  In this short note we show two completely opposite methods of constructing
entangled states. Given a bipartite state $\gamma\in M_k\otimes M_k$, define
$\gamma_S=(Id+F)\gamma (Id+F)$, $\gamma_A=(Id-F)\gamma(Id-F)$, where $F\in
M_k\otimes M_k$ is the flip operator. In the first method, entanglement is a
consequence of the inequality
$\text{rank}(\gamma_S)<\sqrt{\text{rank}(\gamma_A)}$. In the second method,
there is no correlation between $\gamma_S$ and $\gamma_A$. These two methods
show how diverse is quantum entanglement.
  We prove that any bipartite state $\gamma\in M_k\otimes M_k$ satisfies
  $\displaystyle SN(\gamma)\geq\max \left\{
\frac{\text{rank}(\gamma_L)}{\text{rank}(\gamma)},
\frac{\text{rank}(\gamma_R)}{\text{rank}(\gamma)}, \frac{SN(\gamma_S)}{2},
\frac{SN(\gamma_A)}{2} \right\},$ where $SN(\gamma)$ stands for the Schmidt
number of $\gamma$ and $\gamma_L,\gamma_R$ are the marginal states of $\gamma$.
  We also present a family of PPT states in $M_k\otimes M_k$, whose members
have Schmidt number equal to $n$, for any given $1\leq n\leq
\left\lceil\frac{k-1}{2}\right\rceil$. This is a new contribution to the open
problem of finding the best possible Schmidt number for PPT states.
