[proofplan] We first write $X$ inside [affine space](/page/Affine%20Space) by generators for its vanishing ideal and use the Jacobian criterion to express smoothness as a rank condition on the [Jacobian matrix](/page/Jacobian%20Matrix). That rank condition is detected by the nonvanishing of minors, so it defines an open subset of $X$. A standard generic [regularity theorem](/theorems/2750) for finitely generated reduced algebras over a perfect field gives at least one smooth point, and algebraically closed fields are perfect. Finally, irreducibility turns nonempty openness into density. [/proofplan]

[step:Represent the smooth locus by a Jacobian rank condition] Let \begin{align*}R:=k[x_1,\dots,x_n]\end{align*} be the [polynomial ring](/page/Polynomial%20Ring), and let \begin{align*}I(X)\trianglelefteq R\end{align*} be the vanishing ideal of $X$. Since $X$ is irreducible, $I(X)$ is prime. Choose generators \begin{align*}I(X)=(f_1,\dots,f_r)\end{align*} with $f_i\in R$ for each $1\le i\le r$. Let \begin{align*}d:=\dim X\end{align*} and set \begin{align*}c:=n-d.\end{align*} For each point $p\in X$, define the Jacobian matrix at $p$ by \begin{align*}J_p:=\left(\frac{\partial f_i}{\partial x_j}(p)\right)_{1\le i\le r,\ 1\le j\le n}.\end{align*} If $c=0$, then $\dim X=n$. Since $I(X)$ is a prime ideal of $k[x_1,\dots,x_n]$ of height $n-\dim X=0$, it follows that $I(X)=(0)$, so $X=\mathbb A_k^n$. Affine space is smooth at every point, hence $X_{\mathrm{sm}}=X$, which is nonempty, open, and dense. Assume now that $c\ge 1$. The ideal $I(X)$ is the full vanishing ideal of $X$ and is generated by $f_1,\dots,f_r$. Since $X$ is irreducible of dimension $d$, the only irreducible component of $X$ is $X$ itself, with dimension $d$, so every point $p\in X$ lies on an irreducible component of dimension $d$ and no component through $p$ has larger dimension. Also, the codimension of $X$ in $\mathbb A_k^n$ is $n-d=c$. Therefore the hypotheses of [citetheorem:9444] apply to the generators $f_1,\dots,f_r$ of $I(X)$. Hence, for every $p\in X$, \begin{align*}p\in X_{\mathrm{sm}} \iff \operatorname{rank}J_p=n-d=c.\end{align*} [/step]

[step:Convert the rank condition into a Zariski-open subset]Let $\mathcal M$ be the finite set of all $c\times c$ minors of the matrix \begin{align*}J:=\left(\frac{\partial f_i}{\partial x_j}\right)_{1\le i\le r,\ 1\le j\le n}\end{align*} with entries in $R$. For each minor $M\in\mathcal M$, define the principal open subset of $X$ \begin{align*} D_X(M):=\{p\in X:M(p)\ne 0\}. \end{align*} The function $M:X\to k$ is regular because $M$ is the restriction to $X$ of a polynomial in $R$, so $D_X(M)$ is open in the Zariski topology on $X$. For $p\in X$, the matrix $J_p$ has rank at least $c$ exactly when at least one $c\times c$ minor is nonzero at $p$. By the Jacobian criterion from the previous step, smoothness is equivalent to $\operatorname{rank}J_p=c$. Since this criterion identifies the smooth points exactly, we obtain \begin{align*} X_{\mathrm{sm}}=\bigcup_{M\in\mathcal M}D_X(M). \end{align*} Thus $X_{\mathrm{sm}}$ is open in $X$ as a finite union of open subsets.[/step]

[guided]The goal of this step is to turn the abstract condition “$p$ is smooth” into equations and inequalities in the Zariski topology. The Jacobian criterion has already reduced smoothness to the rank condition \begin{align*} p\in X_{\mathrm{sm}} \iff \operatorname{rank}J_p=c. \end{align*} Rank conditions are algebraic because matrix rank is detected by minors. Define $\mathcal M$ to be the finite set of all $c\times c$ minors of \begin{align*} J:=\left(\frac{\partial f_i}{\partial x_j}\right)_{1\le i\le r,\ 1\le j\le n}. \end{align*} Each element $M\in\mathcal M$ is a polynomial in $k[x_1,\dots,x_n]$, since it is a determinant of a matrix whose entries are polynomials. Therefore its restriction to $X$ is a regular function. For such a minor, define \begin{align*} D_X(M):=\{p\in X:M(p)\ne 0\}. \end{align*} This is open in the Zariski topology on $X$, because if \begin{align*} V(M):=\{q\in \mathbb A_k^n:M(q)=0\} \end{align*} denotes the affine hypersurface cut out by $M$, then its complement in $X$ is \begin{align*} X\cap V(M)=\{p\in X:M(p)=0\}, \end{align*} which is closed in $X$. Now fix $p\in X$. The matrix $J_p$ has a nonzero $c\times c$ minor exactly when its rank is at least $c$. The Jacobian criterion in the previous step says that the smooth points are precisely the points where the Jacobian rank is exactly $c$. Hence the points detected by the nonvanishing of these minors are exactly the smooth points: \begin{align*} X_{\mathrm{sm}}=\bigcup_{M\in\mathcal M}D_X(M). \end{align*} Because $\mathcal M$ is finite and each $D_X(M)$ is open in $X$, this union is open. The important point is that openness comes from a nonvanishing condition, not from solving equations: nonvanishing of at least one determinant is precisely what the Zariski topology treats as open.[/guided]

[step:Use generic regularity to show the smooth locus is nonempty] The coordinate ring \begin{align*} k[X]:=R/I(X) \end{align*} is a finitely generated reduced $k$-algebra, and it is an [integral domain](/page/Integral%20Domain) because $I(X)$ is prime. Since $k$ is algebraically closed, $k$ is perfect. By the generic regularity theorem for finitely generated reduced algebras over a perfect field, the regular locus in $\operatorname{Spec} k[X]$ contains a nonempty open subset. This nonempty open subset contains a closed point because $k[X]$ is a finitely generated algebra over the [algebraically closed field](/page/Algebraically%20Closed%20Field) $k$ and $X$ is nonempty. Let $p\in X$ be the corresponding closed point, and let $\mathcal O_{X,p}$ denote the local ring of $X$ at $p$. The chosen point lies in the regular locus, so $\mathcal O_{X,p}$ is a regular local ring. By the definition of smooth point in the theorem statement, $p\in X_{\mathrm{sm}}$. Therefore \begin{align*} X_{\mathrm{sm}}\ne\varnothing. \end{align*} [/step]

[step:Deduce density from irreducibility] We have proved that $X_{\mathrm{sm}}$ is a nonempty open subset of $X$. Since $X$ is irreducible as a [topological space](/page/Topological%20Space), [citetheorem:9409] applies and gives that every nonempty open subset of $X$ is dense in $X$. Applying this to $U:=X_{\mathrm{sm}}$ gives \begin{align*} \overline{X_{\mathrm{sm}}}=X, \end{align*} where $\overline{X_{\mathrm{sm}}}$ denotes the closure of $X_{\mathrm{sm}}$ in the topological space $X$. Thus the smooth locus is a dense open subset of $X$, as required. [/step]