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.

  • $\begingroup$ Maybe I have misunderstood your question, so I will leave this as a comment. A bilinear map $\beta_K : U\times V\to W$ is (by universal property) the same as a linear map $\beta_K: U\otimes_K V \to W$. Also note that the extension-by-scalars of a tensor product is the tensor product of extensions-by-scalars. Anyway, then I think that if $\beta: X \to W$ is surjective, then any field extension is surjective, and conversely. (con't) $\endgroup$ – Theo Johnson-Freyd May 18 '10 at 23:14
  • $\begingroup$ (con't) Indeed: for finite-dimensional spaces we can simply pick bases and then test whether $\beta$ is surjective by some elementary row operations, and those operations are defined over the minimum field that contains the matrix coefficients for $\beta$. In particular, the whole point of extending by scalars is that you don't change the matrix coefficients. So I think the answer to Question A is "all field extensions" and to Question B is "yes", but not for deep reasons. $\endgroup$ – Theo Johnson-Freyd May 18 '10 at 23:16
  • $\begingroup$ So, since I think the question is trivial, I probably misinterpreted it. Or maybe you mean to be asking about extensions of rings? Then it is certainly nontrivial. $\endgroup$ – Theo Johnson-Freyd May 18 '10 at 23:17
  • 2
    $\begingroup$ @Theo: A bilinear map induces a map on the tensor product, but they are not the same. In particular, the latter can be surjective even if the former is not. $\endgroup$ – Bjorn Poonen May 19 '10 at 1:50
  • $\begingroup$ @Theo: Not every element of the tensor product space is decomposable, mathoverflow.net/questions/23478/… $\endgroup$ – Victor Protsak May 19 '10 at 2:10

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.

  • $\begingroup$ Nice! How to eliminate refering to MAGMA? $\endgroup$ – Victor Protsak May 19 '10 at 2:19
  • $\begingroup$ Awesome! I will think a bit about question A and eventually ask for something precise. $\endgroup$ – Xandi Tuni May 19 '10 at 17:47
  • $\begingroup$ @Victor: I used MAGMA only to compute a Weierstrass model for the Jacobian of the curve, and then to compute its group of rational points. The curve turns out to be 20A2 in Cremona's labelling, so the group of rational points can be looked up in tables instead of using MAGMA (it's Z/6Z, and the six rational points are the ones where one of u_1,u_2,u_3 is 0). This eliminates all but the computation of the Weierstrass model, and that too shouldn't be too difficult to do by hand. $\endgroup$ – Bjorn Poonen May 20 '10 at 3:20

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