[proofplan] The forward direction is formal: a birational map $\varphi : V \dashrightarrow W$ with rational inverse $\psi$ induces a pullback $\varphi^* : \mathcal{O}_W(\eta_W) \to \mathcal{O}_V(\eta_V)$ with inverse $\psi^*$. The reverse direction is the substantive content. After choosing standard affine charts $V_0^{\mathrm{aff}} \subset V$ and $W_0^{\mathrm{aff}} \subset W$, [Function Field Finitely Generated](/theorems/2142) gives explicit generators: $\mathcal{O}_V(\eta_V) = k(X_1/X_0, \ldots, X_n/X_0)$ and $\mathcal{O}_W(\eta_W) = k(Y_1/Y_0, \ldots, Y_m/Y_0)$. A $k$-isomorphism $\omega : \mathcal{O}_W(\eta_W) \to \mathcal{O}_V(\eta_V)$ sends each generator $Y_j/Y_0$ to a rational function $h_j$ in the affine coordinates of $V$; homogenising the tuple $[1 : h_1 : \cdots : h_m]$ produces a rational map $\Phi_\omega : V \dashrightarrow W$. Applying the same construction to $\omega^{-1}$ produces $\Phi_{\omega^{-1}} : W \dashrightarrow V$. The identity $\Phi_{\omega^{-1}} \circ \Phi_\omega = \mathrm{id}_V$ is deduced by showing pullback is faithful on rational maps between irreducible projective varieties — a claim that follows from the explicit recovery of a rational map from its pullback via generators of the affine function field. [/proofplan]

[step:Forward direction: a birational equivalence induces a $k$-isomorphism of function fields] Suppose $V$ and $W$ are birational. By definition, there exist rational maps \begin{align*} \varphi : V \dashrightarrow W, \qquad \psi : W \dashrightarrow V \end{align*} with $\psi \circ \varphi = \mathrm{id}_V$ and $\varphi \circ \psi = \mathrm{id}_W$ as rational maps (equality on dense open subsets). A rational map $\varphi : V \dashrightarrow W$ between irreducible projective varieties is dominant when its image is Zariski-dense in $W$. Here $\varphi$ is dominant: if its image $\varphi(U_\varphi)$ (defined on a dense open $U_\varphi \subset V$) were contained in a proper closed subset $Z \subsetneq W$, then $\psi \circ \varphi$ would be defined on the dense open set $U_\varphi \cap \varphi^{-1}(U_\psi)$ of $V$, and its image would lie in $\psi(U_\psi \cap Z)$, which is contained in a proper closed subset of $V$ (because $\psi$ restricted to its domain of definition sends the proper closed $U_\psi \cap Z \subsetneq U_\psi$ to a constructible subset of $V$, whose closure is a proper closed subset: if its closure were all of $V$ then $U_\psi \cap Z$ would be dense in $U_\psi$, forcing $Z \cap U_\psi = U_\psi$ by taking closures in $W$, hence $Z = W$, contradiction). But $\psi \circ \varphi = \mathrm{id}_V$ has image $V$, not a proper subset — contradiction. So $\varphi$ is dominant, and symmetrically $\psi$ is dominant. A dominant rational map induces by pullback a $k$-algebra homomorphism \begin{align*} \varphi^* : \mathcal{O}_W(\eta_W) &\to \mathcal{O}_V(\eta_V), \\ h &\mapsto h \circ \varphi, \end{align*} well-defined because dominance guarantees the composition is not identically undefined on any nonempty open set in $W$'s function field description. Pullback is contravariantly functorial: $(\psi \circ \varphi)^* = \varphi^* \circ \psi^*$ for composable dominant rational maps, and $\mathrm{id}_V^* = \mathrm{id}_{\mathcal{O}_V(\eta_V)}$. Applying this to $\psi \circ \varphi = \mathrm{id}_V$ and $\varphi \circ \psi = \mathrm{id}_W$: \begin{align*} \varphi^* \circ \psi^* = (\psi \circ \varphi)^* = \mathrm{id}_{\mathcal{O}_V(\eta_V)}, \qquad \psi^* \circ \varphi^* = (\varphi \circ \psi)^* = \mathrm{id}_{\mathcal{O}_W(\eta_W)}. \end{align*} Hence $\varphi^*$ and $\psi^*$ are mutually inverse $k$-algebra isomorphisms. [/step]

[step:Reverse direction: choose standard affine charts and exhibit explicit generators] Now suppose $\omega : \mathcal{O}_W(\eta_W) \xrightarrow{\sim} \mathcal{O}_V(\eta_V)$ is a $k$-algebra isomorphism. After permuting the homogeneous coordinates of $\mathbb{P}^n_k$ and $\mathbb{P}^m_k$ (a $k$-linear automorphism), we may assume \begin{align*} V \not\subset \{X_0 = 0\}, \qquad W \not\subset \{Y_0 = 0\}. \end{align*} By the [Affine Cover of Projective Space](/theorems/2129) and [Affine Pieces Are Affine Varieties](/theorems/2138), the standard affine pieces \begin{align*} V_0^{\mathrm{aff}} := V \cap \{X_0 \neq 0\} \subset \mathbb{A}^n_k, \qquad W_0^{\mathrm{aff}} := W \cap \{Y_0 \neq 0\} \subset \mathbb{A}^m_k \end{align*} are nonempty irreducible affine varieties (irreducibility: they are nonempty open subsets of irreducible spaces). By [Function Field Finitely Generated](/theorems/2142) — specifically, the $k$-algebra isomorphism between $\mathcal{O}_V(\eta_V)$ and the function field $k(V_0^{\mathrm{aff}})$ of the affine piece, given by dehomogenisation at $X_0$ — \begin{align*} \mathcal{O}_V(\eta_V) = k(\xi_1, \ldots, \xi_n), \qquad \xi_i := X_i / X_0, \\ \mathcal{O}_W(\eta_W) = k(\zeta_1, \ldots, \zeta_m), \qquad \zeta_j := Y_j / Y_0. \end{align*} [/step]

[step:Construct the rational map $\Phi_\omega : V \dashrightarrow W$ by homogenising $\omega(\zeta_j)$]For $j = 1, \ldots, m$, the image \begin{align*} h_j := \omega(\zeta_j) \in \mathcal{O}_V(\eta_V) = k(\xi_1, \ldots, \xi_n) \end{align*} is a rational function in the affine coordinates of $V_0^{\mathrm{aff}}$. Write \begin{align*} h_j = \frac{f_j(\xi_1, \ldots, \xi_n)}{g_j(\xi_1, \ldots, \xi_n)}, \qquad f_j, g_j \in k[T_1, \ldots, T_n], \quad g_j \neq 0 \text{ in } k[V_0^{\mathrm{aff}}]. \end{align*} Since each $g_j$ is nonzero in $k[V_0^{\mathrm{aff}}]$, which is an integral domain (as $V_0^{\mathrm{aff}}$ is irreducible), the product $g := g_1 g_2 \cdots g_m$ is nonzero in $k[V_0^{\mathrm{aff}}]$. Setting \begin{align*} \tilde f_j := f_j \cdot \prod_{l \neq j} g_l \in k[T_1, \ldots, T_n], \end{align*} we have \begin{align*} h_j = \tilde f_j / g, \qquad j = 1, \ldots, m. \end{align*} Let $D := \max(\deg g, \max_j \deg \tilde f_j)$. Define the homogenisations \begin{align*} G(X_0, \ldots, X_n) &:= X_0^D \, g(X_1/X_0, \ldots, X_n/X_0) \in k[X_0, \ldots, X_n], \\ F_j(X_0, \ldots, X_n) &:= X_0^D \, \tilde f_j(X_1/X_0, \ldots, X_n/X_0) \in k[X_0, \ldots, X_n], \end{align*} which are homogeneous of degree $D$ by [Homogenisation and Dehomogenisation](/theorems/2140); moreover $G \notin I(V)$, since dehomogenising at $X_0 = 1$ recovers $g \neq 0$ in $k[V_0^{\mathrm{aff}}]$. Define \begin{align*} \Phi_\omega : V &\dashrightarrow \mathbb{P}^m_k, \\ [X_0 : X_1 : \cdots : X_n] &\mapsto [G : F_1 : \cdots : F_m]. \end{align*} The components are homogeneous of common degree $D$ and not all identically zero on $V$ (since $G \notin I(V)$), so the formula defines a well-defined rational map on the dense open subset $V \setminus V(G) \subset V$.[/step]

[guided]The construction of $\Phi_\omega$ is the algebraic-to-geometric translation that converts a function-field isomorphism into a rational map. Let us trace why each step is necessary. \textbf{Why dehomogenise first?} The function field $\mathcal{O}_V(\eta_V)$ is built from \emph{homogeneous} quotients, but the isomorphism $\mathcal{O}_V(\eta_V) \cong k(\xi_1, \ldots, \xi_n)$ from [Function Field Finitely Generated](/theorems/2142) lets us work in the simpler affine fraction field $k(\xi_1, \ldots, \xi_n)$. Each $\omega(\zeta_j)$ becomes a rational function $h_j$ in $n$ variables, a routine algebraic object. \textbf{Why a common denominator?} Without it, the candidate map $[1 : h_1 : \cdots : h_m]$ has different denominators in different slots, and rescaling individual slots is not allowed in $\mathbb{P}^m_k$. Multiplying through by $g_1 \cdots g_m$ gives a single denominator $g$; the slots become $[g : \tilde f_1 : \cdots : \tilde f_m]$. We used here that $k[V_0^{\mathrm{aff}}]$ is a domain (because $V_0^{\mathrm{aff}}$ is irreducible), so the product of nonzero elements is nonzero. \textbf{Why homogenise?} A rational map $V \dashrightarrow \mathbb{P}^m_k$ from a projective variety must be given by homogeneous polynomials of common degree. Multiplying by $X_0^D$ as in [Homogenisation and Dehomogenisation](/theorems/2140) ensures that $G$ and all $F_j$ have degree $D$; dehomogenising at $X_0 = 1$ recovers $g$ and $\tilde f_j$. \textbf{Domain of definition.} Since $G \not\equiv 0$ on $V$, the set $V \setminus V(G)$ is a nonempty open subset of $V$, and on this set the formula gives a well-defined point of $\mathbb{P}^m_k$.[/guided]

[step:Show that the image of $\Phi_\omega$ lies in $W$] On the dense open set $V \setminus V(G \cdot X_0)$, we have $G \neq 0$ and $X_0 \neq 0$, so $\Phi_\omega$ lands in $\{Y_0 \neq 0\}$. Dehomogenising, the affine coordinates of the image are \begin{align*} (F_1 / G, \ldots, F_m / G) = (\tilde f_j / g)_j = (h_1, \ldots, h_m) = (\omega(\zeta_1), \ldots, \omega(\zeta_m)) \end{align*} by [Homogenisation and Dehomogenisation](/theorems/2140). It remains to verify that $(h_1, \ldots, h_m) \in W_0^{\mathrm{aff}}$ as a rational map $V_0^{\mathrm{aff}} \dashrightarrow \mathbb{A}^m_k$. Every polynomial $P \in I(W_0^{\mathrm{aff}}) \subset k[T_1, \ldots, T_m]$ satisfies $P(\zeta_1, \ldots, \zeta_m) = 0$ in $\mathcal{O}_W(\eta_W)$ (since $\zeta_j$ are the affine coordinates of $W_0^{\mathrm{aff}}$). Applying the $k$-algebra homomorphism $\omega$: \begin{align*} P(h_1, \ldots, h_m) = P(\omega(\zeta_1), \ldots, \omega(\zeta_m)) = \omega(P(\zeta_1, \ldots, \zeta_m)) = \omega(0) = 0. \end{align*} Hence the affine image lies in $V(I(W_0^{\mathrm{aff}})) = W_0^{\mathrm{aff}}$. Taking projective closures, the image of $\Phi_\omega$ lies in $\overline{W_0^{\mathrm{aff}}}^{\,\mathbb{P}^m_k}$. We verify this closure recovers $W$: since $W_0^{\mathrm{aff}} = W \cap \{Y_0 \neq 0\}$ is a nonempty open subset of the irreducible $W$, it is dense in $W$, so $\overline{W_0^{\mathrm{aff}}}^{\,\mathbb{P}^m_k} = W$ by [Projective Closure of an Affine Variety](/theorems/2141) (hypothesis: $W_0^{\mathrm{aff}}$ is a nonempty open subset of the irreducible projective variety $W$, hence dense in $W$ — verified). So $\Phi_\omega : V \dashrightarrow W$ is a well-defined rational map, and it is dominant: its image, computed via $(h_1, \ldots, h_m)$, must contain a Zariski-dense subset of $W_0^{\mathrm{aff}} \subset W$ — we will deduce this explicitly in Step 6 from the fact that $\Phi_\omega^* = \omega$ is an isomorphism. [/step]

[step:Construct $\Phi_{\omega^{-1}} : W \dashrightarrow V$ symmetrically] Apply the construction of Steps 3–4 to $\omega^{-1} : \mathcal{O}_V(\eta_V) \to \mathcal{O}_W(\eta_W)$. Setting \begin{align*} \tilde h_i := \omega^{-1}(\xi_i) \in \mathcal{O}_W(\eta_W) = k(\zeta_1, \ldots, \zeta_m), \qquad i = 1, \ldots, n, \end{align*} and clearing denominators and homogenising as above yields a rational map $\Phi_{\omega^{-1}} : W \dashrightarrow V$ with image in $V$, by the same argument. [/step]

[step:Identify $\Phi_\omega^* = \omega$ and deduce that $\Phi_{\omega^{-1}} \circ \Phi_\omega = \mathrm{id}_V$ via faithfulness of pullback]We first verify that the pullback of $\Phi_\omega$ on function fields is exactly $\omega$: \begin{align*} \Phi_\omega^* : \mathcal{O}_W(\eta_W) \to \mathcal{O}_V(\eta_V), \qquad \zeta_j \mapsto h_j = \omega(\zeta_j). \end{align*} Indeed, $\zeta_j = Y_j / Y_0$ as a rational function on $W$; composing with $\Phi_\omega$ substitutes $Y_0 \mapsto G$ and $Y_j \mapsto F_j$, giving $F_j / G = h_j$. Since $\{\zeta_j\}$ generate $\mathcal{O}_W(\eta_W)$ as a $k$-algebra (Step 2), and a $k$-algebra homomorphism is determined by its values on a generating set, $\Phi_\omega^* = \omega$. Symmetrically, $\Phi_{\omega^{-1}}^* = \omega^{-1}$. Both $\Phi_\omega$ and $\Phi_{\omega^{-1}}$ are dominant, because their pullbacks $\omega, \omega^{-1}$ are isomorphisms (injective $k$-algebra maps $\mathcal{O}_W(\eta_W) \to \mathcal{O}_V(\eta_V)$ correspond to rational maps whose image is not contained in a proper closed subset; see the forward argument in Step 1). By functoriality of pullback for composable dominant rational maps: \begin{align*} (\Phi_{\omega^{-1}} \circ \Phi_\omega)^* = \Phi_\omega^* \circ \Phi_{\omega^{-1}}^* = \omega \circ \omega^{-1} = \mathrm{id}_{\mathcal{O}_V(\eta_V)} = (\mathrm{id}_V)^*. \end{align*} [claim:Pullback is faithful on dominant rational maps between irreducible projective varieties] [proof] Let $\alpha, \beta : V \dashrightarrow V'$ be dominant rational maps between irreducible projective varieties with $\alpha^* = \beta^* : \mathcal{O}_{V'}(\eta_{V'}) \to \mathcal{O}_V(\eta_V)$. We show $\alpha = \beta$ as rational maps. Choose affine charts $V_0^{\mathrm{aff}} \subset V$ and $V_0^{\prime\,\mathrm{aff}} \subset V'$ as in Step 2, with generators $\xi_1, \ldots, \xi_n$ of $\mathcal{O}_V(\eta_V)$ (the affine coordinates $X_i/X_0$ on $V_0^{\mathrm{aff}}$) and $\zeta_1', \ldots, \zeta_{m'}'$ of $\mathcal{O}_{V'}(\eta_{V'})$ (the affine coordinates $X_i'/X_0'$ on $V_0^{\prime\,\mathrm{aff}}$). On the dense open subset $U_\alpha \cap U_\beta \cap \alpha^{-1}(V_0^{\prime\,\mathrm{aff}}) \cap \beta^{-1}(V_0^{\prime\,\mathrm{aff}}) \subset V$ (nonempty because each factor is open and dense by dominance), both $\alpha$ and $\beta$ take values in $V_0^{\prime\,\mathrm{aff}}$. For $p$ in this open set, the $j$-th affine coordinate of $\alpha(p)$ is $\zeta_j'(\alpha(p)) = (\alpha^* \zeta_j')(p)$, and similarly for $\beta$. Since $\alpha^* \zeta_j' = \beta^* \zeta_j'$ for all $j$, the affine coordinates of $\alpha(p)$ and $\beta(p)$ agree, so $\alpha(p) = \beta(p)$ for $p$ in this dense open set. Hence $\alpha = \beta$ as rational maps. [/proof] [/claim] Applying the claim: $(\Phi_{\omega^{-1}} \circ \Phi_\omega)^* = (\mathrm{id}_V)^*$ and both $\Phi_{\omega^{-1}} \circ \Phi_\omega$ and $\mathrm{id}_V$ are dominant rational maps $V \dashrightarrow V$ (the composition of dominants is dominant: if the image of a composition lay in a proper closed set, pulling back to $V$ via $\Phi_\omega$ would force one of the factors to fail dominance). Hence \begin{align*} \Phi_{\omega^{-1}} \circ \Phi_\omega = \mathrm{id}_V. \end{align*} Symmetrically, $\Phi_\omega \circ \Phi_{\omega^{-1}} = \mathrm{id}_W$. Hence $\Phi_\omega$ is a birational equivalence with inverse $\Phi_{\omega^{-1}}$, completing the reverse direction. Combined with Step 1, this establishes the biconditional.[/step]

[guided]The climactic ingredient is the faithfulness of pullback. Both sides of the equation $(\Phi_{\omega^{-1}} \circ \Phi_\omega)^* = (\mathrm{id}_V)^*$ are $k$-algebra endomorphisms of $\mathcal{O}_V(\eta_V)$; they are equal (both are $\mathrm{id}$). We need to pass from equality of pullbacks to equality of the rational maps themselves. \textbf{Why does faithfulness hold?} A rational map is determined by its effect on a dense open subset. On that subset, where both maps take values in an affine chart of the target, the affine coordinates of the image are exactly the pullbacks of the target's affine coordinate functions. Two rational maps with equal pullbacks therefore give the same affine coordinates of the image at every point in the common dense open subset, forcing pointwise equality on that subset, hence equality as rational maps. This is the place where the proof genuinely uses the projective / irreducible / algebraically-closed setup: irreducibility ensures affine charts are dense (otherwise "dense open subset" could be empty), and the correspondence between affine coordinates and function-field generators is provided by [Function Field Finitely Generated](/theorems/2142).[/guided]