Surjectivity of bilinear forms. It is not uncommon to describe interesting classes of field extensions by declaring that an extension $L|K$ belongs to that class if some type of problem with $K$-coefficiens has a property over $L$ if and only if it has the same property over $K$. I wonder about the following variant:

Question A: For which field extensions $L|K$ the following is true?: Given finite dimensional $K$-vector spaces $U,V,W$, a $K$-bilinear form $\beta_K:U\times V\to W$ is surjective if and only if the corresponding $L$-bilinear form $\beta_L$ obtained by scalar extension is surjective.

Already in characteristic zero an answer to that would be nice. Also, I wonder for which $L|K$ surjectivity of $\beta_K$ implies or is implied by surjectivity of $\beta_L$. 
One can take the geometric point of view: A bilinear form induces a map between associated projective spaces and one asks here for surjectivity of these maps on $K$ or $L$-rational points.
It is not hard to show that if $L$ is the reals or the $p$-adics, then surjectivity of a bilinear form over the rationals implies surjectivity with $L$-coefficients (the argument really uses both, density and local compactness). This is the setting in which the problem originally arised. I was also asking the following

Question B: Given a bilinear form $\beta$ between finite dimensional $\mathbb Q$-vector spaces, is it true that $\beta$ is surjective if and only if for all primes $p$ (including $p=\infty$) the induced $\mathbb Q_p$-bilinear form is surjective.

The answer to that is negative, see Poonen's explicit example below.
 A: The answer to Question B is no, as I'll show below.
Let $U=V=\mathbf{Q}^3$ and $W=\mathbf{Q}^4$.  Define
$$\beta((u_1,u_2,u_3),(v_1,v_2,v_3))=(u_1 v_1,u_2 v_2,u_3 v_3, (u_1+u_2+u_3)(v_1+v_2+v_3)).$$
Claim 1: $\beta$ is not surjective.
Proof: In fact, we will show that $(1,1,1,-1)$ is not in the image.  If it were, then we would have a solution to 
$$(u_1+u_2+u_3)(u_1^{-1}+u_2^{-1}+u_3^{-1})=-1.$$
Clearing denominators leads to an elliptic curve in $\mathbf{P}^2$, but MAGMA shows that all its rational points lie in the lines where some coordinate vanishes.
Claim 2: After base extension to any completion $k$ of $\mathbf{Q}$, the bilinear map $\beta$ becomes surjective.
Proof: Given $(a_1,a_2,a_3,b) \in k^4$, we need to show that it is in the image.  If $a_1=0$, then set $u_1=0$, $u_2=1$, $u_3=1$, $v_2=a_2$, $v_3=a_3$, and then solve for $v_1$ in the remaining constraint.  The same argument works if $a_2=0$ or $a_3=0$.  If $a_1,a_2,a_3$ are all nonzero, then we must find a solution to 
$$(u_1+u_2+u_3)(a_1 u_1^{-1}+a_2 u_2^{-1}+a_3 u_3^{-1})=b.$$ 
Clearing denominators leads to the equation of a projective plane curve
with a smooth $k$-point $(1:-1:0)$, so by the implicit function theorem
there exist nearby $k$-points with $u_1,u_2,u_3$ all nonzero,
which gives us the solution we needed. $\square$
If you want a reasonable answer to Question A, I'd suggest that you try to make it more focused.
