[proofplan] A birational equivalence $\varphi : V \dashrightarrow W$ with inverse $\psi : W \dashrightarrow V$ restricts to mutually inverse isomorphisms on a common Zariski-open subset $U \subset V$ and $\varphi(U) \subset W$. By [Smooth Points Form an Open Dense Subset](/theorems/2149), the smooth loci $V_{\mathrm{sm}}$ and $W_{\mathrm{sm}}$ are open dense, hence intersect $U$ and $\varphi(U)$ respectively. Picking a common smooth point $p$ where both $\varphi$ and $\psi$ are regular, the chain rule of [Differentials Are Well-Defined and Chain Rule Holds](/theorems/2150) applied to $\psi \circ \varphi = \mathrm{id}_V$ and $\varphi \circ \psi = \mathrm{id}_W$ produces mutually inverse linear maps between $T_{V,p}$ and $T_{W, \varphi(p)}$. Smoothness identifies $\dim T_{V,p} = \dim V$ and $\dim T_{W, \varphi(p)} = \dim W$, so the dimensions of the tangent spaces agree as vector spaces, hence the dimensions of the varieties agree. [/proofplan] [step:Fix a common smooth open subset compatible with the birational equivalence] By hypothesis, there exist rational maps $\varphi : V \dashrightarrow W$ and $\psi : W \dashrightarrow V$ such that \begin{align*} \psi \circ \varphi = \mathrm{id}_V \quad \text{and} \quad \varphi \circ \psi = \mathrm{id}_W \end{align*} on Zariski-open dense subsets of $V$ and $W$ respectively. Let $U_\varphi \subset V$ be the (open dense) domain of definition of $\varphi$ and $U_\psi \subset W$ the (open dense) domain of definition of $\psi$. Set \begin{align*} U := U_\varphi \cap \varphi^{-1}(U_\psi) \subset V, \end{align*} which is open in $V$ (as the preimage of an open set under a morphism on $U_\varphi$) and non-empty: it is the largest open subset on which both $\varphi$ and $\psi \circ \varphi$ are defined. Replacing $U$ by $U \cap V_{\mathrm{sm}}$, where $V_{\mathrm{sm}}$ is the smooth locus of $V$, and replacing $\varphi(U)$ by its intersection with $W_{\mathrm{sm}}$, we may assume: \begin{itemize} \item $\varphi : U \to W$ is a morphism with image in $W_{\mathrm{sm}}$, \item $\psi$ is defined on $\varphi(U) \subset W_{\mathrm{sm}}$, \item $\psi \circ \varphi = \mathrm{id}_U$ holds as morphisms on $U$. \end{itemize} Both $V_{\mathrm{sm}}$ and $W_{\mathrm{sm}}$ are non-empty open dense subsets by [Smooth Points Form an Open Dense Subset](/theorems/2149) (applied to the irreducible varieties $V$ and $W$, with $k$ algebraically closed of characteristic zero), so the intersections above remain open dense, and in particular non-empty. Pick a point $p \in U$ — this exists by non-emptiness — and set $q := \varphi(p) \in W_{\mathrm{sm}}$. [/step] [step:Apply the chain rule to obtain mutually inverse differentials] By [Differentials Are Well-Defined and Chain Rule Holds](/theorems/2150), each rational map has a well-defined differential at any point where it is defined, and the chain rule \begin{align*} d(\beta \circ \alpha)_p = d\beta_{\alpha(p)} \circ d\alpha_p \end{align*} holds wherever both compositions and differentials make sense. The identity morphism $\mathrm{id}_V : V \to V$ has differential $d(\mathrm{id}_V)_p = \mathrm{id}_{T_{V,p}}$: if $\mathrm{id}_V$ is expressed by the polynomials $f_i = X_i$, then $J(\mathrm{id}_V)(p) = I_n$ is the identity matrix, and acts as the identity on $T_{V,p} \subset k^n$. Similarly $d(\mathrm{id}_W)_q = \mathrm{id}_{T_{W,q}}$. Applying the chain rule to $\psi \circ \varphi = \mathrm{id}_V$ at $p$ and $\varphi \circ \psi = \mathrm{id}_W$ at $q$: \begin{align*} d\psi_q \circ d\varphi_p &= d(\psi \circ \varphi)_p = d(\mathrm{id}_V)_p = \mathrm{id}_{T_{V,p}}, \\ d\varphi_p \circ d\psi_q &= d(\varphi \circ \psi)_q = d(\mathrm{id}_W)_q = \mathrm{id}_{T_{W,q}}. \end{align*} Hence $d\varphi_p : T_{V,p} \to T_{W,q}$ and $d\psi_q : T_{W,q} \to T_{V,p}$ are mutually inverse linear maps. In particular, $d\varphi_p$ is a linear isomorphism of $k$-vector spaces. [guided] The chain rule is the only mechanism we have for transferring linear-algebraic information between $V$ and $W$. We apply it to the relation $\psi \circ \varphi = \mathrm{id}_V$, which holds on the open dense subset $U \subset V$ — but the chain rule for differentials is a pointwise statement at $p$, so we only need the equality of morphisms on a neighbourhood of $p$, which is automatic. The differential of the identity morphism is the identity linear map. To verify: the identity is expressed by the polynomial map $X = (X_1, \ldots, X_n) \mapsto (X_1, \ldots, X_n)$, whose Jacobian is the $n \times n$ identity matrix. This identity matrix restricts to the identity on $T_{V,p} \subset k^n$. (Note: the differential is independent of the polynomial expression by the previous theorem, so this calculation is unambiguous.) Plugging into the chain rule: \begin{align*} d\psi_q \circ d\varphi_p = d(\psi \circ \varphi)_p = d(\mathrm{id}_V)_p = \mathrm{id}_{T_{V,p}}. \end{align*} Symmetrically, \begin{align*} d\varphi_p \circ d\psi_q = \mathrm{id}_{T_{W,q}}. \end{align*} Two linear maps with both compositions equal to identity are mutually inverse — this is a basic fact in linear algebra (an inverse on one side equals an inverse on the other for square matrices, and here we will see that the maps are indeed between equidimensional spaces). The conclusion $d\varphi_p$ is a $k$-linear isomorphism is the key output of this step. Why did we need to choose $p$ smooth and in the domain of $\psi \circ \varphi$? Smoothness is needed in the next step to translate $\dim T_{V,p}$ into $\dim V$. The domain condition is needed for the chain rule to apply. [/guided] [/step] [step:Translate tangent space dimensions into variety dimensions] By the definition of smoothness, at a smooth point $p$ of an irreducible variety $V$, \begin{align*} \dim_k T_{V,p} = \dim V. \end{align*} Similarly $\dim_k T_{W,q} = \dim W$ since $q \in W_{\mathrm{sm}}$. The linear isomorphism $d\varphi_p : T_{V,p} \to T_{W,q}$ from Step 2 implies these dimensions are equal: \begin{align*} \dim V = \dim_k T_{V,p} = \dim_k T_{W,q} = \dim W. \end{align*} [/step]