Let $C$, $Q \in \mathbb{R}[x\_0,\dots,x\_n]$ be homogeneous of degrees $3$ and $2$ respectively. Consider the scheme $V$ in $\mathbb{P}^n$ defined by
$$ V \; : \; C=Q=0$$.
Suppose


* $V$ is integral (over $\mathbb{C}$);
* $Q$ is indefinite of full rank $n+1$.


**Question:** Show that $V$ has a smooth real point.


**Note:** It is easy to construct a counterexamples without the integrality assumption. For example, take $C=x\_0^3+x\_1 Q$. Then every point on $V$ has multiplicity $3$.



Smooth real points on the intersection of a quadric and a cubic
I think that this is wrong for $n = 3$:


Let $C \subset {\mathbb P}^3\_{\mathbb R}$ be a 
generic cubic surface not containing $(0:0:0:1)$ and take
$$Q \colon x^2 + y^2 + z^2 = \varepsilon w^2$$
with $\varepsilon > 0$ sufficiently small (the coordinates are $x,y,z,w$).
Then $C({\mathbb R}) \cap Q({\mathbb R})$ is empty, but for most choices
of $C$ (and $\varepsilon$) $V = C \cap Q$ will be integral over $\mathbb C$.


For larger $n$, the same kind of example works.


If the signature of $Q$ is $(2,2)$ instead of $(3,1)$, then the statement
is true: $Q$ is isomorphic over $\mathbb R$ to
${\mathbb P}^1 \times {\mathbb P}^1$ and $V$ gets mapped to a curve
of type $(3,3)$, which will always have lots of real points.
So perhaps one needs an additional assumption on the signature of $Q$.


