transcendence base

**Stiger** · January 10th 2009, 08:56 AM

Let

be the quotien field of

Exhibit a transcendence base

for

over

, and express

explicitly as an algebraic extension over

.

**NonCommAlg** · January 10th 2009, 10:35 AM

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 $f=x_1^2 + x_2^2 + x_3^2 - 1$ and $v_j=x_j + <f>, \ \ 1 \leq j \leq 5.$ the claim is that $S=\{v_2,v_3,v_4,v_5 \}$ is a transcendence basis for $F=\mathbb{C}(S,v_1)$ over $\mathbb{C}$ : since $v_1^2+v_2^2 +v_3^2-1=0,$ the element $v_1$ is

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

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

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

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