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For a field $k$, I know that $k(x_1,\cdots,x_n)/k(s_1,\cdots,s_n)$ is a finite Galois extension with Galois group $S_n$ where $s_i$ is an elementary symmetric polynomial. Thus its dimension is $n!$.

What is its base?

Edit: base -> basis

Edit2:I want an explicit example of a basis. If its proof why it is a basis is complicated then I want to see how the basis represents some concrete examples of polynomials like $x_1.$

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What do you mean by «its base»? –  Mariano Suárez-Alvarez Jan 6 '13 at 3:30
    
He likely wants to know what a basis of $k(x_1,\dots,x_n)$ looks like as a $k(s_1,\dots,s_n)$ vector space. –  JSchlather Jan 6 '13 at 3:35
    
Now I'm interested in which elements generate a normal basis. –  anon Jan 6 '13 at 4:36
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2 Answers

An easy way to find a basis is to find some basis for each extension $k(s_1,\ldots,s_n) = K \subset K(x_1) \subset K(x_1,x_2) \subset \ldots \subset K(x_1,\ldots,x_n)$, and compose them.

Since the degree of $K \subset K(x_1)$ is $n$, $(1,x_1,\ldots,x_1^{n-1})$ is a basis of $K(x_1)$ over $K$. Similarly, since the degree of $K(x_1) \subset K(x_1,x_2)$ is $n-1$, $(1,x_2,\ldots,x_2^{n-2})$ is a basis of $K(x_1,x_2)$ over $K(x_1)$. Therefore, the family $\{x_1^{d_1}x_2^{d_2} ; \text{ where } 0 \le d_1 < n , 0 \le d_2 < n-1\}$ is a basis of $K(x_1,x_2)$ over $K$.

Repeat this process until you obtain the basis $\{\prod_{i=1}^n x_i^{d_i} ; \text{ where } \forall i, 0 \le d_i \le n-i \}$ of $K(x_1,\ldots,x_n)$ over $K$

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Great! Thank you very much. –  Tom Jan 10 '13 at 1:59
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Here's an idea that should work: Pick an element $f \in k(x_1, x_2, \ldots, x_n)$ which is not stabilized by any non-identity element of $S_n$. Then the orbit of $f$ under $S_n$ has $n!$ elements. This orbit ought to be a basis.

Edit: This doesn't work; see comments below. On the other hand, by the normal basis theorem, there exists an $f$ for which this works. So we need to put more restrictions on $f$...

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what about $x_1+2x_2+\cdots+nx_n$? –  Olivier Bégassat Jan 6 '13 at 3:45
    
@Olivier I don't see an obvious reason why that element doesn't work. –  Ted Jan 6 '13 at 3:56
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Consider two $n$-cycles $\sigma,\tau$ that aren't powers of each other, for instance $\sigma=(123\cdots n)$ and possibly $\tau=(n\cdots 321)$, then setting $x=x_1+\cdots +nx_n$, we have $$x+\sigma\cdot x+\cdots +\sigma^{n-1}\cdot x=x+\tau\cdot x+\cdots +\tau^{n-1}x=\frac{n(n+1)}{2}(x_1+\cdots+x_n)$$ If $n$ is prime, $\sigma^i$ are distinct from the $\tau^j$ unless $i=j=0$, and thus the points in the orbit aren't linearly independent. I'm not 100% sure, tell me if this fails. –  Olivier Bégassat Jan 6 '13 at 4:10
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Yeah, I think you're right Olivier. So my idea is too simple. –  Ted Jan 6 '13 at 4:18
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Probably this?:$x_1x_2^2\cdots x_n^n$ –  Tom Jan 6 '13 at 6:50
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