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Timeline for Surjectivity of bilinear forms.

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May 19, 2010 at 17:53 history edited Xandi Tuni CC BY-SA 2.5
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May 19, 2010 at 17:49 vote accept Xandi Tuni
May 19, 2010 at 2:10 comment added Victor Protsak @Theo: Not every element of the tensor product space is decomposable, mathoverflow.net/questions/23478/…
May 19, 2010 at 1:54 answer added Bjorn Poonen timeline score: 11
May 19, 2010 at 1:50 comment added Bjorn Poonen @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.
May 18, 2010 at 23:17 comment added Theo Johnson-Freyd 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.
May 18, 2010 at 23:16 comment added Theo Johnson-Freyd (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.
May 18, 2010 at 23:14 comment added Theo Johnson-Freyd 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)
May 18, 2010 at 21:00 history edited Xandi Tuni CC BY-SA 2.5
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May 18, 2010 at 19:53 history asked Xandi Tuni CC BY-SA 2.5