Curve through a point on a variety Let $k$ be a not-necessarily algebraically closed field of characteristic zero. Let $X$ be a positive-dimensional projective variety over $k$. Let $x$ be a closed point on $X$. Does there exist a curve over $k$ on $X$ which contains this point?
Variety = geometrically integral quasi-projective $k$-scheme
Curve = $1$-dimensional variety.
What about the special case where $x$ is a $k$-rational point? 
I can blow-up $X$ at $x$ and take the image in $X$ of an effective ample Cartier divisor via this blow-up and reason by induction. But I'm afraid this doesn't give me a geometrically connected curve passing through $x$.
 A: Your idea is good. Let $X'\to X$ be the blowup along $x$. Then $X'$ is projective, geometrically integral and of dimension $\dim X>1$. Embed $X'$ in some $\mathbb P^n_k$. 
When $k$ is infinite, by Jouanolou, "Théorèmes de Bertini et applications" (Progress in Maths), Corollaire 6.11 (2)+(3), there exists a hyperplane $H$ such that $H\cap X'$ is geometrically integral. 
When $k$ is finite, the existence of such a hypersurface $H$ is proved in Poonen "Bertini theorem over finite fields", Ann. Math. (2000), Proposition 2.7. 
Now the image of $H\cap X'$ in $X$ is a geometrically integral closed subscheme of $X$ passing throught $x$ of dimension $<\dim X$. By induction we find a geometrically  integral curve in $X$ passing through $x$. 
Edit In fact through any closed finite subset $Z$ of $X$, it passes a geometrically integral curve in $X$. The proof is the same, we just blowup $X$ along $Z$ instead of $x$.  
A: See Liu, Exercise 8.1.5. If a connected $k$-variety has a $k$-rational point, it is geometrically connected (Liu, Exercise 3.2.11).
