Why does Shamir's Secret Sharing Scheme need a finite field?
The simplest answer is probably to give an example of information leaked when using Shamir's secret sharing over the integers. Assume that we construct a low degree example, defining $q$ to be a linear polynomial with $q(0)=D$ and $q(1)=a_1$. By interpolation you find that: $$q(x)=(a_1-D)x+D.$$ 

Assume that you are given the share corresponding to evaluation at $2$, i.e. $q(2)$. You can see that $q(2)=2a_1-D$. Since $a_1$ and $D$ are integers, given this single share, you learn the parity of $D$.