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What is a good reference for the following statement:

The transcendence degree of a separably generated field extension $K/L$ is equal to the dimension of the $K$-vector space of derivations $D:K\to K$ that kill $L$.

This statement is used in the very beginning of Mumford's book "Complex projective varieties".

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    There is an excellent discussion of separating transcendence bases in Matsumura's book "Commutative Algebra". See in particular section 27. (Note that Matsumura's "Commutative Algebra" is distinct from his "Commutative Ring Theory", although they have some significant overlap with one another.)2012-10-16

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I don't know a reference, but here is a proof:

  • Choose $x_1,\ldots,x_n \in L$ such that the $x_i$ are algebraically independent over $L$, and $K$ is separable over $L(x_1,\ldots,x_n)$.

  • Check that any $K$-valued derivation $D$ on $L(x_1,\ldots,x_n)$ extends uniquely to $K$. (To this end, if $x \in K$, write $f(x) = 0$ for some separable polynomial $f \in L(x_1,\ldots, x_n)[x]$, and applying $D$ to both sides, deduce that $f'(x)Dx+ (Df)(x) = 0$ (where $Df$ denotes the result of applying $D$ to the coefficients of $f(x)$), hence $Dx = - (Df)(x)/f'(x).$ (This is where separability is used, to be sure that $f'(x) \neq 0.$))

  • Now you are reduced to the case $K = L(x_1,\ldots,x_n)$, in which case the space of such $D$ is spanned over $L$ by the derivitations $\partial_{x_i}$, as is easily checked.

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    @aglea$n$er: Dear aglea$n$er, You're welcome. Regards,2012-10-17
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You can find that result as Theorem 23.12 in page 215 of Patrick Morandi's excellent book Field and Galois Theory. That section in fact is dedicated to derivations and differentials and even though it's been a while since I read that section of the book, I remember it to be nicely written, just as the whole book.

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    Adrian, thank you for the link to the book! Indeed it seems to be very clearly written.2012-10-16