Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $x_1.$

share|cite|improve this question
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

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 $K$

share|cite|improve this answer
Great! Thank you very much. – Tom Jan 10 '13 at 1:59

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 $f$...

share|cite|improve this answer
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
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
Yeah, I think you're right Olivier. So my idea is too simple. – Ted Jan 6 '13 at 4:18
Probably this?:$x_1x_2^2\cdots x_n^n$ – Tom Jan 6 '13 at 6:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.