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 $k$?
Transcendence degree for a $k$-algebra which is an integral domain
3
$\begingroup$
abstract-algebra
commutative-algebra
field-theory
-
3Yes: http://mathoverflow.net/questions/752$1$9 – 2012-10-28
1 Answers
-2
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.
-
0It means, that they are algebraically independent over $k$ and for every $r \in R$ the elements $u_1, ..., u_n$ are dependent. – 2012-10-29