[proofplan] The strategy is to reduce the counting problem to a property of polynomial roots. By the [Primitive Element Theorem](/theorems/???), any number field $L$ is simple over $\mathbb{Q}$: $L = \mathbb{Q}(\alpha)$ for some $\alpha \in L$. This gives a canonical isomorphism $L \cong \mathbb{Q}[x]/(p_\alpha)$, where $p_\alpha$ is the minimal polynomial of $\alpha$. A field embedding $\sigma: L \hookrightarrow \mathbb{C}$ fixing $\mathbb{Q}$ is uniquely determined by the image of $\alpha$ (equivalently, the image of the generator $x$ of the quotient), and this image must be a root of $p_\alpha$ in $\mathbb{C}$. We then check that (a) each root of $p_\alpha$ in $\mathbb{C}$ produces a well-defined embedding and different roots produce different embeddings, so that the set of embeddings is in bijection with the set of complex roots of $p_\alpha$; (b) $p_\alpha$ has exactly $n = \deg p_\alpha$ distinct roots in $\mathbb{C}$, using separability of $L/\mathbb{Q}$ (a characteristic-zero extension is automatically separable) and the [Fundamental Theorem of Algebra](/theorems/???). [/proofplan] [step:Write $L$ as a simple extension $L = \mathbb{Q}(\alpha)$ via a primitive element] By the [Primitive Element Theorem](/theorems/???), any finite separable extension of fields is simple (generated by a single element). We apply this to the extension $L/\mathbb{Q}$. Its hypotheses: 1. **Finite extension**: $[L:\mathbb{Q}] = n < \infty$ by the definition of a number field. 2. **Separable extension**: since $\operatorname{char}(\mathbb{Q}) = 0$, every algebraic extension of $\mathbb{Q}$ is [separable](/pages/???) (the minimal polynomial of any element has distinct roots in a splitting field — no multiple roots can appear in characteristic zero). Both hold. The theorem produces $\alpha \in L$ such that $L = \mathbb{Q}(\alpha)$. Let $p_\alpha \in \mathbb{Q}[x]$ be the monic minimal polynomial of $\alpha$ over $\mathbb{Q}$; by the [tower law](/theorems/???), $[L:\mathbb{Q}] = [\mathbb{Q}(\alpha):\mathbb{Q}] = \deg p_\alpha$, so $\deg p_\alpha = n$. [guided] Our goal is to count field embeddings $\sigma: L \hookrightarrow \mathbb{C}$. To make this counting tractable, we want a concrete description of $L$ as a quotient of a polynomial ring. Every number field is a simple extension: there exists $\alpha \in L$ with $L = \mathbb{Q}(\alpha)$. This is the content of the [Primitive Element Theorem](/theorems/???), which states that every finite separable field extension is simple. **Verifying the hypotheses:** - **Finite extension**: $[L:\mathbb{Q}] = n < \infty$ is part of the definition of "number field" — it is a finite-degree field extension of $\mathbb{Q}$. - **Separable extension**: $\mathbb{Q}$ has characteristic $0$, and every algebraic extension of a characteristic-$0$ field is separable. Recall the definition: an algebraic extension $L/K$ is separable if every element's minimal polynomial has distinct roots in a splitting field. A minimal polynomial $p$ has a multiple root iff $\gcd(p, p') \neq 1$ in a polynomial ring. In characteristic $0$, $\deg p' = \deg p - 1$, and if $p$ is irreducible then either $p' = 0$ (impossible since $\deg p' = \deg p - 1 \geq 0$ only if $p$ is constant) or $\gcd(p, p')$ has degree $< \deg p$, hence $=0$ by irreducibility of $p$. So no multiple roots — every algebraic extension of $\mathbb{Q}$ is separable. Both hypotheses are satisfied; apply the theorem to get $\alpha \in L$ with $L = \mathbb{Q}(\alpha)$. Let $p_\alpha \in \mathbb{Q}[x]$ be the minimal polynomial of $\alpha$ over $\mathbb{Q}$ — the unique monic polynomial of least positive degree vanishing at $\alpha$. By the [tower law](/theorems/???): \begin{align*} n = [L:\mathbb{Q}] = [\mathbb{Q}(\alpha):\mathbb{Q}]. \end{align*} And $[\mathbb{Q}(\alpha):\mathbb{Q}] = \deg p_\alpha$ because $\{1, \alpha, \alpha^2, \ldots, \alpha^{\deg p_\alpha - 1}\}$ is a $\mathbb{Q}$-basis of $\mathbb{Q}(\alpha) = \mathbb{Q}[\alpha]$ (any higher power $\alpha^k$ is reduced using $p_\alpha(\alpha) = 0$). So \begin{align*} \deg p_\alpha = n. \end{align*} [/guided] [/step] [step:Identify $L$ with $\mathbb{Q}[x]/(p_\alpha)$ via the evaluation map] Consider the evaluation homomorphism \begin{align*} \varepsilon: \mathbb{Q}[x] &\to L, \\ f(x) &\mapsto f(\alpha). \end{align*} $\varepsilon$ is a $\mathbb{Q}$-algebra homomorphism. Its image is $\mathbb{Q}[\alpha] = \mathbb{Q}(\alpha) = L$ (the second equality holds because $\alpha$ is algebraic over $\mathbb{Q}$, so $\mathbb{Q}[\alpha]$ is already a field). Its kernel is $\ker \varepsilon = \{f : f(\alpha) = 0\}$. By the [Minimal Polynomial Divides Any Annihilating Polynomial](/theorems/1569), $f(\alpha) = 0 \iff p_\alpha \mid f$; hence $\ker \varepsilon = (p_\alpha)$, the principal ideal generated by $p_\alpha$ in $\mathbb{Q}[x]$. By the first isomorphism theorem, $\varepsilon$ descends to a $\mathbb{Q}$-algebra isomorphism \begin{align*} \overline{\varepsilon}: \mathbb{Q}[x]/(p_\alpha) &\xrightarrow{\;\sim\;} L, \\ [x] &\mapsto \alpha. \end{align*} [guided] Now we produce an explicit description of $L$ as a quotient of a polynomial ring. Consider the evaluation map \begin{align*} \varepsilon: \mathbb{Q}[x] &\to L, \\ f(x) &\mapsto f(\alpha). \end{align*} By the universal property of the polynomial ring, $\varepsilon$ is a $\mathbb{Q}$-algebra homomorphism (it preserves sums, products, and scalar multiplication by $\mathbb{Q}$; evaluation at a single element is the prototypical example). **Image of $\varepsilon$**: by definition, $\operatorname{im} \varepsilon = \mathbb{Q}[\alpha]$, the $\mathbb{Q}$-algebra generated by $\alpha$ — all polynomial expressions in $\alpha$ with $\mathbb{Q}$-coefficients. Since $\alpha$ is algebraic over $\mathbb{Q}$, the ring $\mathbb{Q}[\alpha]$ is in fact already a field (for any nonzero $\beta \in \mathbb{Q}[\alpha]$, one can write $\beta^{-1}$ as a polynomial in $\alpha$ using $p_\alpha(\alpha) = 0$ via the extended Euclidean algorithm on $p_\alpha$ and the polynomial expressing $\beta$). Hence $\mathbb{Q}[\alpha] = \mathbb{Q}(\alpha) = L$. **Kernel of $\varepsilon$**: $\ker \varepsilon = \{f \in \mathbb{Q}[x] : f(\alpha) = 0\}$. By [Minimal Polynomial Divides Any Annihilating Polynomial](/theorems/1569), the condition $f(\alpha) = 0$ is equivalent to $p_\alpha \mid f$ (using that $p_\alpha$ is the minimal polynomial). Therefore \begin{align*} \ker \varepsilon = \{f \in \mathbb{Q}[x] : p_\alpha \mid f\} = (p_\alpha), \end{align*} the ideal of $\mathbb{Q}[x]$ generated by $p_\alpha$. **Apply the first isomorphism theorem**: a ring homomorphism factors through its kernel to give an isomorphism from the quotient to the image: \begin{align*} \overline{\varepsilon}: \mathbb{Q}[x]/(p_\alpha) \xrightarrow{\;\sim\;} \mathbb{Q}[\alpha] = L, \qquad [f] \mapsto f(\alpha). \end{align*} The class $[x] \in \mathbb{Q}[x]/(p_\alpha)$ maps to $\alpha \in L$ under $\overline{\varepsilon}$. This $\mathbb{Q}$-algebra isomorphism is the key tool: it reduces questions about embeddings of $L$ to questions about ring homomorphisms out of $\mathbb{Q}[x]/(p_\alpha)$, which are governed by the polynomial $p_\alpha$. Since $p_\alpha$ is irreducible in $\mathbb{Q}[x]$ (any factorisation would violate minimality of $\deg p_\alpha$), the ideal $(p_\alpha)$ is maximal in the PID $\mathbb{Q}[x]$, and the quotient $\mathbb{Q}[x]/(p_\alpha)$ is indeed a field — consistent with $L$ being a field. [/guided] [/step] [step:Parametrise embeddings $L \hookrightarrow \mathbb{C}$ by the images of $\alpha$, which are roots of $p_\alpha$ in $\mathbb{C}$] [claim:Embeddings out of $\mathbb{Q}[x]/(p_\alpha)$ correspond to complex roots of $p_\alpha$] The map \begin{align*} \Phi: \operatorname{Hom}_{\mathbb{Q}\text{-alg}}(\mathbb{Q}[x]/(p_\alpha), \mathbb{C}) &\to \{\beta \in \mathbb{C} : p_\alpha(\beta) = 0\}, \\ \sigma &\mapsto \sigma([x]) \end{align*} is a bijection. [/claim] [proof] **Well-defined**: for $\sigma \in \operatorname{Hom}_{\mathbb{Q}\text{-alg}}(\mathbb{Q}[x]/(p_\alpha), \mathbb{C})$, let $\beta := \sigma([x]) \in \mathbb{C}$. Apply $\sigma$ to the identity $p_\alpha([x]) = 0$ in $\mathbb{Q}[x]/(p_\alpha)$ (which holds because $p_\alpha \in (p_\alpha)$): by $\mathbb{Q}$-algebra homomorphism property (preserving sums, products, and $\mathbb{Q}$-coefficients), \begin{align*} p_\alpha(\beta) = p_\alpha(\sigma([x])) = \sigma(p_\alpha([x])) = \sigma(0) = 0. \end{align*} So $\beta$ is a root of $p_\alpha$ in $\mathbb{C}$. **Injective**: a $\mathbb{Q}$-algebra homomorphism out of $\mathbb{Q}[x]/(p_\alpha)$ is determined by its value on the generator $[x]$: for any class $[f]$, $\sigma([f]) = f(\sigma([x]))$. So if $\sigma_1, \sigma_2$ agree on $[x]$, they agree everywhere. Hence $\Phi$ is injective. **Surjective**: let $\beta \in \mathbb{C}$ be a root of $p_\alpha$. Define \begin{align*} \tilde \sigma_\beta: \mathbb{Q}[x] &\to \mathbb{C}, \\ f(x) &\mapsto f(\beta), \end{align*} the evaluation-at-$\beta$ homomorphism (a $\mathbb{Q}$-algebra homomorphism by universal property of $\mathbb{Q}[x]$). Its kernel contains $p_\alpha$ because $p_\alpha(\beta) = 0$, hence contains the ideal $(p_\alpha)$. By the universal property of the quotient, $\tilde \sigma_\beta$ descends to a $\mathbb{Q}$-algebra homomorphism \begin{align*} \sigma_\beta: \mathbb{Q}[x]/(p_\alpha) \to \mathbb{C}, \qquad [f] \mapsto f(\beta), \end{align*} with $\sigma_\beta([x]) = \beta$. So $\Phi(\sigma_\beta) = \beta$, proving surjectivity. [/proof] Composing $\Phi$ with the isomorphism $\overline{\varepsilon}: \mathbb{Q}[x]/(p_\alpha) \xrightarrow{\sim} L$ from the previous step, we obtain a bijection \begin{align*} \Psi: \operatorname{Hom}_{\mathbb{Q}\text{-alg}}(L, \mathbb{C}) &\to \{\beta \in \mathbb{C} : p_\alpha(\beta) = 0\}, \\ \sigma &\mapsto \sigma(\alpha). \end{align*} A $\mathbb{Q}$-algebra homomorphism $L \to \mathbb{C}$ is a ring homomorphism fixing $\mathbb{Q}$. Since $L$ is a field, any such nonzero ring homomorphism is automatically injective (the kernel is an ideal of $L$, either $0$ or $L$; fixing $\mathbb{Q} \neq 0$ rules out $L$). Hence $\operatorname{Hom}_{\mathbb{Q}\text{-alg}}(L, \mathbb{C})$ coincides with the set of field embeddings $\sigma: L \hookrightarrow \mathbb{C}$ (which by convention fix $\mathbb{Q}$). Therefore \begin{align*} \#\{\sigma: L \hookrightarrow \mathbb{C}\} = \#\{\beta \in \mathbb{C} : p_\alpha(\beta) = 0\}. \end{align*} [guided] We now set up the bijection that reduces counting embeddings to counting roots of the minimal polynomial. **Stated abstractly**: for any commutative ring $R$ and polynomial $g \in R[x]$, $R$-algebra homomorphisms from $R[x]/(g)$ to any $R$-algebra $S$ are in natural bijection with roots of $g$ in $S$. This follows from two universal properties: 1. $R[x]$ is the free commutative $R$-algebra on one generator: $R$-algebra homomorphisms $R[x] \to S$ correspond to elements $\beta \in S$ (the image of the generator $x$), via $f(x) \mapsto f(\beta)$. 2. Quotient by an ideal $I \subseteq R$: $R$-algebra homomorphisms $R[x]/I \to S$ correspond to $R$-algebra homomorphisms $R[x] \to S$ that vanish on $I$, i.e., send every $g \in I$ to $0$. Combining: $R$-algebra homomorphisms $R[x]/(g) \to S$ correspond to $\beta \in S$ with $g(\beta) = 0$. We apply this with $R = \mathbb{Q}$, $g = p_\alpha$, $S = \mathbb{C}$: \begin{align*} \operatorname{Hom}_{\mathbb{Q}\text{-alg}}(\mathbb{Q}[x]/(p_\alpha), \mathbb{C}) \;\longleftrightarrow\; \{\beta \in \mathbb{C} : p_\alpha(\beta) = 0\}. \end{align*} For transparency, the claim writes out the verification: - **Well-defined forward map**: $\Phi(\sigma) = \sigma([x])$. Why is this a root of $p_\alpha$? Because $p_\alpha([x]) = [p_\alpha(x)] = 0$ in $\mathbb{Q}[x]/(p_\alpha)$ (since $p_\alpha \in (p_\alpha)$), and applying the $\mathbb{Q}$-algebra homomorphism $\sigma$ gives $p_\alpha(\sigma([x])) = 0$ — here we use that $\sigma$ commutes with polynomial expressions with $\mathbb{Q}$-coefficients (which is exactly the content of being a $\mathbb{Q}$-algebra homomorphism). - **Injectivity**: every element of $\mathbb{Q}[x]/(p_\alpha)$ is a polynomial expression in $[x]$ with $\mathbb{Q}$-coefficients; $\sigma$ is determined by its value on $[x]$. - **Surjectivity**: given a root $\beta \in \mathbb{C}$ of $p_\alpha$, evaluation at $\beta$ defines a $\mathbb{Q}$-algebra homomorphism $\mathbb{Q}[x] \to \mathbb{C}$ that vanishes on $(p_\alpha)$ (since $p_\alpha(\beta) = 0$), hence descends to a $\mathbb{Q}$-algebra homomorphism $\sigma_\beta: \mathbb{Q}[x]/(p_\alpha) \to \mathbb{C}$ with $\sigma_\beta([x]) = \beta$. Composing with the isomorphism $\overline{\varepsilon}: \mathbb{Q}[x]/(p_\alpha) \xrightarrow{\sim} L$ (which carries $[x]$ to $\alpha$), we get a bijection \begin{align*} \Psi: \operatorname{Hom}_{\mathbb{Q}\text{-alg}}(L, \mathbb{C}) \xrightarrow{\sim} \{\beta \in \mathbb{C} : p_\alpha(\beta) = 0\}, \qquad \sigma \mapsto \sigma(\alpha). \end{align*} **A subtlety — $\mathbb{Q}$-algebra homomorphisms versus field embeddings**: A field embedding $\sigma: L \hookrightarrow \mathbb{C}$ is an injective ring homomorphism. In the current setting, a "field embedding" is understood to **fix $\mathbb{Q}$**, equivalently to be a $\mathbb{Q}$-algebra homomorphism. Conversely, any nonzero $\mathbb{Q}$-algebra homomorphism $\sigma: L \to \mathbb{C}$ is automatically a field embedding: - Injectivity: $\ker \sigma$ is an ideal of $L$. Since $L$ is a field, its only ideals are $\{0\}$ and $L$. But $\sigma(1) = 1 \neq 0$ (since $\sigma$ is a $\mathbb{Q}$-algebra homomorphism), so $\ker \sigma \neq L$, hence $\ker \sigma = \{0\}$. - Fixing $\mathbb{Q}$: by the definition of $\mathbb{Q}$-algebra homomorphism. So: field embeddings $\sigma: L \hookrightarrow \mathbb{C}$ $\;=\;$ $\mathbb{Q}$-algebra homomorphisms $\sigma: L \to \mathbb{C}$. Therefore \begin{align*} \#\{\sigma: L \hookrightarrow \mathbb{C}\} = \#\{\beta \in \mathbb{C} : p_\alpha(\beta) = 0\}. \end{align*} The problem is reduced to counting complex roots of $p_\alpha$. [/guided] [/step] [step:Count the complex roots of $p_\alpha$ using algebraic closure and separability] The polynomial $p_\alpha \in \mathbb{Q}[x]$ has degree $n$. By the [Fundamental Theorem of Algebra](/theorems/???), $\mathbb{C}$ is algebraically closed, so $p_\alpha$ splits over $\mathbb{C}$: \begin{align*} p_\alpha(x) = \prod_{i=1}^n (x - \alpha_i), \qquad \alpha_1, \ldots, \alpha_n \in \mathbb{C} \text{ (with multiplicity)}. \end{align*} Thus $p_\alpha$ has exactly $n$ roots in $\mathbb{C}$ counted with multiplicity. We now show the roots are all distinct. Since $\operatorname{char}(\mathbb{Q}) = 0$, the extension $L/\mathbb{Q}$ is [separable](/pages/???): every irreducible polynomial over $\mathbb{Q}$ has distinct roots in any splitting field. The minimal polynomial $p_\alpha$ is irreducible in $\mathbb{Q}[x]$ (it is monic of minimal degree annihilating $\alpha$, and any factorisation $p_\alpha = g h$ with $\deg g, \deg h < \deg p_\alpha$ would give $g(\alpha)h(\alpha) = 0$, hence $g(\alpha) = 0$ or $h(\alpha) = 0$ in the domain $L$, contradicting minimality of $\deg p_\alpha$). Therefore its roots $\alpha_1, \ldots, \alpha_n$ in $\mathbb{C}$ are all distinct, i.e. \begin{align*} \#\{\beta \in \mathbb{C} : p_\alpha(\beta) = 0\} = n. \end{align*} Combining with the bijection $\Psi$ from the previous step: \begin{align*} \#\{\sigma: L \hookrightarrow \mathbb{C}\} = n = [L:\mathbb{Q}], \end{align*} completing the proof. [guided] It remains to count the complex roots of $p_\alpha \in \mathbb{Q}[x]$, a polynomial of degree $n$. Two facts are needed: 1. $p_\alpha$ has $n$ roots in $\mathbb{C}$ counted with multiplicity. 2. The roots are in fact distinct (no repeated roots). **Step A: counting with multiplicity.** By the [Fundamental Theorem of Algebra](/theorems/???), $\mathbb{C}$ is algebraically closed: every non-constant polynomial in $\mathbb{C}[x]$ has a root in $\mathbb{C}$. As a consequence, any polynomial $g \in \mathbb{C}[x]$ of degree $d$ splits completely over $\mathbb{C}$ into $d$ linear factors (iterate: peel off a root, divide, repeat). Applied to $p_\alpha$, viewed as an element of $\mathbb{C}[x] \supseteq \mathbb{Q}[x]$: \begin{align*} p_\alpha(x) = \prod_{i=1}^n (x - \alpha_i) \qquad \text{for some } \alpha_i \in \mathbb{C}. \end{align*} The $\alpha_i$ are the roots of $p_\alpha$ in $\mathbb{C}$, counted with multiplicity. **Step B: distinctness.** The $\alpha_i$ might a priori coincide (be repeated roots). We rule this out using separability. **Irreducibility of $p_\alpha$ in $\mathbb{Q}[x]$**: if $p_\alpha = g h$ in $\mathbb{Q}[x]$ with $1 \leq \deg g, \deg h < \deg p_\alpha$, then in the integral domain $L$ we have $g(\alpha) h(\alpha) = p_\alpha(\alpha) = 0$, so $g(\alpha) = 0$ or $h(\alpha) = 0$. Either contradicts the minimality of $\deg p_\alpha$. Hence $p_\alpha$ is irreducible in $\mathbb{Q}[x]$. **Separability of $p_\alpha$**: a polynomial is [separable](/pages/???) if it has no multiple roots in a splitting field. Over a field of characteristic $0$ (like $\mathbb{Q}$), every irreducible polynomial is separable: - Argument: if $p_\alpha$ has a multiple root $\alpha_j$ in its splitting field $M$, then $p_\alpha$ and its formal derivative $p_\alpha' \in \mathbb{Q}[x]$ share the root $\alpha_j$ in $M$. Hence $\gcd(p_\alpha, p_\alpha')$ in $\mathbb{Q}[x]$ (computed via the Euclidean algorithm, which is the same in $\mathbb{Q}[x]$ and $M[x]$) has degree $\geq 1$. - $p_\alpha$ is irreducible of degree $n$, so any divisor of $p_\alpha$ of degree $\geq 1$ is an associate of $p_\alpha$ itself, i.e. has degree exactly $n$. So $\gcd(p_\alpha, p_\alpha')$ has degree $n$. - But $\deg p_\alpha' = n - 1$ (taking formal derivative drops the degree by one, and $n - 1 \geq 0$ for $n \geq 1$; in characteristic $0$ the leading term does not vanish). So the gcd has degree $\leq n - 1 < n$, contradiction. Hence $p_\alpha$ has $n$ **distinct** roots in $\mathbb{C}$. **Combining**: \begin{align*} \#\{\beta \in \mathbb{C} : p_\alpha(\beta) = 0\} = n. \end{align*} And by the bijection $\Psi: \operatorname{Hom}_{\mathbb{Q}\text{-alg}}(L, \mathbb{C}) \xrightarrow{\sim} \{\beta \in \mathbb{C} : p_\alpha(\beta) = 0\}$ from the previous step: \begin{align*} \#\{\sigma: L \hookrightarrow \mathbb{C}\} = n = [L:\mathbb{Q}]. \end{align*} Concretely, the $n$ embeddings are \begin{align*} \sigma_i: L &\hookrightarrow \mathbb{C}, \\ \alpha &\mapsto \alpha_i, \qquad i = 1, \ldots, n, \end{align*} each determined by sending the primitive element $\alpha$ to the $i$-th root of its minimal polynomial. **A failure mode to note**: if $\mathbb{Q}$ were replaced by a field of positive characteristic $p$ and $L/\mathbb{Q}$ were not separable, then $p_\alpha$ could have repeated roots, and the count of embeddings into a suitable algebraic closure would be strictly less than $[L:\mathbb{Q}]$. The equality "$\#$ embeddings $ = n$" is a statement about separable extensions specifically, and it holds for all number fields because characteristic $0$ forces separability. [/guided] [/step]