/theorems/1570), which states: $\alpha \in \mathcal{O}_L \iff p_\alpha \in \mathbb{Z}[x]$. Its hypotheses are exactly "$L$ is a number field and $p_\alpha$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$" — both standing. Direction ($\Rightarrow$) of that criterion, applied to $\alpha \in \mathcal{O}_L$ from Step B, gives $p_\alpha \in \mathbb{Z}[x]$. This completes the proof of (2) $\Leftrightarrow$ (3). Combined with the known equivalence (1) $\Leftrightarrow$ (2) from the [Integrality Criterion via Minimal Polynomial](/theorems/1570), all three conditions are equivalent. A remark on structure: the proof is asymmetric. The forward direction $p_\alpha \in \mathbb{Z}[x] \Rightarrow \chi_\alpha \in \mathbb{Z}[x]$ uses only that $\mathbb{Z}[x]$ is a ring, together with $(\ast)$. The reverse direction does *not* go through "factor $\chi_\alpha$ as $p_\alpha^r$ in $\mathbb{Z}[x]$" — in principle that would work via [Gauss's Lemma](/theorems/???), but it is easier to go through the definition of algebraic integer: Cayley-Hamilton gives a monic integer annihilating polynomial for $\alpha$ directly from $\chi_\alpha$, so $\alpha \in \mathcal{O}_L$ and then $p_\alpha \in \mathbb{Z}[x]$ follows from the known equivalence (1) $\Leftrightarrow$ (2). [/guided] [/step] [step:Deduce $N_{L/\mathbb{Q}}(\alpha), \operatorname{tr}_{L/\mathbb{Q}}(\alpha) \in \mathbb{Z}$ when $\alpha \in \mathcal{O}_L$] By the definition of norm and trace as coefficients of the characteristic polynomial: writing \begin{align*} \chi_\alpha(x) = x^n + c_{n-1} x^{n-1} + \cdots + c_1 x + c_0, \end{align*} we have \begin{align*} \operatorname{tr}_{L/\mathbb{Q}}(\alpha) = -c_{n-1}, \qquad N_{L/\mathbb{Q}}(\alpha) = (-1)^n c_0. \end{align*} Assume $\alpha \in \mathcal{O}_L$. By the equivalence (1) $\Leftrightarrow$ (3) just established, $\chi_\alpha \in \mathbb{Z}[x]$, so every coefficient $c_i \in \mathbb{Z}$. In particular $c_{n-1}, c_0 \in \mathbb{Z}$, and hence \begin{align*} \operatorname{tr}_{L/\mathbb{Q}}(\alpha) = -c_{n-1} \in \mathbb{Z}, \qquad N_{L/\mathbb{Q}}(\alpha) = (-1)^n c_0 \in \mathbb{Z}, \end{align*} since $\mathbb{Z}$ is closed under negation and multiplication by $\pm 1$. [guided] We now harvest the corollary. The trace and norm of an element $\alpha \in L$ with respect to the extension $L/\mathbb{Q}$ are defined as the trace and determinant of the $\mathbb{Q}$-linear endomorphism $m_\alpha: L \to L$: \begin{align*} \operatorname{tr}_{L/\mathbb{Q}}(\alpha) := \operatorname{tr}(m_\alpha), \qquad N_{L/\mathbb{Q}}(\alpha) := \det(m_\alpha). \end{align*} By the relation between a characteristic polynomial and its coefficients — valid for any endomorphism of an $n$-dimensional vector space — if \begin{align*} \chi_\alpha(x) = \det(xI - m_\alpha) = x^n + c_{n-1} x^{n-1} + \cdots + c_1 x + c_0, \end{align*} then the trace is $-c_{n-1}$ (negative of the coefficient of $x^{n-1}$) and the determinant is $(-1)^n c_0$ (up to the sign $(-1)^n$, the constant term). Explicitly, $c_{n-1} = -\operatorname{tr}(m_\alpha)$ and $c_0 = \det(-m_\alpha) = (-1)^n \det(m_\alpha)$. Now assume the hypothesis $\alpha \in \mathcal{O}_L$. By the three-way equivalence just proved, $\chi_\alpha \in \mathbb{Z}[x]$, meaning every coefficient $c_i$ is an integer. In particular $c_{n-1}, c_0 \in \mathbb{Z}$. Since $\mathbb{Z}$ is closed under negation and under multiplication by units $\pm 1$: \begin{align*} \operatorname{tr}_{L/\mathbb{Q}}(\alpha) = -c_{n-1} \in \mathbb{Z}, \qquad N_{L/\mathbb{Q}}(\alpha) = (-1)^n c_0 \in \mathbb{Z}. \end{align*} This concludes the proof of the entire theorem. A final observation: in fact all $n$ coefficients of $\chi_\alpha$ are elementary symmetric functions of the conjugates of $\alpha$ (as we will see in the [Norm and Trace via Embeddings](/theorems/1577) theorem), and the argument above shows that *all* elementary symmetric functions of the conjugates of an algebraic integer are integers. The trace and norm are merely the "first" and "last" of these. [/guided] [/step]