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Let $k$ be a subfield of two fields $E$ and $F$. If a field extension $L/E$ is purely transcendental, is it true that the field extension $F.L/F.E$ obtained by taking the free composite of field extensions with $F$ on both sides by $F$ is also purely transcendental ?

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    $\begingroup$ I am unsure about the definition of free composite. The tensor product of $E$ and $F$ is in general not a domain. What do you mean by field of fractions ? $\endgroup$ – user18119 Jan 21 '13 at 9:02
  • $\begingroup$ You are right, it is more complicated than i thought. I edited. $\endgroup$ – Louis La Brocante Jan 22 '13 at 9:02
  • $\begingroup$ The closest thing to a definition of the free composite I can found comes from this mathoverflow question. $\endgroup$ – JSchlather Jan 23 '13 at 6:30
  • $\begingroup$ The free composite may just be defined as the minimal field extension which contains the two. However it seems that in the comment of @JacobSchlather people do as if the tensor product was a domain... As anybody an idea if it is really the case ? (perhaps there should be some assumptions on the extensions) $\endgroup$ – Louis La Brocante Jan 25 '13 at 15:21
  • $\begingroup$ The tensor product of extensions is not a domain in general (take $E=F$ finite over $k$ of degree $>1$, then $E\otimes_k F$ is not a domain). What could be reasonable is to take the residue field of the tensor product at some minimal prime ideal. $\endgroup$ – user18119 Jan 25 '13 at 16:30

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