[proofplan] The strategy is to express the smooth locus as the locus where the Jacobian matrix of a generating set of $I(V)$ has maximal rank, then deduce both openness and non-emptiness from this description. Openness is immediate: the locus where the rank drops is cut out by the vanishing of certain minors, which are polynomial conditions and hence Zariski-closed. Non-emptiness is substantive: we reduce to a hypersurface model via [Noether Normalisation](/theorems/2152) and [Primitive Generator FF](/theorems/2153), observe that smoothness is shared on the common dense open subset of two birationally equivalent irreducible varieties (dimensions match by [Birational Varieties Same Dim](/theorems/2151)), and on the hypersurface model we produce a non-vanishing partial derivative using separability of the finite extension — available because $k$ has characteristic zero. [/proofplan] [step:Fix a generating set of $I(V)$ and the Jacobian description of the tangent space] Since $V \subset \mathbb{A}^n_k$ is affine, its ideal is \begin{align*} I(V) = \langle f_1, \ldots, f_m \rangle \subset k[X_1, \ldots, X_n], \end{align*} finitely generated by the \emph{Hilbert Basis Theorem} (applied to the Noetherian ring $k[X_1, \ldots, X_n]$). Set $d := \dim V$, where $\dim V$ denotes the Krull dimension of the coordinate ring $k[V] = k[X_1, \ldots, X_n]/I(V)$ (equivalently, the transcendence degree of the function field $k(V)$ over $k$). For $p \in V$, the Jacobian matrix is the $m \times n$ matrix \begin{align*} J(p) := \left( \frac{\partial f_j}{\partial X_i}(p) \right)_{1 \leq j \leq m,\ 1 \leq i \leq n} \in k^{m \times n}. \end{align*} The Zariski tangent space $T_{V,p}$ is by definition $\ker J(p) \subset k^n$, so \begin{align*} \dim_k T_{V,p} = n - \operatorname{rank} J(p). \end{align*} A point $p$ is smooth iff $\dim_k T_{V,p} = d$, equivalently $\operatorname{rank} J(p) = n - d$. [/step] [step:Express the non-smooth locus as the vanishing of $(n-d) \times (n-d)$ minors] Let \begin{align*} S := \{ p \in V : \operatorname{rank} J(p) < n - d \}. \end{align*} By standard linear algebra, $\operatorname{rank} J(p) < n - d$ iff every $(n-d) \times (n-d)$ minor of $J(p)$ vanishes. Each such minor is a polynomial in the entries of $J(p)$, which are themselves polynomials in $X_1, \ldots, X_n$ (since $f_j \in k[X_1, \ldots, X_n]$). Hence each minor is a polynomial $g_\alpha \in k[X_1, \ldots, X_n]$ and \begin{align*} S = V \cap \bigcap_\alpha V(g_\alpha), \end{align*} which is Zariski-closed in $V$. Therefore \begin{align*} V_{\mathrm{sm}} := V \setminus S \end{align*} is Zariski-open in $V$. [guided] We translate the rank condition on $J(p)$ into polynomial equations on $p$. Recall: an $m \times n$ matrix $M$ has $\operatorname{rank}(M) < r$ iff every $r \times r$ minor of $M$ vanishes. With $r = n - d$: \begin{align*} \operatorname{rank} J(p) < n - d \iff \det(J(p)|_\alpha) = 0 \text{ for every } (n-d) \times (n-d) \text{ submatrix indexed by } \alpha. \end{align*} For fixed $\alpha = (\alpha_R, \alpha_C)$ with $\alpha_R \subset \{1, \ldots, m\}$ and $\alpha_C \subset \{1, \ldots, n\}$ of sizes $n-d$, the determinant $g_\alpha(p)$ is a polynomial in the $\{\partial f_j / \partial X_i (p) : j \in \alpha_R, i \in \alpha_C\}$; each partial derivative is itself a polynomial in $X_1, \ldots, X_n$, so $g_\alpha$ is a polynomial. The non-smooth locus $S$ is cut out by the $g_\alpha$, hence closed; its complement $V_{\mathrm{sm}}$ is open. Non-emptiness is the next step. [/guided] [/step] [step:Reduce non-emptiness to exhibiting a smooth point on a birational hypersurface model] We now show $V_{\mathrm{sm}}$ is non-empty. We use two reductions. \textbf{Reduction 1 (birational).} If $V$ and $W$ are birationally equivalent irreducible varieties, they share a common Zariski-open subset $U$: concretely, birational equivalence gives rational maps $\varphi: V \dashrightarrow W$ and $\psi: W \dashrightarrow V$ with $\psi \circ \varphi = \mathrm{id}_V$ and $\varphi \circ \psi = \mathrm{id}_W$, so there exist dense open $U_V \subset V$ and $U_W \subset W$ such that $\varphi|_{U_V} : U_V \xrightarrow{\sim} U_W$ is an isomorphism. Since $U_V$ is open in $V$ and $U_W$ is open in $W$, and smoothness is a local property (the Jacobian/tangent-space description depends only on a Zariski-open neighbourhood of $p$), a point $q \in U_W$ is smooth as a point of $W$ iff the corresponding point $p = \psi(q) \in U_V$ is smooth as a point of $V$. In particular, $V_{\mathrm{sm}} \neq \varnothing \iff W_{\mathrm{sm}} \neq \varnothing$ when $V$ and $W$ are birational irreducible varieties. By [Birational Varieties Same Dim](/theorems/2151), birational equivalence preserves dimension, so the hypersurface model we produce below will have dimension $d = \dim V$. \textbf{Reduction 2 (hypersurface model).} By [Noether Normalisation](/theorems/2152), there exist $x_1, \ldots, x_d \in k(V)$ algebraically independent over $k$ such that $k(V)$ is finite over $k(x_1, \ldots, x_d)$; by [Primitive Generator FF](/theorems/2153), there exists $y \in k(V)$ with \begin{align*} k(V) = k(x_1, \ldots, x_d, y). \end{align*} Let $p(Y) \in k(x_1, \ldots, x_d)[Y]$ be the minimal polynomial of $y$ over $k(x_1, \ldots, x_d)$. Clearing denominators (multiplying by a nonzero element of $k[x_1, \ldots, x_d]$), we obtain an irreducible polynomial \begin{align*} f(X_1, \ldots, X_d, Y) \in k[X_1, \ldots, X_d, Y] \end{align*} whose image in $k(x_1, \ldots, x_d)[Y]$ is (a nonzero scalar multiple of) $p(Y)$. The affine hypersurface \begin{align*} W := V(f) \subset \mathbb{A}^{d+1}_k \end{align*} is irreducible (since $f$ is irreducible) and has function field $k(W) = k(x_1, \ldots, x_d)(y) = k(V)$. Hence $V$ and $W$ are birationally equivalent. By Reduction 1, it suffices to exhibit a smooth point on $W$. [/step] [step:Use separability to produce a non-vanishing partial derivative on the hypersurface] The hypersurface $W = V(f) \subset \mathbb{A}^{d+1}_k$ has $\dim W = d$, so at $p \in W$ smoothness is equivalent to \begin{align*} \operatorname{rank} J(p) = (d+1) - d = 1, \end{align*} i.e. the $1 \times (d+1)$ row $J(p) = (\partial f/\partial X_1(p), \ldots, \partial f / \partial X_d(p), \partial f / \partial Y(p))$ is nonzero. [claim:$\partial f / \partial Y \neq 0$ as a polynomial in $k[X_1, \ldots, X_d, Y]$] [proof] View $f$ as an element of $k(x_1, \ldots, x_d)[Y]$ by inverting the nonzero scalar used in clearing denominators; the resulting polynomial is a nonzero scalar multiple of the minimal polynomial $p(Y)$ of $y$ over $L := k(x_1, \ldots, x_d)$. Because $L$ has characteristic zero, the extension $L(y)/L$ is separable. Equivalently, $p(Y)$ is a separable polynomial over $L$ — its formal derivative $p'(Y) \in L[Y]$ is nonzero (and coprime to $p$). If $p'(Y) = 0$ in $L[Y]$, then $p(Y)$ would have to be a polynomial in $Y^q$ for $q = \operatorname{char} L = 0$ — impossible since $\operatorname{char} L = 0$. Hence $p'(Y) \neq 0$ in $L[Y]$. Since $f$ and $p$ agree up to a nonzero scalar in $L[Y]$, so do their formal $Y$-derivatives: $(\partial f / \partial Y) \in L[Y]$ is a nonzero scalar multiple of $p'(Y)$, hence nonzero in $L[Y]$. A fortiori $\partial f / \partial Y \neq 0$ as an element of $k[X_1, \ldots, X_d, Y]$ (since the map $k[X_1, \ldots, X_d, Y] \to L[Y]$ given by inverting nonzero elements of $k[X_1, \ldots, X_d]$ is injective — it is localisation at a nonzero multiplicative set in a domain). [/proof] [/claim] [claim:$\partial f / \partial Y$ does not vanish identically on $W$] [proof] Suppose for contradiction $\partial f / \partial Y$ vanishes on $W = V(f)$. Since $W$ is irreducible with ideal $I(W) = (f)$ (as $f$ is irreducible and $W = V(f)$), we would have $f \mid \partial f / \partial Y$ in $k[X_1, \ldots, X_d, Y]$. Viewing both in $L[Y]$ with $L = k(x_1, \ldots, x_d)$: $\deg_Y f = \deg p$ and $\deg_Y (\partial f / \partial Y) = \deg p - 1 < \deg p$ (since $\partial f / \partial Y \neq 0$ by the previous claim and the derivative of a nonzero polynomial drops degree by one in characteristic zero). A nonzero polynomial of smaller $Y$-degree cannot be divisible by $f$ in $L[Y]$, hence cannot be divisible in $k[X_1, \ldots, X_d, Y]$ either (since the inclusion $k[X_1, \ldots, X_d, Y] \hookrightarrow L[Y]$ is a localisation, hence flat, and divisibility is preserved). Contradiction. [/proof] [/claim] Since $\partial f / \partial Y$ is a nonzero polynomial not vanishing identically on the irreducible $W$, the set $V(\partial f / \partial Y) \cap W$ is a proper Zariski-closed subset of $W$, so \begin{align*} W \setminus V(\partial f / \partial Y) \subset W \end{align*} is a nonempty Zariski-open subset. At every point $p$ of this open set, $\partial f / \partial Y (p) \neq 0$, so $J(p) \neq 0$, so $p$ is a smooth point of $W$. [guided] We need a point of $W$ at which the gradient of $f$ is nonzero. Since $W = V(f)$ in $\mathbb{A}^{d+1}_k$, the Jacobian is the row $\nabla f(p)$, and smoothness reduces to $\nabla f(p) \neq 0$. The strategy: show $\partial f / \partial Y$ does not vanish identically on $W$. Then the open set $W \setminus V(\partial f / \partial Y)$ is nonempty (by irreducibility of $W$), and every point in it is smooth. Why is $\partial f / \partial Y \neq 0$ as a polynomial? This is where characteristic zero enters. Inverting the nonzero scalar used in clearing denominators, we can view $f$ in $L[Y]$ where $L = k(x_1, \ldots, x_d)$, and there $f$ agrees with the minimal polynomial $p(Y)$ of $y$ over $L$ up to a nonzero scalar. The extension $L(y)/L$ is finite, hence separable (in characteristic zero all finite extensions are separable), hence $p'(Y) \neq 0$ in $L[Y]$. Scaling, $\partial f / \partial Y \neq 0$ in $L[Y]$, hence in $k[X_1, \ldots, X_d, Y]$. Why does $\partial f / \partial Y$ not vanish on $W$? If it did, then $f$ would divide $\partial f / \partial Y$ in $k[X_1, \ldots, X_d, Y]$ (because $I(W) = (f)$). But in $L[Y]$, $\deg_Y (\partial f/\partial Y) < \deg_Y f$, so divisibility in $L[Y]$ fails, hence divisibility in $k[X_1, \ldots, X_d, Y]$ fails. (We use flatness of the localisation $k[X_1, \ldots, X_d, Y] \hookrightarrow L[Y]$ to transfer divisibility.) So the smooth locus of $W$ is nonempty. By Step 3 (Reduction 1), the smooth locus of $V$ is also nonempty. [/guided] [/step] [step:Conclude] By Step 2, $V_{\mathrm{sm}}$ is open in $V$. By Steps 3–4, $V_{\mathrm{sm}}$ is nonempty (the hypersurface model $W$ has a smooth point, and smoothness is preserved on the common open subset shared by $V$ and $W$). Since $V$ is irreducible, any nonempty open subset is dense. Hence $V_{\mathrm{sm}}$ is a nonempty open dense subset of $V$. [/step]