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:



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> **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.
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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



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> **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.
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The answer to that is negative, see Poonen's explicit example below.



Surjectivity of bilinear forms.
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.


