Disconnecting projective space Fix a product of projective spaces (say over R) $X=\mathbb{P}^{n_1}\times \ldots \times \mathbb{P}^{n_k}$ and a polynomial $f$ which is homogeneous of degree $d_i$ in the coordinates of the i-th space. Consider the hypersurface $H$ defined by $f$, and suppose that real points have codimension 1. Is it true that real points of $U=X \setminus H$ are always disconnected?  If not, which are relevant characteristics of $f$ to look at?
Incase they have higher codimension, is it always connected?
It seems to me that complex points always have codimension 1. Is this true? In this case, are complex points of $X\H$ disconnected?
Lastly, I wonder how to test if $f$ yields a hypersurface of real codimension  1. One way would be to see if it is non singular; in this case it has the right codimension, right?
Thank you, Andrea
 A: First question : it really depends. For example, the simplest case is $\Bbb RP^1 \times \Bbb RP^1 \cong T$ where a curve can disconnect or not the torus, depending if it is homologically trivial or not.
Second question : $M$ is a manifold and $Z$ a submanifold of codimension at least two, then $M \backslash Z$ is always connected if $M$ is connected. In particular : a connected complex algebraic variety minus a subvariety of smaller dimension is always connected. It is in general not simply connected and gives an interesting topological invariant.
I think the answer to your third question is yes but I am not sure of a reference. 
The answer to the last question is negative, notice that $x^2 + y^2 + z^2 = 0$ is a smooth empty curve in $\Bbb RP^2$. In fact, the locus of $f_1^2 + \dots +  f_k^2 = 0$ is exactly the intersection of the zero locus of $f_i$ for $i=1, \dots, k$. So a generic choice of $f_i$ should give you a smooth algebraic variety of codimension $k$, defined by one equation. 
Finally, you didn't asked it but let me say that complex hypersurfaces in $\Bbb P^n$ are connected, and the proof of this is non-trivial (see Lefschetz hyperplane theorem).
