1. ## transcendence base

Let be the quotien field of
Exhibit a transcendence base for over , and express explicitly as an algebraic extension over .

2. Originally Posted by Stiger
Let be the quotien field of
Exhibit a transcendence base for over , and express explicitly as an algebraic extension over .
let $\displaystyle f=x_1^2 + x_2^2 + x_3^2 - 1$ and $\displaystyle v_j=x_j + <f>, \ \ 1 \leq j \leq 5.$ the claim is that $\displaystyle S=\{v_2,v_3,v_4,v_5 \}$ is a transcendence basis for $\displaystyle F=\mathbb{C}(S,v_1)$ over $\displaystyle \mathbb{C}$: since $\displaystyle v_1^2+v_2^2 +v_3^2-1=0,$ the element $\displaystyle v_1$ is

algebraic over $\displaystyle \mathbb{C}(S)$ and thus $\displaystyle F$ is algebraic over $\displaystyle \mathbb{C}(S).$ so to prove that $\displaystyle S$ is a trancendence basis for $\displaystyle F$ over $\displaystyle \mathbb{C},$ we only need to show that $\displaystyle S$ is algebraically independent over $\displaystyle \mathbb{C}.$ so suppose that

the elements of $\displaystyle S$ are algebraically dependent over $\displaystyle \mathbb{C}.$ then there exists $\displaystyle 0 \neq g \in \mathbb{C}[y_1, y_2, y_3, y_4]$ such that $\displaystyle g(v_2,v_3,v_4,v_5)=0,$ i.e. $\displaystyle g(x_2,x_3,x_4,x_5) \in <f>,$ which is obviously impossible because

every non-zero element of $\displaystyle <f>$ involves $\displaystyle x_1$ but $\displaystyle g$ does not. $\displaystyle \Box$