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If one has an extension of domains $D_1\subset D_2$ such that the extension of fields of fractions $K_{D_2}/K_{D_1}$ is transcendental of degree $r$ (with $K_{D_i}$ the field of fractions of $D_i$), is it always possible to find $r$ algebraically independent elements over $K_{D_1}$ that lie in $D_2$?

Obviously you can take $r$ algebraically independent elements in $D_2$; you can even take them with the same denominator; let's say $u_1/v,\ldots,u_r/v$. Now, I haven't been able to prove if the $u_i$ are algebraically independent over $D_1$...

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    It's not true that the $u_i$ be necessarily algebraically independent over $K(D_1).$ Take the case of $D_1 = \mathbb{Q},$ $D_2 = \mathbb{Q}[X],$ $v = X$ and $u_1 = 1.$2011-07-27
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    You're absolutely right... thanks a lot!2011-07-27

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Yes. The claim can be proved by an induction on $n$ where $0\leq n \leq r$ for the statements:

$P_n:$ there exist $n$ algebraically independent elements of $D_2$ over $K(D_1).$

The case where $n=0$ is trivial.

Thus, sssume we have proven $P_n$ for some $n

Assume every element of $D_2$ is algebraic over $k(D_1').$ Then so too is every quotient of elements from $D_2.$ It follows $$0 = deg(K(D_2)/k(D_1')) = deg(K(D_2)/K(D_1)) - deg(K(D_1')/K(D_1)) = r -n > 0, $$

a contradiction.

We conclude that there must exist an element of $D_2$ transcendental over $K(D_1').$ Hence, $P_{n+1}$ holds. The claim follows.

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    Great answer, thanks!2011-07-27