Are smooth varieties locally isomorphic to the affine space? A smooth $n$-dimensional manifold is locally isomorphic to $\mathbb{R}^n$. I am wondering if the analogous statement for smooth algebraic varieties is also true.

Let $X$ be an $n$-dimensional connected smooth variety, and $p \in X$ an arbitrary point. Is there an open subset $U \subseteq X$ such that $p \in U$ and $U \cong \mathbb{A}^n$?

Thank you in advance!
 A: Definitely not.  You should think of smoothness as an analytic property, not an algebraic one.  So a smooth variety over $\mathbb{C}$ locally looks like $\mathbb{C}^n$ in the analytic topology, but usually not in the Zariski one.
There are already lots of counterexamples in dimension $1$.  No smooth curve of genus $>0$ can have an open subset isomorphic to an open subset of $\mathbb{A}^1$.  This is because smooth, projective curves are isomorphic if and only if they are birational.  In other words, a smooth projective curve is determined, up to isomorphism, by its behavior on any open set.  And every smooth curve is a smooth projective curve with finitely many points missing.
The closest notion to the one you're talking about is rationality.  But this is a very, very restricted class of things.  In practice people are more interested in things like unirational varieties, those with dominant maps from rational varieties.
To get an idea of just how difficult the situation is in algebraic geometry, compared with differential geometry, look at the Jacobian conjecture.  Given a polynomial map $\mathbb{C}^2\to\mathbb{C}^2$ whose Jacobian determinant is nowhere vanishing (i.e. a nonzero constant), must the map be an isomorphism?  This algebraic version of the inverse function theorem is a famously difficult unsolved problem, even after it was reduced to the case where the polynomials have degree $3$.  It is very, very difficult to import intuition and theorems from the theory of manifolds, particularly when inverse functions are involved (because we don't care when functions exist, we care whether they arise from polynomials).
