Proof. If \(\langle SR \rangle = S\), we can simply take \(S' = S\) and \(R' = R\), so assume \(\langle SR \rangle \neq S\). Then \(SR \not\subseteq S\). Choose \(e \in S\) and \(r \in R\) such that \(er \in SR\setminus S\). Define subspaces \(S_1\) and \(R_1\) as follows: \[\begin{align} S_1 &= S + eR, \\ R_1 &= R \cap (Se^{-1}). \end{align}\] We claim that the following conditions hold:

1. \(\langle S_1 R_1\rangle \subseteq \langle SR\rangle \subseteq A\),

2. \(1 \in R_1 \subseteq R\),

3. \(\dim S_1 + \dim R_1 = \dim S + \dim R\),

4. \(S_1 \subseteq A\) and \(\dim S < \dim S_1\).

Conditions (a) and (b) follow directly from the definitions of \(S\), \(R\), \(S_1\), and \(R_1\).

To verify (c), we first apply the dimension of a sum formula for vector subspaces: \[\begin{align} \dim S_1 &= \dim(S + eR) \\ &= \dim S + \dim eR - \dim(S \cap eR) \\ &= \dim S + \dim R - \dim(S \cap eR). \end{align}\] Now observe that the mapping \(x \mapsto ex\) defines a linear isomorphism from \(R \cap (S e^{-1})\) to \(S \cap eR\), since \(e \neq 0\). This implies \[\dim R_1 = \dim(R \cap (S e^{-1})) = \dim(S \cap eR),\] so \[\dim S_1 + \dim R_1 = \dim S + \dim R,\] confirming (c).

Concerning (d), the inclusion \(S_1 \subseteq A\) follows from (a) and the fact that \(1 \in R_1\). For the dimension inequality, note that \(S_1\) properly contains \(S\), since \(e r \in S_1 \setminus S\).

All four conditions have been verified. Now if \(\langle S_1 R_1 \rangle \neq S_1\), we repeat the above, substituting \(S_1\) for \(S\) and \(R_1\) for \(R\). The process continues until we obtain nonzero subspaces \(S_m\) and \(R_m\) such that \(\langle S_m R_m \rangle = S_m\). Termination is guaranteed, since \(\dim A < \infty\) and the sequence of subspaces \(S_1, S_2, \ldots\) strictly increases in dimension. Let \(S' = S_m\) and \(R' = R_m\). These sets clearly satisfy the conditions in the statement. ◻

Proof. The desired inequality is equivalent to \(|[\ell] \cap P| \;\ge \;\ell - d\), which obviously holds if \(d \geq \ell\). Suppose \(d < \ell\), and assume for a contradiction that \(|[\ell] \cap P| < \;\ell - d\). Then \[|P| = |[\ell] \cap P| + |([n]\setminus [\ell]) \cap P| < \ell -d + n - \ell = n - d,\] contrary to the assumption that \(|P| = n-d\). The lemma is proved. ◻

Proof. Assume that \(A\) is partially matched to \(B\) with defect \(d\). Let \(S \subseteq A\) and \(R \subseteq B\oplus K\) be nonzero \(K\)-subspaces with \(\langle SR \rangle = S\). Choose a basis \(\{ a_1, \ldots , a_{\ell} \}\) for \(S\), and extend this to a basis \(\mathcal{A}=\{a_1,\dots,a_\ell,a_{\ell+1},\dots,a_n\}\) for \(A\). Since \(A\) is partially matched to \(B\) with defect \(d\), there exists a basis \(\mathcal{B}=\{b_1,\dots,b_n\}\) of \(B\) such that \(\mathcal{A}\) is partially matched to \(\mathcal{B}\) with defect \(d\). Thus there exists a subset \(P\) of \([n]\) of size \(n-d\) such that \[a_i^{-1}A\cap B \;\subseteq\; \langle b_1,\dots,b_{i-1},b_{i+1},\dots,b_n\rangle \quad\text{for each } i\in P.\]

Now, we have \[\begin{align} \label{eq1} \bigcap_{i\in[\ell]} \big(a_i^{-1}A \cap B\big) &\subseteq \bigcap_{i\in[\ell] \cap P} \big(a_i^{-1}A \cap B\big) \\ &\subseteq \bigcap_{i\in[\ell] \cap P} \langle b_1, \dots, b_{i-1}, b_{i+1}, \dots, b_n\rangle, \nonumber \end{align}\tag{1}\] where the last two intersections are understood to equal \(B\) if \([\ell] \cap P\) is empty. Clearly \[\dim \! \left( \bigcap_{i\in[\ell] \cap P} \langle b_1, \dots, b_{i-1}, b_{i+1}, \dots, b_n\rangle \right) = n - |[\ell] \cap P|.\] Combining this with (1 ), we get \[\dim \bigcap_{i\in[{\ell}]}(a_i^{-1}A\cap B) \leq n-|[\ell]\cap P|.\]

On the other hand, since \(\langle SR \rangle = S\), we have \(a_i R \subseteq S\) for each \(i \in [\ell]\), and hence \[R \subseteq a_i^{-1} S \subseteq a_i^{-1} A.\] Therefore \[R \cap B \;\subseteq\; \bigcap_{i\in[\ell]} \big(a_i^{-1} A \cap B\big).\] Thus, according to Lemma 1, \[\dim (R\cap B)\leq \dim \bigcap_{i\in[{\ell}]}(a_i^{-1}A\cap B) \leq n-\ell+d,\] which implies the desired inequality \[\dim S-d\leq \dim \big(B/(B\cap R)\big).\]

Conversely, assume that the condition in the statement involving \(S\) and \(R\) holds. We will show that \(A\) is partially matched to \(B\) with defect \(d\). Let \(\mathcal{A} = \{a_1,\dots,a_n\}\) be a basis for \(A\), and let \(J \subseteq [n]\) be nonempty. Set \[S = \langle a_i : i \in J \rangle, \quad T = \bigcap_{i \in J} (a_i^{-1} A \cap B), \quad R = T + K,\] and note that the sum \(T + K\) is direct. We claim that \(\dim T \leq n - |J| + d\). To verify this, we consider two cases.

Case 1: \(\langle SR \rangle = S\). By hypothesis, \[\dim S - d \;\leq\; \dim \big(B/(B\cap R)\big).\] Since \(T \subseteq B\), \(R = T \oplus K\), and \(1 \notin B\), it follows that \(T = B \cap R\). Therefore \[\dim T \;=\; \dim(B \cap R) \;\leq\; \dim B - \dim S + d \;=\; n - |J| + d,\] as claimed.

Case 2: \(\langle SR \rangle \neq S\). We wish to apply Proposition 8. By construction, we have \(ST\subseteq A\). Hence, since \(R=T\oplus K\), \[SR=S(T\oplus K) \subseteq ST + SK \subseteq \langle ST\cup S\rangle \subseteq A.\] Therefore, by Proposition 8, there exist nonzero \(K\)-subspaces \(S'\) and \(R'\) of \(L\) such that

• \(\langle S'R' \rangle = S' \subseteq A\),

• \(1 \in R' \subseteq R \subseteq B \oplus K\),

• \(\dim S' + \dim R' = \dim S + \dim R\).

By applying our hypothesis to \(S'\) and \(R'\), which we are able to do on account of (a) and (b), we obtain \[\dim S' - d \;\leq\; \dim \big(B/(B\cap R')\big).\] Now, it is easy to see that \(R' = (B \cap R')\oplus K\). Hence \(\dim (B \cap R') = \dim R' - 1\), and so we have the inequality \[\dim S'- d \;\leq\; n - \dim R' + 1,\] which, combined with (c) and the fact that \(\dim R = \dim T + 1\), yields \[\dim T \;\leq\; n - |J|+d,\] as claimed.

So in both cases we have \(\dim T \leq n - |J|+d\). Passing to the annihilator in the dual space \(B^*\), we obtain, by way of one of the annihilator identities, \[\dim T^\perp \geq |J|-d,\] which leads, using the other annihilator identity, to \[\dim \sum_{i \in J} (a_i^{-1} A \cap B)^\perp \geq |J| - d.\] By Theorem 6, the family \[\mathcal{W} = \big( (a_1^{-1} A \cap B)^\perp, \dots, (a_n^{-1} A \cap B)^\perp \big)\] admits a free partial transversal with defect \(d\). Thus there is a subset \(P\) of \([n]\) and a linearly independent subset \(\{ f_i : i \in P \}\) of \(B^*\), both of size \(n - d\), such that \[\begin{align} \label{eq2} f_i \in (a_i^{-1} A \cap B)^\perp \quad \text{for each } i \in P. \end{align}\tag{2}\]

Extend \(\{ f_i : i \in P \}\) to a basis \(\mathcal{F} = \{ f_1, \dots, f_n \}\) of \(B^*\), and let \(\mathcal{B} = \{ b_1, \dots, b_n \}\) be the basis of \(B\) dual to \(\mathcal{F}\). We show that \(\mathcal{A}\) is partially matched to \(\mathcal{B}\) with defect \(d\). Since \(f_i(b_j) = \delta_{ij}\), it follows that \[\ker f_i = \langle b_1, \dots, b_{i-1}, b_{i+1}, \dots, b_n \rangle \quad \text{for each } i \in P.\] Combining this with 2 yields \[a_i^{-1} A \cap B \subseteq \langle b_1, \dots, b_{i-1}, b_{i+1}, \dots, b_n \rangle \quad \text{for each } i \in P.\] Hence \(\mathcal{A}\) is partially matched to \(\mathcal{B}\) with defect \(d\). We have thus shown that \(A\) is partially matched to \(B\) with defect \(d\), completing the proof. ◻

Proof. Define \[D=\langle\, b \in B : [K(b):K] \le n \,\rangle.\] Let \(S \subseteq A\) and \(R \subseteq B \oplus K\) be nonzero \(K\)-subspaces with \(\langle S R \rangle = S\). We claim that \(B \cap R \subseteq D\). Indeed, if \(x \in L \setminus D\), then \(x \notin B\) or \([K(x):K] > n\), the latter implying \(x \notin R\) by Proposition 10.

Therefore \(\dim (B \cap R) \leq d\), and so \[\dim S - d \;\le\; n-d \;\le\; \dim B - \dim(B\cap R) \;=\; \dim \big( B / (B\cap R) \big).\] By Theorem 9, \(A\) is partially matched to \(B\) with defect \(d\). ◻

Proof. Let \(d = \delta(A,B)\) and define \(\mathcal{I}\) to be the set of all differences \[\dim S - \dim \big(B/(B\cap R)\big),\] where \(S\) and \(R\) are as in the statement.

Since \(A\) is partially matched to \(B\) with defect \(d\), it follows from Theorem 9 that for all such \(S\) and \(R\), \[\dim S - d \;\le\; \dim \big(B/(B\cap R)\big).\] Hence \(d \ge \max \mathcal{I}\).

For the reverse inequality, let \(d' = \max \mathcal{I}\). Then for all nonzero subspaces \(S \subseteq A\) and \(R \subseteq B \oplus K\) with \(\langle SR \rangle = S\), we have \[\dim S - d' \;\le\; \dim \big(B/(B\cap R)\big).\] Applying Theorem 9 again, \(A\) is partially matched to \(B\) with defect \(d'\). By the minimality of \(d = \delta(A,B)\), it follows that \(d \le d'\). Combining this with the previous paragraph gives \(d = d'\), as claimed. ◻

Proof. Let \(S\subseteq A\) and \(R\subseteq B\oplus K\) be nonzero \(K\)-subspaces with \(\langle SR\rangle = S\). We claim that \[\dim S - d \;\leq\; \dim \big(B/(B\cap R)\big).\]

First, note that \(R\cap S = \{0\}\). Indeed, if there existed a nonzero \(a \in R\cap S\), then by Proposition 10 we would have \(aK(a) \subseteq S\), implying \(1 \in S \subseteq A\), contrary to our assumption that \((A + B) \cap K = \{0\}\).

Using the above, the dimension formula for the sum of two subspaces, and the fact that \(S + (B\cap R) \subseteq A + B\), we get \[\begin{align} \dim S + \dim (B \cap R) = \dim \big( S + (B \cap R) \big) &\leq \dim (A+B) \\ &= \dim A + \dim B - \dim (A\cap B) \\ &= \dim B + d. \end{align}\] Therefore \[\dim S - d \;\leq\; \dim B - \dim (B\cap R) = \dim \big(B/(B\cap R)\big).\] By Theorem 9, \(A\) is partially matched to \(B\) with defect \(d\). ◻

Proof. Since \(B\) is a \(d\)-defective Chowla subspace, there exists a subspace \(B_{0} \subseteq B\) with \(\dim B_{0} \geq n-d\) such that for every nonzero \(x \in B_{0}\), \[[K(x) : K] \geq n+1.\] Let \(S \subseteq A\) and \(R \subseteq B \oplus K\) be nonzero \(K\)-subspaces with \(\langle S R \rangle = S\). We claim that \(R \cap B_{0} = \{ 0 \}\). Indeed, if this did not hold, we could choose a nonzero \(x \in R \cap B_{0}\), and then, by Proposition 10, we would have \([K(x) : K] \leq n\) since \(x \in R\). On the other hand, since \(x \in B_{0}\), we would have \([K(x) : K] \geq n+1\), a contradiction.

Bearing in mind that \(B\cap R\) and \(B_{0}\) are contained in \(B\) and that \((B\cap R)\cap B_{0}=\{0\}\), we have \[\dim(B\cap R)+\dim B_{0}\le n.\] It follows that \[\begin{align} \dim S - d \leq n - d \leq \dim B_{0} \leq n - \dim (B \cap R) =\dim \big(B/(B\cap R)\big). \end{align}\] Applying Theorem 9, we conclude that \(A\) is partially matched to \(B\) with defect \(d\). ◻

Proof. We first prove (i). Fix \(\mathcal{A}=\{a_1,\dots,a_n\}\) and write \(T_i=T_i(\mathcal{A})\). Let \(\mathcal{B}=\{b_1,\dots,b_n\}\) be a basis of \(B\) and let \(\{f_1,\dots,f_n\}\) be the corresponding dual basis of \(B^*\). For each \(i\), the containment \[T_i \subseteq \langle b_1,\dots,b_{i-1},b_{i+1},\dots,b_n\rangle\] holds if and only if \(f_i\) vanishes on \(T_i\), i.e., \[f_i\in T_i^\perp \subseteq B^*.\] Thus the existence of a basis \(\mathcal{B}\) for which \(\mathcal{A}\) is partially matched to \(\mathcal{B}\) with defect \(d\) is equivalent to the existence of a basis \(\{f_1,\dots,f_n\}\) of \(B^*\) such that at least \(n-d\) of the vectors \(f_i\) belong to the corresponding subspaces \(T_i^\perp\).

Let \(U_i = T_i^\perp\). Applying Theorem 6 to the family \((U_1,\dots,U_n)\) in \(B^*\), we find that the minimum number of indices that must be discarded in order for there to exist a linearly independent system \(f_i\in U_i\) from the remaining indices equals \[\max_{J\subseteq [n]} \Bigl(|J|-\dim \sum_{i\in J}U_i \Bigr).\] Using the annihilator identities given above Theorem 9, we obtain \[\begin{align} |J|-\dim \sum_{i\in J}U_i = |J|-\dim \sum_{i\in J}T_i^\perp &= |J|-\dim\Bigl(\bigcap_{i\in J}T_i\Bigr)^\perp \\ &= |J|-\Bigl(n-\dim \bigcap_{i\in J}T_i \Bigr) \\ &= |J|+\dim \bigcap_{i\in J}T_i - n. \end{align}\] Maximizing over \(J\) yields \(\Omega(\mathcal{A})\). Therefore, for the fixed basis \(\mathcal{A}\), the least integer \(d \geq 0\) such that \((U_1,\dots,U_n)\) has a free partial transversal with defect \(d\) is \(\Omega(\mathcal{A})\). It now follows from the first paragraph above that \(\Omega(\mathcal{A})\) is the least integer \(d \geq 0\) such that \(\mathcal{A}\) is partially matched to an ordered basis of \(B\) with defect \(d\). Thus (i) is proved.

For (ii), let \(t = \delta(A,B)\). By definition, \(t\) is the least integer \(d\) such that for every ordered basis \(\mathcal{A}\) of \(A\) there exists an ordered basis of \(B\) witnessing partial matchability with defect \(d\). By (i), the least defect required for a given \(\mathcal{A}\) equals \(\Omega(\mathcal{A})\), hence \(t\) must satisfy \(t \ge \Omega(\mathcal{A})\) for all \(\mathcal{A}\), so \(t\ge \max_{\mathcal{A}}\Omega(\mathcal{A})\). Conversely, if \(t'=\max_{\mathcal{A}}\Omega(\mathcal{A})\), then \(t' \ge \Omega(\mathcal{A})\) for each ordered basis \(\mathcal{A}\), and so (i) supplies, for each \(\mathcal{A}\), an ordered basis of \(B\) witnessing partial matchability with defect \(t'\) (see Remark 5). Hence \(\delta(A,B) = t \le t'\). Therefore \(\delta(A,B)=\max_{\mathcal{A}}\Omega(\mathcal{A})\). ◻

Proof. Assume that \(\delta(A,B)>\ell\). Then \(A\) is not partially matched to \(B\) with defect \(\ell\). From Theorem 9, there exist nonzero \(K\)-subspaces \(S\subseteq A\) and \(R\subseteq B\oplus K\) such that \(\langle SR\rangle = S\) and \[\label{eq:thm41-ineq} \dim S >\dim \bigl(B/(B\cap R)\bigr)+\ell.\qquad{(1)}\] Let \(R_0 = B\cap R\). Then \(R_0\neq \{0\}\), since otherwise \(\dim\big(B/(B\cap R)\big)=\dim B=\dim A \geq \dim S\), contradicting ?? .

By Proposition 10, for every \(a\in S\) we have \(aK(R)\subseteq S\). Since \(R_0\subseteq R\), it follows that \(K(R_0)\subseteq K(R)\), hence \(aK(R_0)\subseteq S\) for all \(a\in S\). Therefore \[S=\sum_{a\in S} aK(R_0).\] Choose \(K\)-subspaces \(Y\) of \(A\) and \(Z\) of \(B\) such that \(A=S\oplus Y\) and \(B=R_0\oplus Z\). Using \(\dim \bigl(B/(B\cap R)\bigr)=\dim B-\dim R_0\), inequality ?? becomes \[\dim S>\dim B-\dim R_0+\ell.\] Since \(\dim A=\dim B\), this implies \[\dim Y=\dim A-\dim S<\dim R_0-\ell,\] which is the desired conclusion.

Conversely, assume that \(A\) and \(B\) admit decompositions as in the statement. Define \[\widetilde{R}=R\oplus K \;\subseteq\;B\oplus K .\] Since \(1\notin B\) and \(R \subseteq B\), it follows that \(B\cap \widetilde{R}=R\). Also, we have \(\langle S\widetilde{R}\rangle=S\), on account of the equation \(S = \sum_{a\in S} aK(R)\) and the fact that \(1 \in \widetilde{R}\).

The inequality \(\dim Y < \dim R-\ell\) implies \[\dim S+\dim R-\ell > \dim S+\dim Y = \dim A = \dim B .\] Therefore \[\dim S-\ell > \dim B-\dim R = \dim(B/R) = \dim\bigl(B/(B\cap \widetilde{R})\bigr).\] By Theorem 3.3, \(A\) is not partially matched to \(B\) with defect \(\ell\), and hence \(\delta(A,B)>\ell\). This completes the proof. ◻

Proof. Assume that there exist \(n\)-dimensional \(K\)-subspaces \(A,B\subseteq L\) with \(1\notin B\) such that \(\delta(A,B)>\ell\). By Theorem 13, there exists a nonzero \(K\)-subspace \(R\subseteq B\) such that \(A\) and \(B\) admit decompositions \[A = S \oplus Y, \qquad B = R \oplus Z,\] where \[S=\sum_{a\in S} aK(R)\] and \(\dim Y < \dim R - \ell\). Define \[\widetilde{R} = R + K \subseteq B \oplus K .\] Then \(\langle S\widetilde{R}\rangle = S\), so Proposition 10 applies to the pair \((S,\widetilde{R})\). In particular, \[[K(R):K] = [K(\widetilde{R}):K] \leq n <\infty.\] Define \(F=K(R)\) and write \([F:K]=m\). Reapplying Proposition 10, we obtain \(\dim S = mq\) for some integer \(q\ge 1\). Set \(r=\dim Y\). Then \[r=\dim Y=\dim A-\dim S=n-mq.\]

Moreover, since \(R\subseteq B\) and \(1\notin B\), we have \(1\notin R\), so \(R\) is a proper \(K\)-subspace of \(F\), which implies \(\dim R\le m-1\). Combining this with \(\dim Y<\dim R-\ell\) yields \[r<\dim R-\ell\le (m-1)-\ell,\] and hence \(r\le m-\ell-2\). Therefore, for each \(j\in [\ell+1]\), we have \[n+j=mq+(r+j)\] with \[1\le r+j\le (m-\ell-2)+(\ell+1)=m-1.\] So the remainder of \(n+j\) modulo \(m\) is never \(0\), i.e., \[m\nmid(n+j)\qquad\text{for all }j\in[\ell+1].\] Finally, since \(q\ge 1\), it follows that \([F:K] = m\le \dim_K S\le \dim_K A=n\). Thus \(F\) is the desired intermediate field.

Conversely, assume that there exists an intermediate field \(K\subsetneq F\subseteq L\) with \(m=[F:K]\le n\) and \(m\nmid(n+j)\) for all \(j\in[\ell+1]\). Write \(n=mq+r\) using integers \(q, r\) with \(0\le r<m\). The non-divisibility condition is equivalent to \[r+j\not\equiv 0 \pmod m \qquad \text{for all } j\in[\ell+1].\] Now, among the integers \(1,2,\dots,m\) there is a unique value \(j_0\) such that \(r+j_0\equiv 0\pmod m\), namely, \(j_0=m-r\). Hence \(m-r\notin[\ell+1]\), i.e., \[m-r>\ell+1,\] and therefore \[r\le m-\ell-2.\]

Choose an \(F\)-subspace \(S\subseteq L\) of \(F\)-dimension \(q\); then \(\dim_K S=mq\). Next choose a \(K\)-subspace \(Y\subseteq L\) with \(\dim_K Y=r\) and \(S\cap Y=\{0\}\), and set \(A=S + Y\). Then \(\dim_K A=mq+r=n\). Pick a \(K\)-hyperplane \(R\subseteq F\) with \(1\notin R\), so \(\dim_K R=m-1\). Choose a \(K\)-subspace \(Z\subseteq L\) such that \(Z\cap F=\{0\}\) and \(\dim_K Z=n-(m-1)=n-m+1\) (this can be done, since \(n < [L:K]\)), and set \(B=R + Z\). Then \(\dim_K B=n\) and also \(1\notin B\), because if \(1=x+z\) with \(x\in R\subseteq F\) and \(z\in Z\), then \(z=1-x \in F\), hence \(z\in Z\cap F=\{0\}\), forcing \(x=1\), contradicting \(1\notin R\). Since \(K(R)\subseteq F\) and \(S\) is an \(F\)-subspace, it follows that \(S\) is a \(K(R)\)-subspace of \(L\). Put differently, \[S=\sum_{a \in S} aK(R).\]

Finally, using \(r\le m-\ell-2\), we obtain \[\dim_K Y=r\le m-\ell-2<(m-1)-\ell=\dim_K R-\ell.\] Thus \(A\) and \(B\) admit decompositions \(A=S\oplus Y\) and \(B=R\oplus Z\) as in the statement of Theorem 13. We conclude that \(\delta(A,B)>\ell\), completing the proof. ◻

Proof. Fix an \(m\in \mathcal{M}(K,L)\) with \(m \leq n\). Note that \(m \geq 2\). We show first that \(\Delta(n) \geq m - 1 - (n \bmod m)\). Write \(n=mq+r\) using integers \(q,r\) with \(0\le r<m\). If \(r=m-1\), then \(m-r-1=0\) and the inequality here is trivial since \(\Delta(n)\ge 0\) by definition. Assume henceforth that \(r\le m-2\), and define \(\ell = m-r-2\ge 0\). Then the condition \(m\nmid (n+j)\) for all \(j\in[\ell +1]\) holds, so by Theorem 14 there exist \(n\)-dimensional \(K\)-subspaces \(A,B\) of \(L\) with \(\delta(A,B)>\ell\), and therefore \(\delta(A,B)\ge \ell +1=m-r-1\). Thus, by the definition of \(\Delta(n)\), we have \(\Delta(n)\ge m-r-1 = m-1-(n \bmod m)\), as desired.

Maximizing over all admissible \(m\) gives us \[\Delta(n)\ge \max_{\substack{m\in \mathcal{M}(K,L)\\ m\le n}} \Bigl(m-1-\bigl(n\bmod m\bigr)\Bigr).\]

For the reverse inequality, let \(t = \Delta(n)\) and let \(A,B\) be \(n\)-dimensional \(K\)-subspaces of \(L\) with \(1 \notin B\) and \(\delta(A,B) = t\). If \(t = 0\), the desired inequality clearly holds, so assume \(t > 0\). Then \(\delta(A,B)>t-1 \geq 0\), so by Theorem 14 there exists an intermediate field \(K\subsetneq F\subseteq L\) such that, setting \(m_0=[F:K]\), we have \(m_0 \le n\) and \[m_0\nmid (n+j)\qquad\text{for all } j\in[t].\] Write \(n=m_0q'+r'\) using integers \(q',r'\) with \(0\le r'<m_0\). Since \(n+(m_0 -r')=m_0(q'+1)\) is divisible by \(m_0\), the above condition forces \(m_0-r'\notin[t]\), and hence \[t\le m_0-r'-1.\] We have shown that \(\Delta(n) = t \leq m_0-1-\bigl(n\bmod m_0\bigr)\) for a specific \(m_0 \in \mathcal{M}(K,L)\) with \(m_0 \leq n\). It follows that \[\Delta(n)\le \max_{\substack{m\in \mathcal{M}(K,L)\\ m\le n}} \Bigl(m-1-\bigl(n\bmod m\bigr)\Bigr),\] completing the proof. ◻

Proof. This follows easily from Theorem 15, since the nontrivial intermediate fields of the extension \(\mathbb{F}_q \subseteq \mathbb{F}_{q^{N}}\) are exactly the fields of the form \(\mathbb{F}_{q^{m}}\), where \(m \geq 2\) is a divisor of \(N\), and since the degree of \(\mathbb{F}_{q^m}\) over \(\mathbb{F}_q\) is \(m\). ◻