I just realized that my early post had some error in it, so I fix it here:
Now according to this, I think you are able to answer the questions huddling in your mind:
Take the metric
where [tex]a[/tex] is some constant. [tex](1+ax)^2[/tex] has a non-zero second derivative wrt x, but the metric is flat!!
This is not right unless one hits
I made it flat! Look at the example above!
You just need to have the Riemann tensor vanished to get a flat spacetime; this can't always be understood from the metric directly. For instance, the above metric has non-vanishing Christoffel symbols but is not curved since the Riemann tensor is non-zero. These exceptional cases do not let us distinguish flat spacetimes from curved ones. In GR, one uses the geodesic coordinates system to make sure that the metric gets flat locally, because once we understand that Christoffel symbols vanish, then the Riemann tensor is necessarily zero but the inverse is not true always, as you can see!
Only locally ture! Why don't you draw a tiny attention to my notes? I said that Schwarzschild connections (Christoffel symbols) do not vanish so the spacetime is curved everywhere and tends to be Minkowski (flat) at large distances (r-->oo) from the source; leading connections to vanish locally because they are of the dimension 1/r outside the source of gravitational field!