[proofplan] Divide $f$ by the minimal polynomial $p_\alpha$ in the polynomial ring $K[x]$, obtaining $f = p_\alpha h + r$ with remainder $r$ of strictly smaller degree than $p_\alpha$. Evaluate at $\alpha$: since $f(\alpha) = 0$ and $p_\alpha(\alpha) = 0$, the residue satisfies $r(\alpha) = 0$. If $r$ were nonzero, we could scale it to a monic polynomial of degree $< \deg p_\alpha$ killing $\alpha$, contradicting the minimality of $\deg p_\alpha$. Hence $r = 0$ and $p_\alpha \mid f$. [/proofplan] [step:Apply the polynomial division algorithm in $K[x]$ to write $f = p_\alpha h + r$ with $\deg r < \deg p_\alpha$] The ring $K[x]$ is a [Euclidean domain](/pages/???) with respect to the degree function: for any $f \in K[x]$ and any nonzero $g \in K[x]$, there exist unique $h, r \in K[x]$ with \begin{align*} f = g h + r, \qquad \text{and either } r = 0 \text{ or } \deg r < \deg g. \end{align*} The uniqueness uses that the leading coefficient of $g$ is a unit in $K$ — which holds because $K$ is a field, so every nonzero element of $K$ is a unit. We apply this to our situation: $g := p_\alpha$ is a nonzero element of $K[x]$ (it is monic, hence has leading coefficient $1 \neq 0$), so there exist $h, r \in K[x]$ with \begin{align*} f = p_\alpha h + r, \qquad \text{and either } r = 0 \text{ or } \deg r < \deg p_\alpha. \tag{$\dagger$} \end{align*} [guided] To detect divisibility $p_\alpha \mid f$, the most direct route in the polynomial ring $K[x]$ is the division algorithm: given any dividend and a nonzero divisor, we can perform long division and obtain a quotient and a remainder of strictly smaller degree. Our aim is to show the remainder is zero, which is equivalent to divisibility. The division algorithm is valid because $K[x]$ is a [Euclidean domain](/pages/???): the degree function $\deg: K[x] \setminus \{0\} \to \mathbb{Z}_{\geq 0}$ makes $K[x]$ Euclidean, meaning for any $f \in K[x]$ and any nonzero $g \in K[x]$ there exist unique $h, r \in K[x]$ with $f = g h + r$ and either $r = 0$ or $\deg r < \deg g$. The key hypothesis for this construction is that the leading coefficient of $g$ is invertible in $K$ — which is automatic because $K$ is a field and every nonzero element of a field is a unit. We verify the hypotheses of the division algorithm: the divisor $p_\alpha$ is nonzero because it is monic (its leading coefficient is $1 \in K^\times$). Hence, applying the division algorithm with $g = p_\alpha$: \begin{align*} f = p_\alpha h + r, \qquad \text{and either } r = 0 \text{ or } \deg r < \deg p_\alpha. \end{align*} We now have two polynomials $h, r \in K[x]$ satisfying this identity; our task reduces to showing $r = 0$. [/guided] [/step] [step:Evaluate $(\dagger)$ at $x = \alpha$ to deduce $r(\alpha) = 0$] The evaluation map \begin{align*} \operatorname{ev}_\alpha: K[x] &\to L \\ g(x) &\mapsto g(\alpha) \end{align*} is a ring homomorphism, where $L$ is the field extension in which $\alpha$ lives (or any $K$-algebra containing $\alpha$). Applying $\operatorname{ev}_\alpha$ to both sides of $(\dagger)$: \begin{align*} f(\alpha) = p_\alpha(\alpha) \cdot h(\alpha) + r(\alpha). \end{align*} By hypothesis $f(\alpha) = 0$, and by definition of the minimal polynomial $p_\alpha(\alpha) = 0$, so \begin{align*} 0 = 0 \cdot h(\alpha) + r(\alpha) = r(\alpha). \tag{$\ddagger$} \end{align*} [guided] The evaluation map $\operatorname{ev}_\alpha: K[x] \to L$, $g(x) \mapsto g(\alpha)$, is a ring homomorphism because $L$ is a commutative ring containing $K$ and a single element $\alpha$, and evaluation preserves sums and products (this is the [universal property of the polynomial ring](/pages/???): $K[x]$ is the free commutative $K$-algebra on one generator $x$, so $K$-algebra homomorphisms out of $K[x]$ correspond bijectively to choices of image for $x$). Applying $\operatorname{ev}_\alpha$ to $(\dagger)$: \begin{align*} f(\alpha) = p_\alpha(\alpha) \cdot h(\alpha) + r(\alpha). \end{align*} Now we consume the two hypotheses we have about $\alpha$: - $f(\alpha) = 0$ is the hypothesis of the theorem. - $p_\alpha(\alpha) = 0$ by definition of the minimal polynomial: $p_\alpha$ is the (unique) monic polynomial of least degree in $K[x]$ with $p_\alpha(\alpha) = 0$. Substituting both gives $0 = 0 \cdot h(\alpha) + r(\alpha)$, so $r(\alpha) = 0$. The remainder polynomial therefore vanishes at $\alpha$. [/guided] [/step] [step:Rule out $r \neq 0$ by minimality of $\deg p_\alpha$] We show $r = 0$. Suppose for contradiction $r \neq 0$. Let $c \in K^\times$ be the leading coefficient of $r$ (nonzero because $r \neq 0$, and a unit in $K$ because $K$ is a field). Define \begin{align*} \tilde r := c^{-1} r \in K[x]. \end{align*} Then $\tilde r$ is monic (its leading coefficient is $c^{-1} \cdot c = 1$), satisfies $\deg \tilde r = \deg r < \deg p_\alpha$ (scaling by a unit preserves degree), and evaluating gives \begin{align*} \tilde r(\alpha) = c^{-1} \cdot r(\alpha) = c^{-1} \cdot 0 = 0. \end{align*} Thus $\tilde r \in K[x]$ is a monic polynomial of degree strictly less than $\deg p_\alpha$ with $\tilde r(\alpha) = 0$. But by definition $p_\alpha$ is the monic polynomial of *least* degree in $K[x]$ annihilating $\alpha$. This contradicts the minimality of $\deg p_\alpha$. Hence $r = 0$, and $(\dagger)$ becomes $f = p_\alpha h$, i.e., $p_\alpha \mid f$ in $K[x]$. [guided] We want $r = 0$. From $(\ddagger)$ we know $r(\alpha) = 0$; from $(\dagger)$ we know $r = 0$ or $\deg r < \deg p_\alpha$. Suppose for contradiction the second alternative holds, i.e., $r \neq 0$ and $\deg r < \deg p_\alpha$. To leverage this, we want to compare $r$ to $p_\alpha$ via the minimality property of $p_\alpha$. The catch: $p_\alpha$ is characterised as the monic polynomial of least degree killing $\alpha$, but $r$ need not be monic. We **normalise** by rescaling: let $c \in K^\times$ be the leading coefficient of $r$ (nonzero because $r \neq 0$, and a unit because we are in the field $K$). Set \begin{align*} \tilde r := c^{-1} r \in K[x]. \end{align*} Why is this well-defined? Because $c \in K^\times$ has a multiplicative inverse $c^{-1} \in K$ (this uses $K$ being a field — the division would fail over a general ring). Properties of $\tilde r$: - **Monic**: the leading coefficient of $\tilde r$ is $c^{-1} \cdot c = 1$. - **Degree**: $\deg \tilde r = \deg r$ because we scaled by a nonzero scalar, which does not change degree. - **Kills $\alpha$**: $\tilde r(\alpha) = c^{-1} r(\alpha) = c^{-1} \cdot 0 = 0$ (the evaluation map is $K$-linear). So $\tilde r \in K[x]$ is monic, annihilates $\alpha$, and has degree $\deg \tilde r = \deg r < \deg p_\alpha$. But the minimal polynomial $p_\alpha$ is, by definition, the monic polynomial in $K[x]$ of **least degree** with $p_\alpha(\alpha) = 0$. We have exhibited a monic polynomial $\tilde r$ killing $\alpha$ of strictly smaller degree. Contradiction. Therefore $r = 0$. Substituting into $(\dagger)$: \begin{align*} f = p_\alpha h + 0 = p_\alpha h, \end{align*} which is precisely the assertion $p_\alpha \mid f$ in $K[x]$. This completes the proof. An observation: the argument uses the *existence and uniqueness* of the minimal polynomial (so that "the" minimal polynomial is a meaningful object) together with the division algorithm in $K[x]$. The same argument goes through for the minimal polynomial of an operator over any field, the minimal polynomial of an element over any field extension — the structural ingredient is that $K[x]$ is a Euclidean domain and $p_\alpha$ is defined as a monic element of minimal degree satisfying a prescribed condition. [/guided] [/step]