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If $(n-k)\mid\delta$ and $|\mathbb F|>\min\{N_1,N_2\}$ with $$\begin{aligned}
N_1&:=(L+1)(n-k)\binom{(L+1+\frac{\delta}{n-k})n}{(L+1)(n-k)}\\
N_2&:=(L+1)(n-k)(\frac{\delta n}{n-k} +k+1)^{(n-k)(L+1)},\end{aligned}$$ then there exists an $(n,k,\delta)$ complete MDP convolutional code over $\mathbb F$.

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Each fullsize minor of $\mathfrak{H}$ is a polynomial of degree $(L+1)(n-k)$. Moreover, the number of not-trivially zero fullsize minors of $\mathfrak{H}$ is upper bounded by $\binom{L+1+\frac{\delta}{n-k}}{(L+1)(n-k)}$, which is the number of all fullsize minors, as well as by $(\frac{\delta n}{n-k} +k+1)^{(L+1)(n-k)}$ since the index of each chosen column has to lie in an interval with $\frac{\delta n}{n-k} +k+1$ elements (see Lemma \[index\]).

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(i) It depends on the parameters of the code, which of the two bounds $N_1$ or $N_2$ is better, i.e. smaller. For example for $k=n-1$, the second bound is better for $n=2$, for $n=3$ the bounds are identical, and for $n\geq 4$ the first bound is better.\
(ii) For complete MDP, one has $\nu=\frac{\delta}{n-k}$ and hence $L\geq\nu\geq 1$, which implies $\binom{\nu n+k}{\lfloor 1/2(\nu n+k)\rfloor}^{(n-k)(L+1)}\cdot((n-k)(L+1))^{1/2(n-k)(L+1)}\geq \binom{\nu n+k}{\lfloor 1/2(\nu n+k)\rfloor}^{(n-k)(L+1)}\cdot(n-k)(L+1)\geq (L+1)(n-k)(\frac{\delta n}{n-k} +k+1)^{(n-k)(L+1)}$ if $(n,k,\delta)\neq(2,1,1)$. This shows that - unless $(n,k,\delta)=(2,1,1)$ - the bound on the field size presented here is better than the bound obtained by the construction in [@cmdp], which is clearly very weak (which is due to the fact that it 