& 0_{(n-k)\nu-\delta} & 0
\end{array} \right)$ and $H_{\nu-1}=\left(\begin{array}{ccc}
\ast & 0 & \ast \\ 
0_{((n-k)\nu-\delta)\times(\delta-(n-k)(\nu-1))} & I_{(n-k)\nu-\delta} & \ast
\end{array} \right)$, respectively. In this way, one can ensure that the generated code has really the given degree $\delta$.

According to Theorem \[sz\] (a), a polynomial $g$ over $\mathbb F$ with $m$ variables (and $\deg(g)\geq 0$) has at most $\deg(g)\cdot|\mathbb F|^{m-1}$ zeros. Altogether, there are $|\mathbb F|^m$ tuples of points. Therefore, for having at least one of them being not a zero, it is sufficient that $|\mathbb F|^m>\deg(g)\cdot|\mathbb F|^{m-1}$, i.e. $|\mathbb F|>\deg(g)$.

We apply this result to the polynomial $g$ formed by the product of all not trivially zero fullsize minors of $\mathfrak{H}_L$.

Since some of the entries of $H_{\nu}$ and $H_{\nu-1}$ are fixed zeros or ones, one has less variables than the number of entries of the coefficient matrices of $H$ but this has no influence on the result (note that if $\nu>L$, i.e. $\nu=L+1$, and $\lfloor\frac{\delta}{k}\rfloor=0$, for which case we give a better bound in a later section, all entries of $H_{\nu}$ do not occur in the polynomial on which we apply Schwartz-Zippel). What influences Schwartz-Zippel is not the number of variables but the degree of the polynomial $g$. 