  Maximum distance profile (MDP) convolutional codes have the property that
their column distances are as large as possible for given rate and degree.
There exists a well-known criterion to check whether a code is MDP using the
generator or the parity-check matrix of the code.
  In this paper, we show that under the assumption that $n-k$ divides $\delta$
or $k$ divides $\delta$, a polynomial matrix that fulfills the MDP criterion is
actually always left prime. In particular, when $k$ divides $\delta$, this
implies that each MDP convolutional code is noncatastrophic. Moreover, when
$n-k$ and $k$ do not divide $\delta$, we show that the MDP criterion is in
general not enough to ensure left primeness. In this case, with one more
assumption, we still can guarantee the result.
