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Let $F|K$ be a finite field extension. I want to prove that if $E|K$ is an algebraic extension, then the bilinear form $$ F\times F\rightarrow K, (x,y)\mapsto Tr_{F|K}(xy) $$ is nondegenerate if and only if $$ E\otimes_{K} F\times E\otimes_{K} F\rightarrow E, (x,y)\mapsto Tr_{E\otimes_{K}F|E}(xy) $$ is nondegenerate.

I don't know where to start from.

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  • $\begingroup$ Hm. Are you sure you mean $Tr_{E\otimes_KF\mid K}$? $\endgroup$
    – Mariano Suárez-Álvarez
    Oct 14, 2015 at 7:48
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    $\begingroup$ My answer is about $Tr_{E\otimes_KF\mid E}$. If the characteristic of $K$ is zero, for example, the trace form over $K$ of $E\otimes_KF$ is nondegenerate iff $E\otimes_KF$ is a semisimple $K$-algebra, and your hypotheses are not enough to guarantee that. $\endgroup$
    – Mariano Suárez-Álvarez
    Oct 14, 2015 at 8:10
  • $\begingroup$ the usualone, if the first is made up of elements $f_i$, the second one is just the one made up with the corresponding $1\otimes f_i$. $\endgroup$
    – Mariano Suárez-Álvarez
    Oct 14, 2015 at 8:27
  • $\begingroup$ I suggest that you try to prove that $\{1\otimes f_i\}$ is a basis of $E\otimes_KF$ as an $E$-vector space. That $\{f_i\}$ is not a basis of $F$ as an $E$-vector space has nothing to do with this (and in fact does not even make sense, as $F$ is not an $E$-vector space in any sensible way :-) ) $\endgroup$
    – Mariano Suárez-Álvarez
    Oct 14, 2015 at 8:37
  • $\begingroup$ The $L$ is probably a typo. $\endgroup$
    – Mariano Suárez-Álvarez
    Oct 14, 2015 at 8:40

1 Answer 1

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Non-degeneracy of a bilinear form is equivalent to having the matrix of the form with respect to a basis have non-zero determinant.

Show that your two bilinear forms have matrices with respect to appropriate bases of their underlying vector spaces which have the same determinant — in fact, the two matrices can be chosed to be the same matrix!

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