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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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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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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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