that $H$ has generic row degrees since the existence of an MDP convolutional code with generic row degrees over $\mathbb F$ obviously implies the existence of an MDP convolutional code over $\mathbb F$. The genericity of the row degrees implies $\nu=\lceil\frac{\delta}{n-k}\rceil$ and therefore $L= \lfloor\frac{\delta}{n-k}\rfloor\leq\nu$. If $(n-k)\mid \delta$, i.e. $L=\nu$, all row degrees of $H$ are equal to $\nu$ and hence $H_L=H_{\nu}$ does not contain fixed zeros. If $(n-k)\nmid\delta$, i.e. $L=\nu-1$ and all row degrees of $H$ are either equal to $\nu-1$ or equal to $\nu$, $H_L$ does not contain fixed zeros, too.

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There exists an MDP convolutional code with $\delta<k$ over $\mathbb F$ if either

1.  $|\mathbb F|>\binom{(L+1)n-1}{(L+1)(n-k)-1}$ **or**

2.  in the case $L\geq 1$, $|\mathbb F|>S(n,k,\delta)$ with $$\begin{aligned}
    S(n,k,\delta)&:=\sum_{j=n-k+1}^{(n-k)L-1}\binom{n-1}{j-1}\binom{\lfloor\frac{j}{n-k}\rfloor n}{(\lfloor\frac{j}{n-k}\rfloor+1)(n-k)-j}\binom{(\lfloor\frac{j}{n-k}\rfloor+1)n}{n-k}\cdots\binom{Ln}{n-k}+\\
    &+\sum_{j=\max\{(n-k)L,n-k+1\}}^{(n-k)(L+1)-1}\binom{n-1}{j-1}\binom{Ln}{(L+1)(n-k)-j}+\binom{n-1}{(L+1)(n-k)-1}\end{aligned}$$

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For $y\in\{1,\hdots,(L+1)(n-k)\}$, define $\mathcal{H}_L^{(y)}$ as the matrix consisting of the first $y$ rows of $\mathcal{H}_L=\left[\begin{matrix} H_0 & & 0\\ \vdots & \ddots & \\ H_{L} & \cdots & H_0 \end{matrix}\right]$.

We prove via induction with respect to $y$ that if $\mathbb F$ fullfilles condition $1$ or $2$, then it is possible to find values for $H_0,\hdots,H_L$ over $\mathbb F$ such that every fullsize minor of $\mathcal{H}_L^{(y)}$ that is not trivially zero is nonzero.

For 