Linear independence in the algebraic closure of $\mathbb{C}(z)$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T02:09:53Z http://mathoverflow.net/feeds/question/38348 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38348/linear-independence-in-the-algebraic-closure-of-mathbbcz Linear independence in the algebraic closure of $\mathbb{C}(z)$ Alex 2010-09-10T20:11:14Z 2010-09-11T01:21:52Z <p>Fix $N>0$. Let $b_i=(b_{i,1}, b_{i,2}, b_{i,3}, b_{i,4})$, $i=1,\ldots, m$, be distinct 4-tuples of integers with with all $0\leq b_{i,j}&lt; N$. (The zero tuple is disallowed.) </p> <p>Define $w_i=(\prod_{j=1}^4 (z-z_j)^{b_{i,j}})^{\frac1{N}}$. </p> <p>Consider $w_i$ as an element of the following vector space: the algebraic closure of $\mathbb{C}(z)$ over the field $\mathbb{C}(z)$. </p> <p>I believe that Puiseux series can be used to show the linear independence of the $w_i$. Are there any other approaches that might show the linear independence of the $w_i$? </p> <p>More generally, are there any other general techniques for proving linear independence of functions in the given vector space? </p> http://mathoverflow.net/questions/38348/linear-independence-in-the-algebraic-closure-of-mathbbcz/38375#38375 Answer by Felipe Voloch for Linear independence in the algebraic closure of $\mathbb{C}(z)$ Felipe Voloch 2010-09-11T01:21:52Z 2010-09-11T01:21:52Z <p>You can also use Galois theory or monodromy. Take a minimal linear dependence relation and apply the automorphimsm of the algebraic closure that fixes $(z-z_j)^{1/N},j>1$ and fot $j=1$ multiplies the function by an N-th root of unity, thus getting a new relation and you can produce a shorter relation from those two. </p>