[proofplan] We reduce the projective statement to the affine one by passing to a standard affine chart. After choosing coordinates so that $V$ is not contained in the hyperplane $\{X_0 = 0\}$, the open set $V_0^{\mathrm{aff}} := V \cap \{X_0 \neq 0\}$ is a nonempty affine variety in $\mathbb{A}^n_k$, and we exhibit an isomorphism of function fields $\mathcal{O}_V(\eta) \cong \mathcal{O}_{V_0^{\mathrm{aff}}}(\eta)$. The right-hand side is the fraction field of the affine coordinate ring $k[V_0^{\mathrm{aff}}] = k[\bar{x}_1, \ldots, \bar{x}_n]$, where $\bar{x}_i$ is the residue of $X_i/X_0$. Since this fraction field is generated over $k$ by the $n$ elements $\bar{x}_1, \ldots, \bar{x}_n$, it is finitely generated as a field extension of $k$. Transporting along the isomorphism gives the same conclusion for $\mathcal{O}_V(\eta)$. [/proofplan] [step:Choose coordinates so $V$ meets the standard affine chart $\{X_0 \neq 0\}$] Since $V$ is a projective variety in $\mathbb{P}^n_k$ and the standard affine charts $U_i = \{X_i \neq 0\}$ for $i = 0, 1, \ldots, n$ cover $\mathbb{P}^n_k$ by the [Affine Cover of Projective Space](/theorems/2129), at least one chart $U_i$ has $V \cap U_i \neq \varnothing$. After permuting the homogeneous coordinates (a $k$-linear automorphism of $\mathbb{P}^n_k$ that preserves all relevant structures), we may assume \begin{align*} V \not\subset \{X_0 = 0\}, \qquad \text{equivalently} \qquad V_0^{\mathrm{aff}} := V \cap U_0 \neq \varnothing. \end{align*} By the [Affine Pieces Are Affine Varieties](/theorems/2138), the intersection $V_0^{\mathrm{aff}}$ is an affine variety: namely, the affine variety in $\mathbb{A}^n_k = U_0$ cut out by the dehomogenisations (with respect to $X_0$) of a homogeneous generating set of the ideal $I(V)$. Moreover $V_0^{\mathrm{aff}}$ is irreducible because it is a nonempty open subset of the irreducible space $V$ (an open subset of an irreducible space is irreducible as a topological subspace). Hence $V_0^{\mathrm{aff}}$ has a well-defined function field $\mathcal{O}_{V_0^{\mathrm{aff}}}(\eta)$ at its own generic point. [/step] [step:Identify $\mathcal{O}_{V_0^{\mathrm{aff}}}(\eta)$ as the fraction field generated by $\bar{x}_1, \ldots, \bar{x}_n$ over $k$] Write $A = k[X_1, \ldots, X_n]$ and let $J = I(V_0^{\mathrm{aff}}) \subset A$ be the vanishing ideal of $V_0^{\mathrm{aff}}$ in the affine chart $U_0 \cong \mathbb{A}^n_k$. The affine coordinate ring is \begin{align*} k[V_0^{\mathrm{aff}}] = A / J, \end{align*} and we write $\bar{x}_i := X_i + J \in k[V_0^{\mathrm{aff}}]$ for the residues of the chart coordinates. Since $V_0^{\mathrm{aff}}$ is irreducible, by the [Irreducibility via Prime Ideals](/theorems/2126), $J$ is a prime ideal, so $k[V_0^{\mathrm{aff}}]$ is an integral domain. By definition of the function field of an irreducible affine variety, \begin{align*} \mathcal{O}_{V_0^{\mathrm{aff}}}(\eta) = \operatorname{Frac}(k[V_0^{\mathrm{aff}}]). \end{align*} Every element of this fraction field is a quotient $\bar{f}/\bar{g}$ with $\bar{f}, \bar{g} \in k[V_0^{\mathrm{aff}}]$ and $\bar{g} \neq 0$. Since $\bar{f}$ and $\bar{g}$ are polynomial expressions in the $\bar{x}_i$ with coefficients in $k$, every element of $\operatorname{Frac}(k[V_0^{\mathrm{aff}}])$ is a rational expression in $\bar{x}_1, \ldots, \bar{x}_n$ over $k$. Hence \begin{align*} \mathcal{O}_{V_0^{\mathrm{aff}}}(\eta) = k(\bar{x}_1, \ldots, \bar{x}_n), \end{align*} the field generated over $k$ by the finite set $\{\bar{x}_1, \ldots, \bar{x}_n\}$. In particular, $\mathcal{O}_{V_0^{\mathrm{aff}}}(\eta)$ is a finitely generated field extension of $k$. [/step] [step:Construct a $k$-algebra isomorphism $\Psi : \mathcal{O}_V(\eta) \to \mathcal{O}_{V_0^{\mathrm{aff}}}(\eta)$ via dehomogenisation] Recall that $\mathcal{O}_V(\eta)$ consists of equivalence classes of pairs $(F, G)$ with $F, G \in k[X_0, \ldots, X_n]$ homogeneous of the same degree $d \ge 0$ and $G \notin I(V)$ (the homogeneous vanishing ideal of $V$), modulo the equivalence $(F, G) \sim (F', G')$ if $F G' - F' G \in I(V)$. We write a representative as a formal quotient $F/G$. Define the dehomogenisation map at $X_0$: \begin{align*} \Psi : \mathcal{O}_V(\eta) &\to \mathcal{O}_{V_0^{\mathrm{aff}}}(\eta) \\ \frac{F(X_0, \ldots, X_n)}{G(X_0, \ldots, X_n)} &\mapsto \frac{F(1, X_1, \ldots, X_n) + J}{G(1, X_1, \ldots, X_n) + J}. \end{align*} [guided] We are about to define a map between two function fields and need to verify (i) $\Psi$ is well-defined as a map of sets, (ii) $\Psi$ is a ring homomorphism, (iii) $\Psi$ is a $k$-algebra homomorphism, (iv) $\Psi$ is a bijection. Each verification is small — but missing any one of them leaves the construction incomplete. \textbf{Why dehomogenise at $X_0$?} The function field $\mathcal{O}_V(\eta)$ is built from homogeneous quotients of equal degree $d$. The substitution $X_0 = 1$ has the effect of dividing both numerator and denominator by $X_0^d$ — this is exactly the rescaling that turns a homogeneous degree-$d$ rational function on $\mathbb{P}^n_k$ into a rational function on the affine chart $U_0 = \{X_0 \neq 0\}$. So $\Psi$ is the algebraic incarnation of the geometric restriction $V \dashrightarrow V_0^{\mathrm{aff}}$. \textbf{Well-definedness.} We must verify two things. First, the denominator $G(1, X_1, \ldots, X_n)$ is non-zero in $k[V_0^{\mathrm{aff}}] = A/J$. We argue contrapositively: if $G(1, X_1, \ldots, X_n) \in J = I(V_0^{\mathrm{aff}})$, then $G$ vanishes on $V_0^{\mathrm{aff}}$ as a polynomial in the affine coordinates of $U_0$. But points of $V_0^{\mathrm{aff}}$ correspond exactly to points of $V$ with $X_0 \neq 0$; on such points, $G(X_0, \ldots, X_n)$ vanishes iff $G(1, X_1/X_0, \ldots, X_n/X_0) \cdot X_0^d$ vanishes iff $G(1, X_1/X_0, \ldots, X_n/X_0)$ vanishes (since $X_0 \neq 0$). Hence $G$ vanishes on the dense open set $V_0^{\mathrm{aff}} \subset V$, and by the [Density of Complements](/theorems/2137) (applied to the irreducible $V$ with $X_0$ a nonzero homogeneous polynomial), $V_0^{\mathrm{aff}}$ is dense in $V$, so $G$ vanishes on $V$, contradicting $G \notin I(V)$. Second, $\Psi$ respects the equivalence relation. If $(F, G) \sim (F', G')$, i.e. $FG' - F'G \in I(V)$, then dehomogenising gives $(F G' - F' G)(1, X_1, \ldots, X_n) \in J$ (since the dehomogenisation map $A_{\mathrm{hom}} \to A$, $H \mapsto H(1, X_1, \ldots, X_n)$, sends $I(V)$ into $J$ — this is exactly the content of [Homogenisation and Dehomogenisation](/theorems/2140)). Hence $F(1,\cdot) G'(1,\cdot) - F'(1,\cdot) G(1,\cdot) \in J$, so $F(1,\cdot)/G(1,\cdot) = F'(1,\cdot)/G'(1,\cdot)$ in $\operatorname{Frac}(A/J)$. \textbf{Ring homomorphism.} The map $A_{\mathrm{hom}} \to A$, $H \mapsto H(1, X_1, \ldots, X_n)$, is a ring homomorphism (substitution in a polynomial ring is always a ring homomorphism). Composing with the localisation $A \to \operatorname{Frac}(A/J)$ and using that the formal quotient $F/G$ in $\mathcal{O}_V(\eta)$ corresponds to $F(1,\cdot)/G(1,\cdot)$ in the fraction field, the addition and multiplication formulas for fractions match on both sides, so $\Psi$ is a ring homomorphism. \textbf{$k$-algebra homomorphism.} The constants $c \in k$ are represented in $\mathcal{O}_V(\eta)$ by the pair $(c, 1)$ (homogeneous of degree $0$); these dehomogenise to $c/1 \in \operatorname{Frac}(A/J)$, the same constant in the affine function field. So $\Psi$ is the identity on $k$. \textbf{Surjectivity.} Given $h = \bar{f}/\bar{g} \in \mathcal{O}_{V_0^{\mathrm{aff}}}(\eta)$ with $f, g \in A = k[X_1, \ldots, X_n]$ and $g \notin J$, let $D = \max(\deg f, \deg g)$ and define \begin{align*} F^h(X_0, \ldots, X_n) &:= X_0^D \cdot f(X_1/X_0, \ldots, X_n/X_0), \\ G^h(X_0, \ldots, X_n) &:= X_0^D \cdot g(X_1/X_0, \ldots, X_n/X_0). \end{align*} Both $F^h$ and $G^h$ are homogeneous of degree $D$ (this is the standard homogenisation construction; see [Homogenisation and Dehomogenisation](/theorems/2140)), and $G^h \notin I(V)$ because dehomogenising recovers $g \notin J$. So $F^h/G^h \in \mathcal{O}_V(\eta)$, and dehomogenisation recovers $f/g$ — confirming surjectivity. \textbf{Injectivity.} Suppose $\Psi(F/G) = 0$ in $\operatorname{Frac}(A/J)$, i.e. $F(1, X_1, \ldots, X_n) \in J$. We must show $F \in I(V)$. Let $d = \deg F$. The homogenisation of $F(1, X_1, \ldots, X_n) \in J$ — multiplying through by an appropriate power of $X_0$ to restore degree $d$ — recovers $F$ exactly (because dehomogenisation followed by homogenisation is the identity on degree-$d$ homogeneous polynomials). Since $J = I(V_0^{\mathrm{aff}})$ corresponds (under the homogenisation correspondence of [Homogenisation and Dehomogenisation](/theorems/2140)) to homogeneous polynomials vanishing on $V_0^{\mathrm{aff}}$, and $V_0^{\mathrm{aff}}$ is dense in $V$, we conclude $F$ vanishes on $V$, i.e. $F \in I(V)$. Therefore $F/G = 0$ in $\mathcal{O}_V(\eta)$. [/guided] To verify the construction in the exact mode: the dehomogenisation $H \mapsto H(1, X_1, \ldots, X_n)$ sends $I(V)$ into $J = I(V_0^{\mathrm{aff}})$ by [Homogenisation and Dehomogenisation](/theorems/2140), so $\Psi$ is well-defined on equivalence classes. The denominator $G(1, X_1, \ldots, X_n)$ is non-zero modulo $J$ because if it were zero modulo $J$, then $G$ would vanish on $V_0^{\mathrm{aff}}$, hence on its closure $V$ (density established by [Density of Complements](/theorems/2137) applied to the nonzero homogeneous polynomial $X_0$ on the irreducible variety $V$), contradicting $G \notin I(V)$. The map respects ring operations because polynomial substitution is a ring homomorphism, and it fixes $k$ because constants are degree-$0$ homogeneous polynomials sending to themselves. [/step] [step:Verify $\Psi$ is bijective by exhibiting an inverse via homogenisation] For surjectivity, given $\bar{f}/\bar{g} \in \mathcal{O}_{V_0^{\mathrm{aff}}}(\eta)$ with $f, g \in A$ and $g \notin J$, set $D := \max(\deg f, \deg g)$ and define the homogenisations \begin{align*} F^h(X_0, \ldots, X_n) &:= X_0^D \, f(X_1/X_0, \ldots, X_n/X_0), \\ G^h(X_0, \ldots, X_n) &:= X_0^D \, g(X_1/X_0, \ldots, X_n/X_0). \end{align*} Both are homogeneous of degree $D$ in $k[X_0, \ldots, X_n]$ by [Homogenisation and Dehomogenisation](/theorems/2140), and $G^h \notin I(V)$ (otherwise its dehomogenisation $g$ would lie in $J$, contradicting $g \notin J$). Hence $F^h/G^h \in \mathcal{O}_V(\eta)$, and \begin{align*} \Psi(F^h/G^h) = \frac{F^h(1, X_1, \ldots, X_n)}{G^h(1, X_1, \ldots, X_n)} = \frac{f}{g} = \bar{f}/\bar{g} \quad \text{in } \operatorname{Frac}(A/J). \end{align*} For injectivity, suppose $F/G \in \ker \Psi$, i.e. $F(1, X_1, \ldots, X_n) \in J$. Let $d = \deg F = \deg G$. Homogenising the relation $F(1, X_1, \ldots, X_n) \in J$ to degree $d$ recovers $F$ itself (since dehomogenisation followed by homogenisation to the original degree is the identity on homogeneous polynomials by [Homogenisation and Dehomogenisation](/theorems/2140)). The homogeneous polynomial obtained by homogenising any element of $J$ to a fixed degree vanishes on $V_0^{\mathrm{aff}}$, and since $V_0^{\mathrm{aff}}$ is dense in the irreducible $V$ (by [Density of Complements](/theorems/2137) with the nonzero homogeneous polynomial $X_0$), the homogenisation vanishes on $V$, i.e. $F \in I(V)$. Hence $F/G$ is the zero class in $\mathcal{O}_V(\eta)$, and $\Psi$ is injective. Combining, $\Psi$ is a $k$-algebra isomorphism $\mathcal{O}_V(\eta) \xrightarrow{\sim} \mathcal{O}_{V_0^{\mathrm{aff}}}(\eta)$. [/step] [step:Conclude that $\mathcal{O}_V(\eta)$ is finitely generated over $k$] By Step 2, $\mathcal{O}_{V_0^{\mathrm{aff}}}(\eta) = k(\bar{x}_1, \ldots, \bar{x}_n)$ is generated over $k$ as a field by the $n$ elements $\bar{x}_1, \ldots, \bar{x}_n \in k[V_0^{\mathrm{aff}}]$. By Step 4, the isomorphism $\Psi : \mathcal{O}_V(\eta) \xrightarrow{\sim} \mathcal{O}_{V_0^{\mathrm{aff}}}(\eta)$ is a $k$-algebra isomorphism. The preimages \begin{align*} \xi_i := \Psi^{-1}(\bar{x}_i) \in \mathcal{O}_V(\eta), \qquad i = 1, \ldots, n, \end{align*} are explicitly the residue classes of the rational functions $X_i/X_0$ in the function field $\mathcal{O}_V(\eta)$. Since field isomorphisms preserve generating sets, $\mathcal{O}_V(\eta) = k(\xi_1, \ldots, \xi_n)$, generated over $k$ by the $n$ elements $\xi_1, \ldots, \xi_n$. This is the required finite generation, completing the proof. [/step]