[proofplan] We construct a bijection between $\operatorname{Hom}_K(K(\alpha), E)$ and $\operatorname{Root}_{P_\alpha}(E)$ by sending each $K$-homomorphism $\sigma$ to $\sigma(\alpha)$. The argument has three parts. First, we verify that $\sigma(\alpha)$ is indeed a root of $P_\alpha$ in $E$, using the fact that $\sigma$ fixes $K$ and $P_\alpha$ has coefficients in $K$. Second, we prove injectivity by showing that every element of $K(\alpha)$ is a polynomial expression in $\alpha$ over $K$, so $\sigma$ is determined by $\sigma(\alpha)$. Third, given any root $\beta \in \operatorname{Root}_{P_\alpha}(E)$, we construct a $K$-homomorphism $\sigma_\beta \colon K(\alpha) \to E$ with $\sigma_\beta(\alpha) = \beta$ by defining $\sigma_\beta(g(\alpha)) = g(\beta)$ and verifying well-definedness via divisibility by the minimal polynomial. The cardinality bound $|\operatorname{Hom}_K(K(\alpha), E)| \le \deg P_\alpha$ then follows because $P_\alpha$ has at most $\deg P_\alpha$ roots in $E$. [/proofplan] [step:Show that $\sigma(\alpha)$ is a root of $P_\alpha$ for every $K$-homomorphism $\sigma$] Let $\sigma \in \operatorname{Hom}_K(K(\alpha), E)$. Write the minimal polynomial as \begin{align*} P_\alpha = \sum_{k=0}^{d} c_k t^k \in K[t], \quad d := \deg P_\alpha, \end{align*} where $c_0, \ldots, c_d \in K$ and $c_d = 1$ (since minimal polynomials are monic). Since $\sigma$ is a ring homomorphism that fixes $K$ pointwise, we have $\sigma(c_k) = c_k$ for every $k \in \{0, \ldots, d\}$. Applying $\sigma$ to the relation $P_\alpha(\alpha) = 0$: \begin{align*} 0 = \sigma(0) = \sigma\!\left(\sum_{k=0}^{d} c_k \alpha^k\right) = \sum_{k=0}^{d} \sigma(c_k) \, \sigma(\alpha)^k = \sum_{k=0}^{d} c_k \, \sigma(\alpha)^k = P_\alpha(\sigma(\alpha)). \end{align*} The second equality uses the fact that $\sigma$ is a ring homomorphism (preserving sums and products), and the third equality uses $\sigma(c_k) = c_k$. Therefore $\sigma(\alpha) \in \operatorname{Root}_{P_\alpha}(E)$, so the map \begin{align*} \Phi \colon \operatorname{Hom}_K(K(\alpha), E) &\to \operatorname{Root}_{P_\alpha}(E) \\ \sigma &\mapsto \sigma(\alpha) \end{align*} is well-defined. [guided] We need to verify that $\Phi$ actually lands in $\operatorname{Root}_{P_\alpha}(E)$, i.e., that $\sigma(\alpha)$ is a root of $P_\alpha$ in $E$. The key property we use is that $\sigma$ is a $K$-homomorphism: it is a ring homomorphism $K(\alpha) \to E$ that restricts to the identity on $K$. What does "fixes $K$" buy us? The coefficients $c_0, \ldots, c_d$ of $P_\alpha$ all lie in $K$, so $\sigma$ passes through them unchanged. This means $\sigma$ commutes with evaluating any polynomial with $K$-coefficients: for any $g = \sum_k c_k t^k \in K[t]$, \begin{align*} \sigma(g(\alpha)) = \sigma\!\left(\sum_{k=0}^{d} c_k \alpha^k\right) = \sum_{k=0}^{d} c_k \, \sigma(\alpha)^k = g(\sigma(\alpha)). \end{align*} The first equality is the definition of evaluating $g$ at $\alpha$. The second uses that $\sigma$ preserves addition and multiplication (ring homomorphism) together with $\sigma(c_k) = c_k$ (fixes $K$). The third is the definition of evaluating $g$ at $\sigma(\alpha)$. Applying this to $g = P_\alpha$ and using $P_\alpha(\alpha) = 0$ (since $P_\alpha$ is the minimal polynomial of $\alpha$ over $K$): \begin{align*} P_\alpha(\sigma(\alpha)) = \sigma(P_\alpha(\alpha)) = \sigma(0) = 0. \end{align*} Hence $\sigma(\alpha) \in \operatorname{Root}_{P_\alpha}(E)$. Note how this argument would fail if $\sigma$ did not fix $K$: if $\sigma(c_k) \neq c_k$ for some coefficient, the polynomial that $\sigma(\alpha)$ satisfies would be $\sum_k \sigma(c_k) t^k$, which is a different polynomial from $P_\alpha$. [/guided] [/step] [step:Prove that $\Phi$ is injective by showing $\sigma$ is determined by $\sigma(\alpha)$] Let $\sigma_1, \sigma_2 \in \operatorname{Hom}_K(K(\alpha), E)$ with $\sigma_1(\alpha) = \sigma_2(\alpha)$. Since $\alpha$ is algebraic over $K$ with minimal polynomial $P_\alpha$ of degree $d$, every element of $K(\alpha)$ can be written uniquely as \begin{align*} g(\alpha) = \sum_{k=0}^{d-1} a_k \alpha^k \end{align*} for some $a_0, \ldots, a_{d-1} \in K$. (This follows from the isomorphism $K(\alpha) \cong K[t]/(P_\alpha)$, which identifies $K(\alpha)$ with the $K$-vector space of polynomials of degree at most $d - 1$.) For any such element: \begin{align*} \sigma_1(g(\alpha)) = \sum_{k=0}^{d-1} a_k \, \sigma_1(\alpha)^k = \sum_{k=0}^{d-1} a_k \, \sigma_2(\alpha)^k = \sigma_2(g(\alpha)), \end{align*} where the first and third equalities use the fact that $\sigma_1$ and $\sigma_2$ are $K$-homomorphisms (preserving sums, products, and fixing $K$-coefficients), and the second equality uses the hypothesis $\sigma_1(\alpha) = \sigma_2(\alpha)$. Since every element of $K(\alpha)$ has this form, $\sigma_1 = \sigma_2$ as maps $K(\alpha) \to E$. Therefore $\Phi$ is injective. [guided] Why is $\sigma$ determined by $\sigma(\alpha)$? The essential point is that $K(\alpha)$ is generated by $\alpha$ as a $K$-algebra. More precisely, the structure theorem for simple algebraic extensions gives $K(\alpha) \cong K[t]/(P_\alpha)$, so $\{1, \alpha, \alpha^2, \ldots, \alpha^{d-1}\}$ is a $K$-basis of $K(\alpha)$. Every element of $K(\alpha)$ is therefore a $K$-linear combination of powers of $\alpha$. A $K$-homomorphism $\sigma$ preserves addition, multiplication, and every element of $K$. Once we know $\sigma(\alpha)$, the value of $\sigma$ on any basis element $\alpha^k$ is forced: $\sigma(\alpha^k) = \sigma(\alpha)^k$ (since $\sigma$ preserves multiplication). The value on a general element $\sum_k a_k \alpha^k$ is then forced by additivity and the condition $\sigma(a_k) = a_k$. This argument reveals a broader principle: a $K$-algebra homomorphism from a finitely generated $K$-algebra is determined by its values on the generators. Here $K(\alpha)$ is generated by a single element $\alpha$, so a single value $\sigma(\alpha)$ determines $\sigma$ completely. What if $\alpha$ were transcendental? Then $K(\alpha) \cong K(t)$, the field of rational functions, which is not a finite-dimensional $K$-vector space. A $K$-homomorphism from $K(t)$ into $E$ is still determined by the image of $t$, but that image is no longer constrained to be a root of any polynomial — it can be any element of $E$. The algebraicity of $\alpha$ is what restricts the image to roots of $P_\alpha$. [/guided] [/step] [step:Construct a $K$-homomorphism from each root $\beta \in \operatorname{Root}_{P_\alpha}(E)$ to show surjectivity] Let $\beta \in \operatorname{Root}_{P_\alpha}(E)$. We construct $\sigma_\beta \in \operatorname{Hom}_K(K(\alpha), E)$ with $\sigma_\beta(\alpha) = \beta$. Since $K(\alpha) \cong K[t]/(P_\alpha)$, every element of $K(\alpha)$ has a unique representative as $g(\alpha)$ for some $g \in K[t]$ with $\deg g < d$. Define \begin{align*} \sigma_\beta \colon K(\alpha) &\to E \\ g(\alpha) &\mapsto g(\beta). \end{align*} [claim:Well-definedness of $\sigma_\beta$] The map $\sigma_\beta$ is well-defined: if $g(\alpha) = h(\alpha)$ for $g, h \in K[t]$, then $g(\beta) = h(\beta)$. [/claim] [proof] Suppose $g, h \in K[t]$ satisfy $g(\alpha) = h(\alpha)$. Then $(g - h)(\alpha) = 0$, so $\alpha$ is a root of $g - h \in K[t]$. Since $P_\alpha$ is the minimal polynomial of $\alpha$ over $K$, and the minimal polynomial divides every polynomial in $K[t]$ that vanishes at $\alpha$, there exists $q \in K[t]$ such that \begin{align*} g - h = q \cdot P_\alpha. \end{align*} Evaluating at $\beta$: \begin{align*} (g - h)(\beta) = q(\beta) \cdot P_\alpha(\beta) = q(\beta) \cdot 0 = 0, \end{align*} where $P_\alpha(\beta) = 0$ because $\beta \in \operatorname{Root}_{P_\alpha}(E)$. Therefore $g(\beta) = h(\beta)$. [/proof] We verify that $\sigma_\beta$ is a $K$-homomorphism. For any $g_1, g_2 \in K[t]$: \begin{align*} \sigma_\beta(g_1(\alpha) + g_2(\alpha)) &= \sigma_\beta((g_1 + g_2)(\alpha)) = (g_1 + g_2)(\beta) = g_1(\beta) + g_2(\beta) = \sigma_\beta(g_1(\alpha)) + \sigma_\beta(g_2(\alpha)), \\ \sigma_\beta(g_1(\alpha) \cdot g_2(\alpha)) &= \sigma_\beta((g_1 \cdot g_2)(\alpha)) = (g_1 \cdot g_2)(\beta) = g_1(\beta) \cdot g_2(\beta) = \sigma_\beta(g_1(\alpha)) \cdot \sigma_\beta(g_2(\alpha)). \end{align*} The first equality in each line uses the fact that polynomial addition and multiplication commute with evaluation (i.e., the evaluation map $K[t] \to K(\alpha)$, $g \mapsto g(\alpha)$ is a ring homomorphism). Moreover, for any $c \in K$, the constant polynomial $g(t) = c$ satisfies $g(\alpha) = c$ and $g(\beta) = c$, so $\sigma_\beta(c) = c$. Thus $\sigma_\beta$ fixes $K$ pointwise. Finally, $\sigma_\beta(\alpha) = \beta$ since $\alpha$ corresponds to the polynomial $g(t) = t$, and $g(\beta) = \beta$. Therefore $\sigma_\beta \in \operatorname{Hom}_K(K(\alpha), E)$ and $\Phi(\sigma_\beta) = \sigma_\beta(\alpha) = \beta$. Since $\beta \in \operatorname{Root}_{P_\alpha}(E)$ was arbitrary, $\Phi$ is surjective. [guided] The surjectivity step is the most substantial part of the proof. Given a root $\beta$ of $P_\alpha$ in $E$, we need to build a $K$-homomorphism that sends $\alpha$ to $\beta$. The natural idea is to define $\sigma_\beta$ by "replacing $\alpha$ with $\beta$" in every expression. This works because elements of $K(\alpha)$ are polynomial expressions in $\alpha$: the isomorphism $K(\alpha) \cong K[t]/(P_\alpha)$ tells us that every element of $K(\alpha)$ can be written as $g(\alpha)$ for some $g \in K[t]$, and the representation is unique modulo $P_\alpha$. We simply evaluate the same polynomial at $\beta$ instead. The critical issue is **well-definedness**. The same element of $K(\alpha)$ can be represented by different polynomials: if $g(\alpha) = h(\alpha)$, we need $g(\beta) = h(\beta)$. Why should this be true? The condition $g(\alpha) = h(\alpha)$ means $(g - h)(\alpha) = 0$, which means $P_\alpha$ divides $g - h$ in $K[t]$ (by the defining property of the minimal polynomial: it divides every polynomial in $K[t]$ vanishing at $\alpha$). Writing $g - h = q \cdot P_\alpha$, we evaluate at $\beta$ to get $(g - h)(\beta) = q(\beta) \cdot P_\alpha(\beta)$. This is where we use the hypothesis that $\beta$ is a root of $P_\alpha$: $P_\alpha(\beta) = 0$ forces $(g - h)(\beta) = 0$, so $g(\beta) = h(\beta)$. What would go wrong if $\beta$ were not a root of $P_\alpha$? Then $P_\alpha(\beta) \neq 0$, and the factorisation $g - h = q \cdot P_\alpha$ would give $(g - h)(\beta) = q(\beta) \cdot P_\alpha(\beta)$, which need not be zero. Different polynomial representatives of the same element of $K(\alpha)$ could then map to different values in $E$, and the map would not be well-defined. An equivalent way to see the construction: consider the evaluation homomorphism \begin{align*} \operatorname{ev}_\beta \colon K[t] &\to E \\ g &\mapsto g(\beta). \end{align*} This is a ring homomorphism. Its kernel contains $P_\alpha$ (since $P_\alpha(\beta) = 0$), and therefore contains the ideal $(P_\alpha)$. By the universal property of quotient rings, $\operatorname{ev}_\beta$ factors through $K[t]/(P_\alpha) \cong K(\alpha)$, yielding the desired $K$-homomorphism $\sigma_\beta \colon K(\alpha) \to E$. [/guided] [/step] [step:Conclude the bijection and the cardinality bound] The map $\Phi \colon \operatorname{Hom}_K(K(\alpha), E) \to \operatorname{Root}_{P_\alpha}(E)$ defined by $\sigma \mapsto \sigma(\alpha)$ is well-defined (Step 1), injective (Step 2), and surjective (Step 3). Therefore $\Phi$ is a bijection, and \begin{align*} |\operatorname{Hom}_K(K(\alpha), E)| = |\operatorname{Root}_{P_\alpha}(E)|. \end{align*} Since $P_\alpha \in K[t]$ has degree $d = \deg P_\alpha$, and a polynomial of degree $d$ over a field has at most $d$ roots in any field (by the Factor Theorem applied inductively), we obtain \begin{align*} |\operatorname{Hom}_K(K(\alpha), E)| = |\operatorname{Root}_{P_\alpha}(E)| \le \deg P_\alpha. \end{align*} [/step]