The universal hyperplane. In an mathematical article on the net, i find the following paragraph :
For a closed scheme $ X \subset \mathbb{P}^n $, there is a natural subscheme of the Grassmannian $ \mathbb{G} ( k , \mathbb{P}^n ) $ whose points correspond to the $k$ - planes contained in $ X $, called the Fano scheme $ F_k(X)$.
Moreover, the is a family of $ k$-planes in $X$ defined as a subvariety $ \Phi (X,k) \subset X \times \mathbb{G} (k, \mathbb{P}^n ) $ with nice properties such that the fiber of $ \Phi (X,k) $ over the point $ p $ of the Grassmannian corresponding to a $ k $ - plane $ L $ is $ L \times \{ p \} \subset X \times \{ p \} = X $.
The simplest example of such a family is perhaps the universal hyperplane :
$$ \Phi = \Phi ( \mathbb{P}^{n} , n-1 ) \subset \mathbb{P}^n \times \mathbb{G} ( n-1 , n ) = \mathbb{P}^n \times {\mathbb{P}^{n}}^* $$
If we write $ \mathbb{P}^n = \mathbb{P} V $, and choose dual baseas $ \{ a_i \} $ and $ \{ z_i \} $ for $ V $ and $ V^* $ respectively, then $ \Phi $ is defined by the equation $ \displaystyle \sum_i a_i z_i = 0 $.


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*Questions :


$1)$ Can you explain to me please, why is $ \Phi $ defined by the equation $ \displaystyle \sum_i a_i z_i = 0 $ ? How do we arrive at this equation : $ \displaystyle \sum_i a_i z_i = 0 $ from the beginning ?. 
$2)$ Do we have : $ \Phi(X,k) \subset F_k(X) $ ?
Thanks in advance for your help.
 A: 1) An hyperplane is a subspace $H\subset \mathbb{P}^n$ defined by an equation $\sum a_i z_i$ for some fixed $a_0,...,a_n\in k$. Note that they should not be all zero, and two $(n+1)$-uplets define the same hyperplane iff they are collinear. Thus there is a bijection between points of $\mathbb{P}^n$ and hyperplanes $H\subset \mathbb{P}^n$ by the correspondance : $\underline{a}=[a_0:...:a_n]\mapsto H_{\underline{a}}$. where $H_{\underline{a}}$ is the hyperplane defined by the equation $\sum a_i z_i=0$.
Now, the universal hyperplane is when $a_0,...,a_n$ are still indeterminates, or more precisely $[a_0:...:a_n]$ is not a point, but the whole $\mathbb{P}^n$. So the universal hyperplane is the subspace $\Phi\subset\mathbb{P}^n\times\mathbb{P}^n$ defined by the equation $\sum a_i z_i=0$.
Thus you have maps :
$$ \Phi\overset{i}\hookrightarrow\mathbb{P}^n\times\mathbb{P}^n\xrightarrow{p_1}\mathbb{P}^n$$
such that if $\underline{a}:\operatorname{Spec}k\to\mathbb{P}^n$ is any $k$-point, then you can form the pullback :
$$\require{AMScd}
\begin{CD}
H_{\underline{a}}@>>> \mathbb{P}^n@>>>\operatorname{Spec}k\\
@VVV@VVV@VV\underline{a}V\\
\Phi@>>i>\mathbb{P}^n\times\mathbb{P}^n@>>p_1>\mathbb{P}^n
\end{CD}
$$
and you have that the pullback of the universal hyperplane to any $k$-points $[a_0:...:a_n]$ is the hyperplane defined by the equation $\sum a_iz_i=0$.
A note on the dual : if you want something coordinate-free, you will have to use duals. There is a correspondence between points of $\mathbb{P}(V^*)$ and hyperplanes in $\mathbb{P}(V)$ given by $\varphi\mapsto\ker\varphi$. Thus the universal hyperplane is the subspace $\Phi:\mathbb{P}(V^*)\times\mathbb{P}(V)$ defined by the equation $\varphi(\underline{z})=0$. If you choose basis and dual basis as in your post, you will find the equation $\sum a_i z_i=0$.

2) No $\Phi(X,k)$ is not a subspace of $\mathbb{G}(k,\mathbb{P}^n)$ hence not a subspace of $F_k(X)$. We have $$\Phi(X,k)=\{ (x,P)\in X\times\mathbb{G}(k,\mathbb{P}^n)~|~x\in P\subset X\}$$ So we have instead a diagram :
$$ X\overset{p_1}\leftarrow \Phi(X,k)\overset{p_2}\to F_k(X)\subset\mathbb{G}(k,\mathbb{P}^n)$$
with $p_1,p_2$ being onto. The fiber of $p_1$ over any $x\in X$ is the set of $k$-planes in $X$ going through $x$. The fiber of $p_2$ over any $P\in\mathbb{G}(k,\mathbb{P}^n)$ (seen as a point) is either empty if $P\not\in F_k(X)$ or the set $\{(x,P)\in X\times\{P\}, x\in P\}$.
