This degree is in all cases at most $(L+1)(n-k)$ times the number of not trivially zero fullsize minors of $\mathfrak{H}_L$.

It remains to estimate the degree of the polynomial $g$ from the preceding theorem to get an explicite bound for the field size.

\[M\] \
If $|\mathbb F|>\min\{M_1,M_2,M_3\}$ with $$\begin{aligned}
M_1&:=(L+1)(n-k)\binom{(L+1)n}{(L+1)(n-k)}\\
M_2&:=(L+1)(n-k)\binom{n}{n-k}\binom{n+k}{n-k}\cdots\binom{n+Lk}{n-k}\\
M_3&:=(L+1)(n-k)\sum_{i=n-k}^{n}\binom{i-1}{n-k-1}\binom{2n-i}{n-k}\cdots\binom{2n+(L-1)k-i}{n-k}
$$ then there exists an $(n,k,\delta)$ MDP convolutional code over $\mathbb F$.

 \
To show that $|\mathbb F|>M_i$ for some $i\in\{1,2,3\}$ is sufficient, one has to show that the factor after $(L+1)(n-k)$ in the formulas is an upper bound for the number of not trivially zero fullsize minors of $\mathcal{H}_L$. For $M_1$ this is clear because there we use just the formula for all fullsize minors.\
For $M_2$, we use the condition that we have to choose $n-k$ columns from the first $n$ columns, then $n-k$ columns from the first $2n$ columns without the $n-k$ columns we have already chosen and so on until we end up with choosing $n-k$ columns from $(L+1)n$ columns without the $L(n-k)$ columns we have already chosen.\
For $M_3$, we denote by $i\in\{n-k,\hdots,n\}$ the index of the $(n-k)$-th column we choose. Thus, one has to choose $n-k-1$ columns with smaller index than 