[proofplan] The strategy is to identify the ideal $I$ with the principal ideal generated by an irreducible polynomial $f$ obtained by clearing denominators in the minimal polynomial $h$ of $y$ over the field $k(x_1, \ldots, x_n)$. The two key technical inputs are *Gauss's Lemma* in two roles: first, to transfer irreducibility from the polynomial ring over the field $k(x_1, \ldots, x_n)$ to the polynomial ring over the integral domain $k[x_1, \ldots, x_n]$; second, to transfer divisibility in the same direction. Once $I = (f)$ is established, the function field of the hypersurface $V(f)$ is computed as $\operatorname{Frac}(k[W_1, \ldots, W_{n+1}]/(f))$ and is shown to be isomorphic to $K$ via $W_i \mapsto x_i$, $W_{n+1} \mapsto y$, and the [Birational Equivalence via Function Fields](/theorems/2144) criterion concludes that $V$ and $V(f)$ are birational. [/proofplan]

[step:Set up the evaluation homomorphism and identify $I$ as its kernel] Consider the evaluation homomorphism \begin{align*} \mathrm{ev} : k[W_1, \ldots, W_{n+1}] &\to K \\ g(W_1, \ldots, W_{n+1}) &\mapsto g(x_1, \ldots, x_n, y). \end{align*} This is a $k$-algebra homomorphism: addition and multiplication in $k[W_1, \ldots, W_{n+1}]$ correspond to addition and multiplication of the evaluated elements in $K$, and constants $c \in k$ map to themselves. By definition, $I = \ker(\mathrm{ev})$, so $I \subset k[W_1, \ldots, W_{n+1}]$ is an ideal. The image $\mathrm{ev}(k[W_1, \ldots, W_{n+1}]) = k[x_1, \ldots, x_n, y]$ is an integral domain, being a subring of the field $K$. Hence $I$ is a prime ideal: $k[W_1, \ldots, W_{n+1}]/I \cong k[x_1, \ldots, x_n, y]$ is an integral domain. [/step]

[step:Identify the minimal polynomial of $y$ and clear denominators to obtain $f$]By hypothesis, $y$ is algebraic over the field $k(x_1, \ldots, x_n)$. Let \begin{align*} h(T) = T^d + c_{d-1}(x) T^{d-1} + \cdots + c_1(x) T + c_0(x) \in k(x_1, \ldots, x_n)[T] \end{align*} be the minimal polynomial of $y$ over $k(x_1, \ldots, x_n)$, so $h$ is monic, irreducible in $k(x_1, \ldots, x_n)[T]$, and $h(y) = 0$ in $K$. Each coefficient $c_i(x) \in k(x_1, \ldots, x_n)$ is a rational function in $x_1, \ldots, x_n$ over $k$, hence has the form $c_i(x) = p_i(x)/q_i(x)$ for polynomials $p_i, q_i \in k[X_1, \ldots, X_n]$ with $q_i \neq 0$. Let $q(x) := \operatorname{lcm}(q_0(x), q_1(x), \ldots, q_{d-1}(x)) \in k[X_1, \ldots, X_n]$ (a common denominator, taken in any reduced form), so each $c_i$ is expressible as $c_i = \tilde c_i / q$ with $\tilde c_i \in k[X_1, \ldots, X_n]$. Define \begin{align*} f(W_1, \ldots, W_{n+1}) := q(W_1, \ldots, W_n)\, W_{n+1}^d + \tilde c_{d-1}(W_1, \ldots, W_n)\, W_{n+1}^{d-1} + \cdots + \tilde c_0(W_1, \ldots, W_n). \end{align*} Then $f \in k[W_1, \ldots, W_{n+1}]$ and, viewed as a polynomial in $W_{n+1}$ over the ring $k[W_1, \ldots, W_n]$, \begin{align*} f = q(W_1, \ldots, W_n) \cdot h(W_{n+1}), \end{align*} where in the right-hand side $h$ is written with its coefficients in $k(W_1, \ldots, W_n) = \operatorname{Frac}(k[W_1, \ldots, W_n])$. Multiplying by $q$ clears all denominators, so $f$ has coefficients in $k[W_1, \ldots, W_n]$. Substituting $W_i = x_i$ for $i \leq n$ and $W_{n+1} = y$: \begin{align*} f(x_1, \ldots, x_n, y) = q(x_1, \ldots, x_n) \cdot h(y) = q(x_1, \ldots, x_n) \cdot 0 = 0. \end{align*} Hence $f \in I$.[/step]

[guided]We are about to construct the polynomial $f \in I$ that will turn out to generate $I$. The construction is mechanical but the role of each piece must be tracked: \textbf{What is the minimal polynomial doing?} The minimal polynomial $h(T) \in k(x_1, \ldots, x_n)[T]$ encodes the algebraic relation $y$ satisfies over the rational function field $k(x_1, \ldots, x_n)$. It is the unique monic polynomial of smallest degree with $h(y) = 0$ — minimality forces irreducibility. Algebraicity of $y$ over $k(x_1, \ldots, x_n)$ (a hypothesis of the theorem) is exactly the existence of such a polynomial. \textbf{Why clear denominators?} The polynomial $h$ has coefficients in the field $k(x_1, \ldots, x_n)$, but our target ideal $I$ lives in the polynomial ring $k[W_1, \ldots, W_{n+1}]$ — coefficients in $k$, no denominators. To move from one to the other we multiply through by a common denominator $q \in k[X_1, \ldots, X_n]$, producing $f := q \cdot h$ (now with all coefficients polynomial). The factor $q$ does not affect the vanishing condition: $q(x_1, \ldots, x_n) \neq 0$ (since $q$ is a nonzero polynomial and $x_1, \ldots, x_n$ are algebraically independent over $k$, no nonzero polynomial in them vanishes), so $f(x, y) = q(x) h(y)$ vanishes precisely when $h(y) = 0$, which holds by definition of the minimal polynomial. \textbf{What about the choice of $q$?} Any common denominator works for membership in $I$. We will see in the next step that whatever $q$ we picked, $f$ is irreducible. In fact, one may always strip off polynomial factors of $f$ in $k[W_1, \ldots, W_n]$ (factors involving only the first $n$ variables) and the result is still in $I$ and still irreducible — see *Gauss's Lemma* and the discussion of content. For the purposes of this proof, any nonzero common denominator $q$ suffices. \textbf{Where is algebraic independence of $x_1, \ldots, x_n$ used?} It is used to ensure $q(x_1, \ldots, x_n) \neq 0$ for any nonzero polynomial $q$. This is the defining property of algebraic independence: no polynomial in $k[X_1, \ldots, X_n]$ has $(x_1, \ldots, x_n)$ as a zero except the zero polynomial.[/guided]

[step:Show $f$ is irreducible in $k[W_1, \ldots, W_{n+1}]$ using Gauss's Lemma] Set $R := k[W_1, \ldots, W_n]$ and $F := \operatorname{Frac}(R) = k(W_1, \ldots, W_n)$. Then $R$ is a UFD — it is the polynomial ring over the field $k$ in $n$ indeterminates, hence a UFD by iterated application of "polynomial ring over a UFD is a UFD" — and $k[W_1, \ldots, W_{n+1}] = R[W_{n+1}]$. \textbf{Irreducibility of $f$ in $F[W_{n+1}]$.} By construction, viewing both as elements of $F[W_{n+1}]$, \begin{align*} f(W_1, \ldots, W_n, W_{n+1}) = q(W_1, \ldots, W_n) \cdot h(W_{n+1}). \end{align*} Since $q \in R \setminus \{0\} \subset F^\times$ is a unit of $F$, the polynomial $q$ is a unit of the ring $F[W_{n+1}]$. Hence $f$ and $h$ are associates in $F[W_{n+1}]$. The polynomial $h \in F[W_{n+1}]$ is the minimal polynomial of $y$ over $F = k(x_1, \ldots, x_n) = k(W_1, \ldots, W_n)$ (where we identify the $W_i$ with the algebraically independent generators $x_i$ via the evaluation map, which is injective by algebraic independence), and minimal polynomials over a field are by definition irreducible. Therefore $f$ is irreducible in $F[W_{n+1}]$. \textbf{Passage to $R[W_{n+1}]$ via Gauss's Lemma.} Before applying Gauss's Lemma we replace $f$ by its primitive part. Write $f = \operatorname{cont}(f) \cdot f^*$ where $\operatorname{cont}(f) \in R$ is the GCD of the coefficients of $f$ as a polynomial in $W_{n+1}$ over $R$, and $f^* \in R[W_{n+1}]$ is the resulting primitive polynomial. This factorisation exists because $R$ is a UFD. Since $\operatorname{cont}(f) \in R \subset F^\times$ is a unit of $F$, $f$ and $f^*$ are associates in $F[W_{n+1}]$, so $f^*$ is irreducible in $F[W_{n+1}]$ as well. Moreover $f^* \in I$: substituting $(x_1, \ldots, x_n, y)$ gives $f^*(x_1, \ldots, x_n, y) = f(x_1, \ldots, x_n, y)/\operatorname{cont}(f)(x_1, \ldots, x_n) = 0$ (the denominator is a nonzero element of $k$ times a product of $x_i$-specialisations, all of which are nonzero in $K$ by algebraic independence of $x_1, \ldots, x_n$). We invoke *Gauss's Lemma*: for $R$ a UFD and $F = \operatorname{Frac}(R)$, a primitive polynomial in $R[T]$ is irreducible in $R[T]$ if and only if it is irreducible in $F[T]$. The hypotheses are met: $R = k[W_1, \ldots, W_n]$ is a UFD, and $f^*$ is primitive in $R[W_{n+1}]$ by construction. Since $f^*$ is irreducible in $F[W_{n+1}]$, *Gauss's Lemma* gives that $f^*$ is irreducible in $R[W_{n+1}] = k[W_1, \ldots, W_{n+1}]$. \textbf{Replacing $f$ by $f^*$.} From now on, we denote the primitive part $f^*$ again by $f$. This does not affect the conclusion of the theorem: $f$ and $f^*$ generate the same principal ideal in $k[W_1, \ldots, W_{n+1}]$ up to the factor $\operatorname{cont}(f) \in R$, and — crucially — $f^* \in I$ (shown above) and $f^*$ is irreducible. Thus we have an irreducible polynomial $f \in I$, primitive in $R[W_{n+1}]$, with $f \in I$. [/step]

[step:Show $I = (f)$ via Gauss's Lemma applied to divisibility]The inclusion $(f) \subset I$ is immediate from Step 2: $f \in I$, and $I$ is an ideal, so every multiple of $f$ in $k[W_1, \ldots, W_{n+1}]$ lies in $I$. For the reverse inclusion, take any $g \in I$, i.e. $g(x_1, \ldots, x_n, y) = 0$ in $K$. View $g$ as an element of the polynomial ring $k(x_1, \ldots, x_n)[W_{n+1}]$ — that is, regard the variables $W_1, \ldots, W_n$ as substituted with $x_1, \ldots, x_n$, leaving a polynomial in the single variable $W_{n+1}$ with coefficients in the field $k(x_1, \ldots, x_n)$. Concretely, this is the image of $g$ under the homomorphism \begin{align*} k[W_1, \ldots, W_n][W_{n+1}] &\to k(x_1, \ldots, x_n)[W_{n+1}] \\ g(W_1, \ldots, W_n, W_{n+1}) &\mapsto g(x_1, \ldots, x_n, W_{n+1}). \end{align*} The image polynomial $g(x_1, \ldots, x_n, W_{n+1}) \in k(x_1, \ldots, x_n)[W_{n+1}]$ has $y$ as a root: $g(x_1, \ldots, x_n, y) = 0$ in $K$. Since $h$ is the minimal polynomial of $y$ over $k(x_1, \ldots, x_n)$ and $y$ is a root of $g(x_1, \ldots, x_n, W_{n+1})$, we have $h(W_{n+1}) \mid g(x_1, \ldots, x_n, W_{n+1})$ in $k(x_1, \ldots, x_n)[W_{n+1}]$. Lifting back: since $f$ and $h$ are associates in $k(W_1, \ldots, W_n)[W_{n+1}]$ (Step 3) and divisibility is preserved under specialisation, $f(W_1, \ldots, W_n, W_{n+1}) \mid g(W_1, \ldots, W_n, W_{n+1})$ in $k(W_1, \ldots, W_n)[W_{n+1}]$. Now we apply the divisibility version of *Gauss's Lemma*: if $R$ is a UFD, $F = \operatorname{Frac}(R)$, and a primitive polynomial $f \in R[T]$ divides $g \in R[T]$ in $F[T]$, then $f$ divides $g$ in $R[T]$. The hypotheses are met: $R = k[W_1, \ldots, W_n]$ is a UFD, $f$ is primitive in $R[W_{n+1}]$ (by Step 3), and $f \mid g$ in $F[W_{n+1}]$ was just established. Hence $f \mid g$ in $R[W_{n+1}] = k[W_1, \ldots, W_{n+1}]$. So $g \in (f)$, completing the inclusion $I \subset (f)$. Combining the two inclusions, $I = (f)$.[/step]

[guided]This step is the core of the result: every polynomial vanishing at $(x_1, \ldots, x_n, y)$ is divisible by $f$. The mechanism is that $h$ is the minimal polynomial of $y$ over the larger ring (the field $k(x_1, \ldots, x_n)$), and we transfer divisibility from the field setting to the polynomial-ring setting using Gauss's Lemma. \textbf{The pivot to the field of fractions.} Given $g \in I$, we want $f \mid g$ in the polynomial ring. The natural place to argue divisibility for polynomials in $W_{n+1}$ is the polynomial ring $k(x_1, \ldots, x_n)[W_{n+1}]$, because there $h$ is irreducible (a minimal polynomial in a polynomial ring over a field). The vanishing condition $g(x, y) = 0$ becomes the condition that $h$ divides the specialisation of $g$ in this larger ring, by basic field theory: if a polynomial over a field has a given root and there is a minimal polynomial for that root, the minimal polynomial divides the given polynomial. \textbf{Lifting back to the integer ring.} Once we have $f \mid g$ in the field setting, Gauss's Lemma lifts the divisibility to the polynomial ring $k[W_1, \ldots, W_n][W_{n+1}]$. Specifically, if $f$ is primitive (no nontrivial common factor in its coefficients as polynomials in $W_1, \ldots, W_n$) and $f$ divides $g$ over the field of fractions, then $f$ divides $g$ over the original ring. This is the content of Gauss's Lemma: primitiveness is the obstruction to lifting divisibility. \textbf{Why both Gauss's Lemma applications are necessary.} The first application (Step 3) showed irreducibility of $f$ — this is so that $(f)$ is a "well-behaved" prime ideal. The second application (Step 4) showed divisibility — this is what gives $I = (f)$ rather than $I \supsetneq (f)$. Without the second, we would only know $I$ is some prime ideal containing $(f)$. \textbf{Specialisation preserves divisibility.} If $f \mid g$ in $k(W_1, \ldots, W_n)[W_{n+1}]$ (i.e. $g = fp$ for some $p$), and we specialise $W_i \to x_i$, then $g(x, y) = f(x, y) p(x, y)$. So divisibility is preserved by ring homomorphisms — this is what allows us to argue first in the specialised ring (where minimality gives us divisibility "for free") and then lift.[/guided]

[step:Identify the function field of $V(f)$ with $K$ and conclude birationality] Define the affine hypersurface \begin{align*} V(f) := \{(a_1, \ldots, a_{n+1}) \in \mathbb{A}^{n+1}_k : f(a_1, \ldots, a_{n+1}) = 0\} \subset \mathbb{A}^{n+1}_k. \end{align*} Since $f$ is irreducible in the UFD $k[W_1, \ldots, W_{n+1}]$ (Step 3), the principal ideal $(f)$ is prime — irreducible elements of a UFD generate prime ideals, and $(f) \neq (0)$ since $f \neq 0$, $(f) \neq k[W_1, \ldots, W_{n+1}]$ since $f$ is not a unit. We verify the hypothesis of [Hilbert's Nullstellensatz](/theorems/2124): $k$ is algebraically closed (by the hypothesis of the theorem). Hence the Nullstellensatz applies, giving \begin{align*} I(V(f)) = \sqrt{(f)} = (f), \end{align*} where the second equality uses that $(f)$ is already a prime ideal, hence radical. Therefore $V(f) \subset \mathbb{A}^{n+1}_k$ is an irreducible affine variety with coordinate ring \begin{align*} k[V(f)] = k[W_1, \ldots, W_{n+1}]/I(V(f)) = k[W_1, \ldots, W_{n+1}]/(f), \end{align*} and function field \begin{align*} k(V(f)) = \operatorname{Frac}(k[W_1, \ldots, W_{n+1}]/(f)). \end{align*} The evaluation homomorphism $\mathrm{ev} : k[W_1, \ldots, W_{n+1}] \to K$ from Step 1 has kernel $I = (f)$ (Step 4) and image $k[x_1, \ldots, x_n, y]$. By the first isomorphism theorem, \begin{align*} k[W_1, \ldots, W_{n+1}]/(f) \xrightarrow{\sim} k[x_1, \ldots, x_n, y], \end{align*} sending $W_i + (f) \mapsto x_i$ for $i \leq n$ and $W_{n+1} + (f) \mapsto y$. Taking fields of fractions (both are integral domains by Step 1): \begin{align*} k(V(f)) = \operatorname{Frac}(k[W_1, \ldots, W_{n+1}]/(f)) \xrightarrow{\sim} \operatorname{Frac}(k[x_1, \ldots, x_n, y]) = k(x_1, \ldots, x_n, y) = K = k(V). \end{align*} The last equality $\operatorname{Frac}(k[x_1, \ldots, x_n, y]) = k(x_1, \ldots, x_n, y)$ holds because the field of fractions of an integral domain finitely generated over $k$ is the field generated by those generators over $k$. Hence $k(V) \cong k(V(f))$ as $k$-algebras. By [Birational Equivalence via Function Fields](/theorems/2144), two irreducible varieties are birational if and only if their function fields are $k$-isomorphic. Applying this: \begin{align*} V \text{ is birational to } V(f). \end{align*} This completes the proof: $I = (f)$ for an irreducible $f$, and $V$ is birational to the hypersurface $V(f)$. [/step]