"A spinning top spins much longer because it experiences less frictional torque" is wrong? The above quote was found in my physics textbook, but it struck me as strange because my understanding of friction is that the surface area doesn't matter in calculating the amount of frictional force.
Another question that asked a similar thing on stackexchange was answered basically by saying that a spinning top with a narrow point spins better and longer because of "precession"?
Why does a top spin so well?
So my question is: is the above statement just flat out wrong? Is the reason it spins much longer not because of torque, but because of other properties of a narrow point?
 A: There are two types of friction, static and kinetic friction. Imagine trying to push a table across a carpet. Initially you need to generate some force to get the table to move, but once the table starts moving you may feel it's easier to push it further.
The type of friction that is important for your spinning top example would be the kinetic friction, because it resists the spinning motion of your top, similarly to when you feel some resistance while pushing a table across a carpet. If you let go of the top after you give it some initial velocity, the friction force will cause a torque opposing the top's spinning direction, which will cause the angular velocity of the top to decrease.
Remember that friction is caused between materials because they are pretty rough and jagged at a very small scale, even a perfectly flat glass table may be very rough when you look at it with a powerful microscope. The spinning top's point is also not perfectly smooth so there will inevitably be friction between the top and the glass.
However, if you would spin the top on a much rougher material such as a sponge or in sand, the effects are amplified and the top will slow down much faster.
A: The surface area doesn't matter because $F_a = \mu N$. If the contact area is very small as in a top, the normal is the same, and $F_a$ doesn't change.
But the average distance ($d$) between the center of spin in the ground and the other points of contact (because the "point" of contact indeed has some area) is very small. 
So the resisting (friction) torque, $T = F_a d$ is as small as the contact area tends to a theoretical point. 
A: The amount of frictional force for a linear moving object doesn't depend on its contact surface with the surface it's moving on. As long the object has the same mass (so the normal force stays the same) it doesn't depend on how the object moves over the surface (if there is enough friction available, which depends on both the surface of the object and the underground). The object can have any form. So if we give two objects, differing only in their form, they come at rest at the same distance from a line where we give them the same velocity. Because of the (non-conservative) friction force that's directed opposite to the velocity of the objects.
In the case of a spinning object like the top (let's assume no linear motion for the center of mass and no friction with the surrounding air) one can not say the same. Let's change the surface of the almost point-like bottom of the top into a not-almost-pointlike-bottom (say a cylindrical disc) while leaving the mass of the top the same (of course the top with a cylindrical bottom is not a top anymore, but I use it to make something clear).
When in this case we give the two rotating objects an equal initial rotation, it's clear that the object with the cylindrical disc beneath it will come faster to a full stop than the pointy one (the top).
This means it is indeed the collection of torques (which is bigger in the cylindrical bottom side than in the point-like from of the torque).
So your teacher is right. I don't see what other properties of the narrow point make the rotation of the top decrease.
