Wednesday, November 09, 2005

Algebraic Independence

So, I was going to write a post explaining algebraic independence, but then I decided that it's probably nearly impossible to explain it so that *everyone* understands. Instead, I've decided to post the summary that makes sense to me. If you want more details, feel free to ask.

A subset $X = \{x_1, \ldots, x_n\}$ is \emph{algebraically independent} over a field $F$ if there does \emph{not} exist a polynomial $f$ with coefficients in $F$ such that $f(x_1, \ldots, x_n)=0$. The elements of $X$ do not have to be in $F$.

A transcendental number is algebraically independent over the rationals, $\Q$.

I hope you don't mind the LaTeX... I just cut and pasted from my notes.

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