smooth curves passing through a smooth rational point is dense Let $X$ be an integral scheme of finite type over a field $k$, and $x$ be a smooth rational point on it. Why the union of smooth curves passing through $x$ is dense in $X$?
This is a statement given in Florian Pop - Embedding Problems Over Large Fields, on page 5.
If $k$ is infinite, I may argue as follows: if not, then all smooth curves are contained in a closed subset $Z$ of lower dimension. Write it as union of irreducible components $Z_i$, there must be some tangential direction not contained in any $Z_i$, but along which we can find a smooth curve. However, this does not apply to finite fields.
Specific reasons for finite fields and a uniform proof are both appreciated.
 A: I’m assuming that the statement you want to show is the following one: for every nonempty Zariski open $U \subset X$, there is a connected one-dimensional scheme $C$, smooth over $k$, and an immersion $C \rightarrow X$ through which $x \in X(k)$ factors (curves going through $x$ in the following) such that the image of $C$ in $X$ meets $U$.
Since $X$ is integral, it is enough to show that a sufficiently small open subset of $X$ (containing $x$) satisfies the property. Because $X$ is smooth at $x$, there is an open subset $x \in U \subset X$ such that $U$ is smooth over $k$. Then there is an open subset $x \in V \subset U$ endowed with an étale $k$-map $V \rightarrow \mathbb{A}^r_k$. So we can assume that we have an étale $k$-map $f: X \rightarrow \mathbb{A}^d_k$.
Let $j: C’ \rightarrow \mathbb{A}^d_k$ be a curve going through $f(x)$, and let $C=f^{-1}(C’)$ (considered as a subscheme of $X$). Then $C \rightarrow C’$ is a base change of $f$ so is étale, so $C$ is a smooth one-dimensional $k$-scheme, and $x$ factors through $C$.
Then the connected component $C_1$ of $x$ in $C$ is a curve going through $x$, and $f(C_1)$ is an open subset of $C’$ (since $f: C \rightarrow C’$ is étale, it is open), so it is cofinite in $C’$.
Now, assume that $F \subset X$ is a closed subset such that no smooth curve going through $x$ leaves $F$. Let $T$ be the (closure of the) scheme-theoretic image of $F$ through $f$. Then $\dim{T} \leq \dim{F} < \dim{X}=d$. $C’ \cap T$ is a closed subset of $C’$ which contains $f(C_1)$ (which is co-finite), so it is full, and $C’ \subset T$.
Let’s recap: we have a Zariski-closed proper subset $T \subset \mathbb{A}^d_k$ and a point $f(x) \in T$ such that every $k$-smooth curve through $f(x)$ remains in $T$. Clearly, $d>1$ and we can assume $f(x)=0$.
There exists a nonzero $P \in k[X_1,\ldots,X_d]$ be a polynomial vanishing over $T$. Clearly, $P$ vanishes at all points in $k^d$. Thus $k$ is finite.
Consider a finite field extension $F=k(\alpha)$ of large enough cardinality. Let $Q_2,\ldots,Q_d \in k[T]$ be arbitrary polynomials. Then $P$ vanishes on the image of $(T,Q_2,\ldots,Q_d)$ – so $P(\alpha, Q_2(\alpha),\ldots,Q_d(\alpha))=0$. Thus $P$ vanishes on $\{\alpha\} \times F^{d-1}$. Thus $P$ vanishes on $G \times F^{d-1}$ where $G$ is the set of generators of $F/k$ (it has arbitrarily large cardinality if $F$ is large enough). Taking $F$ (and $G$) large enough wrt the degree of $P$, it follows that $P=0$.

The argument can be seen as follows:

*

*reduce to the case of $X=\mathbb{A}^d_k$ and $x=0$.

*turn this into a commutative algebra statement: if $P \in k[X_1,\ldots,X_d]$ is such that $P(Q_1(t),\ldots,Q_d(t))=0$ for all $(Q_1,\ldots,Q_d)$ such that $k[t]=k[Q_1(t),\ldots,Q_d(t)]$, then $P=0$.

*solve.

Clearly, 3) is the part which still needs a uniform proof. Here it is. Let $p$ be a prime number invertible in $k$. Consider the $d$-uple $Q(T)=(T^{n_1},\ldots,T^{n_d})$ where all the $n_d$ are positive integers and one of them is one. Then $P(Q(T))=0$. Let $T$ run through the $p^{\infty}$-roots of unity (in an algebraic closure of $k$), and $Q$ over all possible $d$-uples, then $Q(T)$ runs through all the $d$-uples of $p^{\infty}$-roots of unity. So $P$ vanishes on $\mu_{p^{\infty}}^d$, and $\mu_{p^{\infty}}$ is infinite, so $P=0$.
