Why does this state have a Schmidt rank of 1? A system is entangled if and only if the Schmidt rank is greater than 1.
Why does this $$\left[\frac{1}{\sqrt{2}}\left(\left|0\right\rangle+\left|1\right\rangle\right)\right]\otimes\left[\frac{1}{\sqrt{2}}\left(\left|0\right\rangle+i\,\left|1\right\rangle\right)\right] = \frac{\left|00\right\rangle+i\,\left|01\right\rangle+\left|10\right\rangle+i\,\left|11\right\rangle}{2}$$
have a Schmidt rank equal to one? or equivalently why is this not entangled? ($\otimes$ denotes the tensor product)
 A: Notice that the state you have written down is in the form $|u\rangle \otimes |v\rangle$.  It is, by definition, an unentangled state because an state is said to be unentangled if and only if it can be written as a tensor product of states.
Now, the question becomes why its schmidt rank is equal to $1$.  Well, consider the following orthonormal bases for the Hilbert space spanned by $\{|0\rangle, |1\rangle\}$
$$
  B_1 = \{|u\rangle, |u'\}, \qquad B_2 = \{|v\rangle, |v'\rangle\}
$$
where
$$
  |u\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle), \qquad |u'\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
$$
and
$$
  |v\rangle = \frac{1}{\sqrt{2}}(|0\rangle + i|1\rangle), \qquad |v'\rangle = \frac{1}{\sqrt{2}}(|0\rangle - i|1\rangle)
$$
The Schmidt decomposition of $|u\rangle\otimes |v\rangle$ in these bases is
$$
  |u\rangle\otimes |v\rangle = 1\cdot|u\rangle\otimes |v\rangle + 0\cdot|u'\rangle\otimes |v\rangle + 0\cdot|u\rangle\otimes |v'\rangle+ 0\cdot|u'\rangle\otimes |v'\rangle
$$
Notice that there is only one Schmidt (non-negative) coefficient in this decomposition, so the Schmidt rank of the state is $1$.  Actually, the following fact is generally true
Fact. A state is unentangled if and only if its Schmidt rank is $1$.
So once you know that the Schmidt rank is 1, then you know that the state can be written as a tensor product of two states, and vice versa.
