Linear [12,9,4]-code over F11 Concerning linear error-correcting codes over finite fields of odd characteristic.
Over $\mathbb F_{13}$ I can use the following Vandermonde-style parity check matrix:
$$H=\begin{pmatrix}1&1&1&1&1&1&1&1&1&1&1&1\\1& 2 &3 &4 &5& 6 &7& 8 &9& 10&11 &12\\1^2& 2^2 &3^2 &4^2 &5^2& 6^2 &7^2& 8^2 &9^2& 10^2&11^2 &12^2\end{pmatrix}$$ 
(How to derive a generator matrix from this can be found in any decent book on error correction)
This code has minimal distance d = n-k+1 = 4, since (Justesen-Hoholdt, Lemma 1.2.3) any collection of three columns of $H$ has determinant $(a-b)(a-c)(b-c)\quad, a,b,c\in \text{second row of  $H$}$, which is nonzero mod 13.
My question ist: Can I construct a [n=12,k=9,d=4] linear code over $\mathbb F_{\boldsymbol{11}}$ as well? Or can one prove that this is not possible?
There are some lists of existing codes online, but they do not cover the case $\mathbb F_{11}$. 

This is possible.
Consider the check matrix
$$H=\begin{pmatrix}1&1&1&1&1&1&1&1&1&1&1&0\\0&1& 2 &3 &4 &5& 6 &7& 8 &9& 10&0\\0^2&1^2& 2^2 &3^2 &4^2 &5^2& 6^2 &7^2& 8^2 &9^2& 10^2&1\end{pmatrix}.$$ 
We need to prove that any three columns are linearly independent over $\Bbb{F}_{11}$. If the triple of columns does not include the last column, then this follows from a Vandermonde argument as in the check matrix you used to produce a $[12,9,4]_{13}$ code. On the other hand, if the triple includes the last column and any two others we can expand the resulting determinant along that last column. We end up with a 2x2 Vandermonde determinant, and the claim follows again.
