[proofplan] The argument is a two-step combination of two named results. First, [Noether Normalisation for Function Fields](/theorems/2152) provides an intermediate field $K_0 = k(x_1, \ldots, x_n)$ with $K_0/k$ purely transcendental and $K/K_0$ finite separable (separability is automatic in characteristic zero). Second, the [Primitive Element Theorem](/theorems/???) applies to the finite separable extension $K/K_0$, producing a single $y \in K$ with $K = K_0(y)$. Substituting $K_0 = k(x_1, \ldots, x_n)$ gives $K = k(x_1, \ldots, x_n, y)$, which is the desired presentation. [/proofplan]

[step:Reduce to the case $K/k$ is transcendental, treating the algebraic case directly] We split on whether $K/k$ is algebraic or transcendental. If $K/k$ is algebraic, then since $K/k$ is finitely generated and algebraic, it is in particular finite. Setting $n = 0$ (so the empty tuple $x_1, \ldots, x_n$ is taken — the algebraically independent set is empty and $K_0 = k$), we apply the [Primitive Element Theorem](/theorems/???) to the finite separable extension $K/k$ (separability is automatic in characteristic zero) to obtain $y \in K$ with $K = k(y)$, which is the required form. If $K/k$ is transcendental, then [Noether Normalisation for Function Fields](/theorems/2152) applies — we proceed in the next step. [/step]

[step:Apply Noether Normalisation to obtain a finite separable subextension] Assume $K/k$ is transcendental. By [Noether Normalisation for Function Fields](/theorems/2152), the hypotheses of which are exactly that $K/k$ is finitely generated transcendental, there exists an intermediate field \begin{align*} k \subset K_0 \subset K \end{align*} such that: \begin{itemize} \item $K_0/k$ is purely transcendental: there exist $x_1, \ldots, x_n \in K_0$ algebraically independent over $k$ with $K_0 = k(x_1, \ldots, x_n)$, where $n \geq 1$ (since $K/k$ is transcendental, the transcendence degree is positive). \item $K/K_0$ is finite, and separable in characteristic zero. \end{itemize} Both conclusions hold by the cited theorem with $n$ equal to the transcendence degree of $K/k$. [/step]

[step:Apply the Primitive Element Theorem to $K/K_0$] The [Primitive Element Theorem](/theorems/???) states: every finite separable field extension is generated by a single element. We verify both hypotheses for the extension $K/K_0$: \begin{itemize} \item $K/K_0$ is finite, by Step 2. \item $K/K_0$ is separable, by Step 2 (using $\operatorname{char}(k) = 0$, which forces $\operatorname{char}(K_0) = 0$ since $k \subset K_0$, and hence every algebraic extension of $K_0$ is separable). \end{itemize} The Primitive Element Theorem therefore produces an element $y \in K$ such that \begin{align*} K = K_0(y). \end{align*} [/step]

[step:Combine to obtain the required presentation]Substituting $K_0 = k(x_1, \ldots, x_n)$ into $K = K_0(y)$: \begin{align*} K = k(x_1, \ldots, x_n)(y) = k(x_1, \ldots, x_n, y), \end{align*} where the last equality is the standard identification — adjoining $y$ to $K_0$ as a sub-extension of $K$ is the same as adjoining $x_1, \ldots, x_n, y$ to $k$. We verify the two required properties: \begin{itemize} \item $x_1, \ldots, x_n$ are algebraically independent over $k$: this is the property of $K_0/k$ being purely transcendental from Step 2. \item $y$ is algebraic over $k(x_1, \ldots, x_n) = K_0$: since $K = K_0(y)$ is finite over $K_0$, the element $y$ is algebraic over $K_0$. \end{itemize} This completes the proof.[/step]

[guided]The proof is a clean composition of two named results, but it is worth tracking how the hypotheses of each are used. \textbf{Why Noether Normalisation?} The starting hypothesis is that $K/k$ is finitely generated. We have no control over the relations between the generators — they could be heavily algebraic or purely transcendental. Noether Normalisation reorganises the extension into a canonical two-step tower: a purely transcendental piece $K_0/k$ on the bottom, and an algebraic (in fact finite separable) piece $K/K_0$ on the top. After this reorganisation, we know exactly what kind of extension we are dealing with at each level. \textbf{Why the Primitive Element Theorem?} The PET requires finiteness and separability. Finiteness comes from Noether Normalisation. Separability comes from characteristic zero — every algebraic extension of a characteristic-zero field is separable, because the minimal polynomial of an algebraic element cannot be a polynomial in $Y^p$ when there is no characteristic $p$. This is the only place the characteristic-zero hypothesis is used directly in the proof structure (Noether Normalisation also uses it, but only to assert separability there too). \textbf{Why finite plus separable?} Without separability, the conclusion of the PET fails: there exist finite inseparable extensions that are not generated by a single element. So separability is genuinely needed. Without finiteness, the PET also fails — an infinite algebraic extension cannot be generated by one element. Both hypotheses are used. \textbf{What if $K/k$ is algebraic?} Then there is no transcendence basis to extract, and we cannot apply Noether Normalisation in the form stated (which requires $K/k$ transcendental). Instead, $K/k$ is itself finite, and we apply PET directly to obtain $y$ with $K = k(y)$, taking $n = 0$ and $K_0 = k$. The conclusion is the degenerate case of the theorem statement: an empty algebraically independent set, with $y$ algebraic over $k$. \textbf{The final substitution.} Iterated field adjunction satisfies $k(x_1, \ldots, x_n)(y) = k(x_1, \ldots, x_n, y)$ — the smallest subfield containing $k$, $x_1, \ldots, x_n$, and $y$ is the same regardless of the order in which we adjoin. This is a basic fact about fields. It allows us to flatten the two-step tower $k \subset K_0 \subset K$ into the single expression $K = k(x_1, \ldots, x_n, y)$.[/guided]