Maps from products of Brauer-Severi varieties and sections Let $X$ be a Brauer-Severi variety over a field $k$ of characteristic $0$. In other words, suppose that $X_{\overline{k}} \cong \mathbb{P}_{\overline{k}}^n$.
I came across a statement that the map $X \times_k X \longrightarrow X$ sending an element of $X \times_k X$ to its first factor having a section (the diagonal map) implies that $X \times_k X \cong \mathbb{P}_X^n$ (the fiber product with $X$). Why is this the case? I tried to show that $X \times_k X$ has the universal property of $\mathbb{P}_X^n$, but I'm not sure why this is clear in the first place. Is there a simple way of thinking about this that I'm missing?
Some further questions:


*

*More generally, what are other situations where maps/projections from a product of Brauer-Severi varieties having a section implies that the product is isomorphic or birational to a "similar" fiber product with projective space?

*Are there suitable varieties $A$ over $k$ such that everything above works when we replace $\mathbb{P}_k^n$ with $A$?

*Vague generalization: When do maps with sections induce isomorphisms with some (fiber) product?
Edit: 
Given the answer below, the third question may be rephrased as follows:
When does the existence of a section imply a map is Zariski locally trivial? (e.g. Brauer-Severi varieties) How does this compare to the topological situation? (e.g. principal bundles with a global section)
 A: This is false. Take $X$ to be a smooth plane conic without a rational point. Consider the surface
$$S = X \times X.$$
Over the algebraic closure this becomes isomorphic to $\mathbb{P}^1 \times \mathbb{P}^1$. Recall that $\mathrm{Pic}(\mathbb{P}^1 \times \mathbb{P}^1) = \mathbb{Z}^2$. Then one easily sees that 
$$\mathrm{Pic}(S) = \langle (2,0), (1,1), (0,2) \rangle.$$
The divisors $(2,0)$ and $(0,2)$ come from the trivial divisor and a closed point of degree $2$ on one factor. The class $(1,1)$ comes from the diagonal embedding. Note that there is no divisor of type $(1,2)$.
Now consider the surface
$$S' = \mathbb{P}^1_k \times X.$$
Here
$$\mathrm{Pic}(S') = \langle (1,0), (0,2) \rangle.$$
This has a divisor of type $(1,2)$, thus $S \not \cong S'$.
What happened?
You seem to be looking for the notion of Brauer-Severi schemes. The theory is similar to that of Brauer-Severi varieties, but over a base scheme rather than a field.
Let $Y$ be a scheme. A Brauer-Severi scheme over $Y$ is a proper morphism $P \to Y$ which is etale locally ismorphic to the trivial projective bundle $\mathbb{P}^n_Y \to Y$ for some $n$.
Then the following holds: $P \to Y$ has a section if and only if it is Zariski locally isomorphic to the trivial projective bundle. This means that $P$ is the projectivisation of some vector bundle on $Y$.
In the above counter-example, one takes $P = X \times X$ and $Y = X$. This has a section hence is the projectivisation of some vector bundle. But it is not the projectivisation of the trivial vector bundle.
