Reading we found this version of the Segre's Lemma of tangents, and we don't understand it. I will put an * follow by a number in the section we don't understand so you can see what are we thinking. Tanks for your help.
PD: We are not English - Speakers, so sorry if there is someting wrong write, and you can also comment how we can improve our writening.
Let $S$ be a set of vectors of $\mathbb(F)_q^k$ such that ever subset of $S$ of size $k$ is a basis for $\mathbb(F)_q^k$.
For each subset $C\subset S$ such that $|C|=k-2$. there is a set $L_C$ de exactly $t = q + 1 - (|S| - k + 2) = q + k - 1 - |S|$ hiperplanes wich contains $C$ and no other vector of $S$. (*1 ¿why? We think is just $|S|-(k-2)$ because there only can have one more element of $S$)
Deffinition 1:
We deffine as $H_C$ to the set of $t$ linear applications with the propiety of $\forall \Sigma\in L_C$, $\exists f\in H_C$ such that $\Sigma=ker(f)$
Deffinition 2
We define as tangent of $C$ to the function $T_C:\mathbb(F)_q^k\rightarrow\mathbb(F)_q$ given for $$ T_C(u)=\prod_{f\in H_C} f(u) $$
Segre's Lemma of Tangents
if $k = 3$ and $x, y, z \in S$ it follows that
$$T_{\left\{ x\right\}}(y) T_{\left\{ y\right\}}(z) T_{\left\{ z\right\}}(x) = (-1)^{ t + 1}T_{\left\{ x\right\}}(z)T_{\left\{ z\right\}}(y)T_{\left\{ y\right\}}(x)$$
Proof
Whit respect to the basis $\left\{ x,y,z\right\}$ the tangent function $T_{\left\{ x\right\}}$ is the evauation of a polynomial of degree t,
$$ \prod (a_{23} X_2 +a_{32} X_3) $$
for somes $a_{ij}$.(*2 how can we obtain the a's)
for all $d = (d_1,d_2,d_3) \in S\backslash\left\{ x,y,z\right\}$ the subspace $\langle x,d \rangle$; defines the ecuation $d_3X_2 -d_2X_3 = 0$ (*3 this is for what? we never hear about this subspace or this ecuations again). The $q-1$ 2-dimensional subspaces wich contains $x$, but don´t contains $y$ or $z$, are defined by the ecuations
$$ X_2-\alpha X_3 = 0$$
With $\alpha\in\mathbb(F)_q^*$. Since the product of the non-zero elements of$\mathbb(F)_q$ is $-1$, it follows that
$$ \prod_{d\in S\backslash\left\{ x,y,z\right\}}\frac{d_2}{d_3}\prod_{X_2,X_3\in L_C}\frac{(-a_{32})}{a_{23}}=-1 $$ (* how can that be an implication of the above)
Observe that $ \prod a_{23} = T_{\left\{ x\right\}}(y) $ and $\prod a_{32} = T_{\left\{ x\right\}}(z)$, and so the above implies that
$$ T_{\left\{ x\right\}}(z)\prod_{d\in S\backslash\left\{ x,y,z\right\}}d_2 = (-1)^{ t + 1}T_{\left\{ x\right\}}(y)\prod_{d\in S\backslash\left\{ x,y,z\right\}}d_3 $$
Multiplying this ecuation with the corresponding for $y$ and for $z$ we obtain
$$ T_{\left\{ x\right\}}(z)T_{\left\{ z\right\}}(y)T_{\left\{ y\right\}}(x)\prod_{d\in S\backslash\left\{ x,y,z\right\}}d_1d_2d_3 = (-1)^{ 3t + 3}T_{\left\{ x\right\}}(y)T_{\left\{ y\right\}}(z)T_{\left\{ z\right\}}(x)\prod_{d\in S\backslash\left\{ x,y,z\right\}}d_1d_2d_3 $$ and we can conclude
$$ (-1)^{t + 1}T_{\left\{ x\right\}}(z)T_{\left\{ z\right\}}(y)T_{\left\{ y\right\}}(x) = T_{\left\{ x\right\}}(y)T_{\left\{ y\right\}}(z)T_{\left\{ z\right\}}(x) $$