aHere is particular case showing that $d(n,k)=d_K(n,k)$ definitely depends on the field $K$ in general, beyond pathologies of finite fields:

$d_{\mathbf{R}}(4,2)=4$

(although $d(4,2)=5$ in the algebraically closed case).

Possibly a similar argument yields $d_K(4,2)=4$ as soon as $K$ has non-squares but I haven't checked details.

Consider the 2-planes in $K^4$ $$A=\{(t,s,0,0)\},\;B=\{(0,0,t,s)\},\; C=\{(t,s,t,s)\},\;D=\{(t,t+2s,t+s,s)\},$$ where $(t,s)$ is understood to range over $K^2$. Let's prove that

$\{A,B,C,D\}$ is a complement repository as soon as $-1$ is not a square in $K$.

Proof:

For the moment, $K$ is an arbitrary field.

For $(e,f)\in K^2\smallsetminus \{(0,0)\}$, define $P_{e,f}$ to be the 2-plane $\{(te,tf,se,sf):(t,s)\in K^2\}$. Write $\mathcal{P}=\{P_{e,f}:(e,f)\neq (0,0)\}$.

Clearly $P_{e,f}$ intersects nontrivially each of $A,B,C$ (taking $s=0$, $t=0$, $t=s$ respectively). I claim that conversely if a 2-plane $P$ intersects nontrivially each of $A,B,C$, then $P\in\mathcal{P}$. This is a simple exercise. Indeed, if $P$ intersects $A,B$ nontrivially, then $P=(P\cap A)\oplus (P\cap B)$ and hence there exists $(c,f,a,b)$ with $(c,f),(a,b)\neq (0,0)$ such that $P=\{(tc,tf,sa,sb):(t,s)\in K^2\}$. If $P\cap C$ is nontrivial, for some $(t,s)\neq (0,0)$ we have $tc=sa$ and $tf=sb$. This means that both $(a,c)$ and $(b,f)$ are collinear to the nonzero vector $(t,s)$. If $(b,f)=(0,0)$ we deduce $P=P_{1,0}$. Otherwise, $(a,c)=e(b,f)$ for some $e\in K$. So we have $P=\{(tef,tf,seb,sb):(t,s)\in K^2\}=P_{e,1}$.

Therefore, the final claim that $\{A,B,C,D\}$ is a complement repository is equivalent to showing that $D\cap P_{e,f}=\{0\}$ for every $(e,f)\in K^2 \smallsetminus\{(0,0)\}$.

Since a basis of $P_{e,f}$ is $((e,f,0,0),(0,0,e,f))$ and a basis of $D$ is $((0,2,1,1),(1,1,1,0))$, this amounts to showing that the determinant $\delta(e,f)=\begin{vmatrix}e & f & 0 & 0\\0 & 0 & e & f\\ 0 & 2 & 1 & 1\\ 1 & 1 & 1 & 0\end{vmatrix}$ is nonzero for all $(e,f)\in K^2\smallsetminus\{(0,0)\}$. The computation yields $\delta(e,f)=e^2+f^2$.

Therefore the non-vanishing holds if and only if $-1$ is not a square in $K$. This means that $\{A,B,C,D\}$ is a complement repository if and only if $-1$ is not a square in $K$.

Here is a particular case showing that $d(n,k)=d_K(n,k)$ definitely depends on the field $K$ in general, beyond pathologies of finite fields:

$d_{\mathbf{R}}(4,2)=4$

(although $d(4,2)=5$ in the algebraically closed case).

Possibly a similar argument yields $d_K(4,2)=4$ as soon as $K$ has non-squares but I haven't checked details.

Consider the 2-planes in $K^4$ $$A=\{(t,s,0,0)\},\;B=\{(0,0,t,s)\},\; C=\{(t,s,t,s)\},\;D=\{(t,t+2s,t+s,s)\},$$ where $(t,s)$ is understood to range over $K^2$. To prove $d_{\mathbf{R}}(4,2)\le 4$ ($d_K(4,2)\ge 4$ is an elementary verification for an arbitrary field $K$), let's prove that

$\{A,B,C,D\}$ is a complement repository as soon as $-1$ is not a square in $K$.

Proof:

For the moment, $K$ is an arbitrary field.

For $(e,f)\in K^2\smallsetminus \{(0,0)\}$, define $P_{e,f}$ to be the 2-plane $\{(te,tf,se,sf):(t,s)\in K^2\}$. Write $\mathcal{P}=\{P_{e,f}:(e,f)\neq (0,0)\}$.

Clearly $P_{e,f}$ intersects nontrivially each of $A,B,C$ (taking $s=0$, $t=0$, $t=s$ respectively). I claim that conversely if a 2-plane $P$ intersects nontrivially each of $A,B,C$, then $P\in\mathcal{P}$. This is a simple exercise. Indeed, if $P$ intersects $A,B$ nontrivially, then $P=(P\cap A)\oplus (P\cap B)$ and hence there exists $(c,f,a,b)$ with $(c,f),(a,b)\neq (0,0)$ such that $P=\{(tc,tf,sa,sb):(t,s)\in K^2\}$. If $P\cap C$ is nontrivial, for some $(t,s)\neq (0,0)$ we have $tc=sa$ and $tf=sb$. This means that both $(a,c)$ and $(b,f)$ are collinear to the nonzero vector $(t,s)$. If $(b,f)=(0,0)$ we deduce $P=P_{1,0}$. Otherwise, $(a,c)=e(b,f)$ for some $e\in K$. So we have $P=\{(tef,tf,seb,sb):(t,s)\in K^2\}=P_{e,1}$.

Therefore, the final claim that $\{A,B,C,D\}$ is a complement repository is equivalent to showing that $D\cap P_{e,f}=\{0\}$ for every $(e,f)\in K^2 \smallsetminus\{(0,0)\}$.

Since a basis of $P_{e,f}$ is $((e,f,0,0),(0,0,e,f))$ and a basis of $D$ is $((0,2,1,1),(1,1,1,0))$, this amounts to showing that the determinant $\delta(e,f)=\begin{vmatrix}e & f & 0 & 0\\0 & 0 & e & f\\ 0 & 2 & 1 & 1\\ 1 & 1 & 1 & 0\end{vmatrix}$ is nonzero for all $(e,f)\in K^2\smallsetminus\{(0,0)\}$. The computation yields $\delta(e,f)=e^2+f^2$.

Therefore the non-vanishing holds if and only if $-1$ is not a square in $K$. This means that $\{A,B,C,D\}$ is a complement repository if and only if $-1$ is not a square in $K$.

Here is particular case showing that $d(n,k)=d_K(n,k)$ definitely depends on the field $K$ in general, beyond pathologies of finite fields:

$d_{\mathbf{R}}(4,2)=4$

(although $d(4,2)=5$ in the algebraically closed case).

Possibly a similar argument yields $d_K(4,2)=4$ as soon as $K$ has non-squares but I haven't checked details.

Consider the 2-planes in $K^4$ $$A=\{(t,s,0,0)\},\;B=\{(0,0,t,s)\},\; C=\{(t,s,t,s)\},\;D=\{(t,t+2s,t+s,t+s)\},$$ where $(t,s)$ is understood to range over $K^2$. Let's prove that

$\{A,B,C,D\}$ is a complement repository as soon as $-1$ is not a square in $K$.

Proof:

For the moment, $K$ is an arbitrary field.

For $(e,f)\in K^2\smallsetminus \{(0,0)\}$, define $P_{e,f}$ to be the 2-plane $\{(te,tf,se,sf):(t,s)\in K^2\}$. Write $\mathcal{P}=\{P_{e,f}:(e,f)\neq (0,0)\}$.

Clearly $P_{e,f}$ intersects nontrivially each of $A,B,C$ (taking $s=0$, $t=0$, $t=s$ respectively). I claim that conversely if a 2-plane $P$ intersects nontrivially each of $A,B,C$, then $P\in\mathcal{P}$. This is a simple exercise. Indeed, if $P$ intersects $A,B$ nontrivially, then $P=(P\cap A)\oplus (P\cap B)$ and hence there exists $(c,f,a,b)$ with $(c,f),(a,b)\neq (0,0)$ such that $P=\{(tc,tf,sa,sb):(t,s)\in K^2\}$. If $P\cap C$ is nontrivial, for some $(t,s)\neq (0,0)$ we have $tc=sa$ and $tf=sb$. This means that both $(a,c)$ and $(b,f)$ are collinear to the nonzero vector $(t,s)$. If $(b,f)=(0,0)$ we deduce $P=P_{1,0}$. Otherwise, $(a,c)=e(b,f)$ for some $e\in K$. So we have $P=\{(tef,tf,seb,sb):(t,s)\in K^2\}=P_{e,1}$.

Therefore, the final claim that $\{A,B,C,D\}$ is equivalent to showing that $D\cap P_{(e,f)}=\{0\}$ for every $(e,f)\in K^2 \smallsetminus\{(0,0)\}$.

Since a basis of $P_{e,f}$ is $((e,f,0,0),(0,0,e,f))$ and a basis of $D$ is $((0,2,1,1),(1,1,1,0))$, this amounts to showing that the determinant $\delta(e,f)=\begin{vmatrix}e & f & 0 & 0\\0 & 0 & e & f\\ 0 & 2 & 1 & 1\\ 1 & 1 & 1 & 0\end{vmatrix}$ is nonzero for all $(e,f)\in K^2\smallsetminus\{(0,0\}$. The computation yields $\delta(e,f)=e^2+f^2$.

Therefore the non-vanishing holds if and only if as $-1$ is not a square in $K$. This means that $\{A,B,C,D\}$ is a complement repository if and only if $-1$ is not a square in $K$.

aHere is particular case showing that $d(n,k)=d_K(n,k)$ definitely depends on the field $K$ in general, beyond pathologies of finite fields:

$d_{\mathbf{R}}(4,2)=4$

(although $d(4,2)=5$ in the algebraically closed case).

Possibly a similar argument yields $d_K(4,2)=4$ as soon as $K$ has non-squares but I haven't checked details.

Consider the 2-planes in $K^4$ $$A=\{(t,s,0,0)\},\;B=\{(0,0,t,s)\},\; C=\{(t,s,t,s)\},\;D=\{(t,t+2s,t+s,s)\},$$ where $(t,s)$ is understood to range over $K^2$. Let's prove that

$\{A,B,C,D\}$ is a complement repository as soon as $-1$ is not a square in $K$.

Proof:

For the moment, $K$ is an arbitrary field.

For $(e,f)\in K^2\smallsetminus \{(0,0)\}$, define $P_{e,f}$ to be the 2-plane $\{(te,tf,se,sf):(t,s)\in K^2\}$. Write $\mathcal{P}=\{P_{e,f}:(e,f)\neq (0,0)\}$.

Clearly $P_{e,f}$ intersects nontrivially each of $A,B,C$ (taking $s=0$, $t=0$, $t=s$ respectively). I claim that conversely if a 2-plane $P$ intersects nontrivially each of $A,B,C$, then $P\in\mathcal{P}$. This is a simple exercise. Indeed, if $P$ intersects $A,B$ nontrivially, then $P=(P\cap A)\oplus (P\cap B)$ and hence there exists $(c,f,a,b)$ with $(c,f),(a,b)\neq (0,0)$ such that $P=\{(tc,tf,sa,sb):(t,s)\in K^2\}$. If $P\cap C$ is nontrivial, for some $(t,s)\neq (0,0)$ we have $tc=sa$ and $tf=sb$. This means that both $(a,c)$ and $(b,f)$ are collinear to the nonzero vector $(t,s)$. If $(b,f)=(0,0)$ we deduce $P=P_{1,0}$. Otherwise, $(a,c)=e(b,f)$ for some $e\in K$. So we have $P=\{(tef,tf,seb,sb):(t,s)\in K^2\}=P_{e,1}$.

Therefore, the final claim that $\{A,B,C,D\}$ is a complement repository is equivalent to showing that $D\cap P_{e,f}=\{0\}$ for every $(e,f)\in K^2 \smallsetminus\{(0,0)\}$.

Since a basis of $P_{e,f}$ is $((e,f,0,0),(0,0,e,f))$ and a basis of $D$ is $((0,2,1,1),(1,1,1,0))$, this amounts to showing that the determinant $\delta(e,f)=\begin{vmatrix}e & f & 0 & 0\\0 & 0 & e & f\\ 0 & 2 & 1 & 1\\ 1 & 1 & 1 & 0\end{vmatrix}$ is nonzero for all $(e,f)\in K^2\smallsetminus\{(0,0)\}$. The computation yields $\delta(e,f)=e^2+f^2$.

Therefore the non-vanishing holds if and only if $-1$ is not a square in $K$. This means that $\{A,B,C,D\}$ is a complement repository if and only if $-1$ is not a square in $K$.