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Let $R$ be an integral domain over a field $k$. Is it true, that $\deg.\mathrm{tr}_k \ \mathrm{Frac}(R)$ is the greatest number of elements of $R$ algebraically independent over $k$?

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Yes: – Georges Elencwajg Oct 28 '12 at 20:44

Every transcendence basis of an algebra is a transcendence basis of its quotient field. Because the set of algebraic elements over subfield is a subfield, and the lowest subfield containing algebra is its quotient field.

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1) I don't think the notion of transcendence basis of an algebra makes sense. Of course you can make up several definitions but it is not clear that they will be equivalent.2) Given such a definition why should a transcendence basis of $R$ be a transcendence basis of $Frac (R)$ ? – Georges Elencwajg Oct 28 '12 at 20:32
It's simple. Let ${u_1, ..., u_n}$ be the transcendence basis of R. Then $k(u_1, ..., u_n)$ is a subfield in $Frac(R)$. Let $F$ be the set of elements, algebraic over $k(u_1, ..., u_n)$. Then $R \subset F$, consequently $Frac(R) = F$. – user46336 Oct 28 '12 at 20:52
Dear user, what does it mean that "$u_1,..., u_n $ is the transcendence basis of $R$" ? – Georges Elencwajg Oct 28 '12 at 21:26
It means, that they are algebraically independent over $k$ and for every $r \in R$ the elements $u_1, ..., u_n$ are dependent. – user46336 Oct 29 '12 at 5:56

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