Is $X\times \mathbb{P}^{n}$ connected for a smooth, proper and connected scheme $X$?  Let $S$ be a base-scheme, for our purposes we can assume it is the spectrum of a field $k$.
Given a smooth, proper and connected $S$-scheme $X$ and an integer $n\geq 1$,
is then the fibre-product $X\times_{S}\mathbb{P}^{n}_{S}$ still connected?
If not, are their any assumptions on $X$ respectively the base scheme $S$ that would guarantee connectedness?
 A: This question is taking ridiculously many attempts to get a clean, correct answer.
First: As inkspot points out, a scheme is connected iff it has no idempotent global sections other than zero and one.  Thus, whether a scheme is connected is determined purely by its ring of global sections.
Second: As Mattia Talpo and Charles Siegel both observe, $X \times_S \mathbb{P}^n_S$ is naturally isomorphic to $\mathbb{P}^n_X$.
Third: The global section ring of $\mathbb{P}^n_X$ is isomorphic to the global section ring of $X$.  To see this, first observe that if $f \colon \mathbb{P}^n_X \to X$ is the obvious morphism, then
$$\Gamma(\mathbb{P}^n_X, \mathcal{O}_{\mathbb{P}^n_X}) \cong \Gamma(X, f_{*}\mathcal{O}_{\mathbb{P}^n_X}).$$
This follows from the definition of $f_{*}$.  Thus, it suffices to show that $f_* \mathcal{O}_{\mathbb{P}^n_X} = \mathcal{O}_X$.  For this, we can work locally on $X$. If $U \subset X$ is an open affine, then our desired statement $f_* \mathcal{O}_{\mathbb{P}^n_U} = \mathcal{O}_U$ is true by standard facts about projective space over an affine scheme.
Conclusion: $X \times_S \mathbb{P}^n_S$ is connected iff $X$ is.  This statement requires no hypotheses on either $X$ or $S$ (not even local Noetherianness).
A: A scheme $Z$ is connected if and only if the only idempotent $f\in H^0(Z,\mathcal O_Z)$ are $f=0,1$. Since $H^0(S,\mathcal O_S)=H^0(S\times Y,\mathcal O_{S\times Y})$ for any proper $k$-variety $Y$ such that $k$ is algebraically closed in $H^0(Y,\mathcal O_Y)$, it follows that $S$ is connected if and only if $S\times Y$ is so. 
A: The product is just $\mathbb{P}^n_X$ (because it is $X\times_S (S\times\mathbb{P}^n_\mathbb{Z})$.  The fibers then of $\mathbb{P}^n_X\to X$ are projective spaces, and thus connected, so all you need is $X$ is connected.
A: To second BCnrd's remark about the need to have assumptions on the surjection, this is a purely topological issue that can occur whether you're looking at schemes or not. For instance, the projection of
$$ [0,1) \times \{0\}\ \cup\ [1,2]\times \{1\} $$
on the first factor has both a connected base and connected source (but is not a proper map).
