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First, is the following: $$f=\frac{3}{5}(x_1^5 + x_2^5 + x_3^5 + x_4^5)-\frac{7}{12}(x_1^2x_2^2 - x_1^2x_3^2-x_1^2x_4^2-x_2^2x_3^2-x_2^2x_4^2-x_3^2x_4^2)$$ a symmetric polynomial? And, if yes, how do you write $f$ as a linear combination of the following standard symmetric polynomials:

$f_1=x_1+x_2+x_3, f_2=x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4, f_3=x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4, f_4=x_1x_2x_3x_4?$

I know of a method to answer this using Groebner basis. However, the program tells me $f$ isn't symmetric, but it seems like it must be symmetric.

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    $\begingroup$ Why do you think it "must be" symmetric. As it is written, It looks like $x_1^2x_2^2$ has a different sign than the other similar terms. $\endgroup$ – GEdgar Jun 28 '16 at 14:08
  • $\begingroup$ @user264885: Definitions are your friends. Not understanding a definition can be a legitimate reason to ask about how to apply a definition to a particular case, as you've done here. $\endgroup$ – hardmath Jun 29 '16 at 2:36
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A polynomial in $n$ variables is symmetric exactly when you can do any permutation on the variables and leave the polynomials unchanged. In other words, you must have $$f(x_1, \dots, x_n) = f(x_{\sigma(1)}, \dots, x_{\sigma(n)})$$ for any permutation $\sigma \in \mathfrak{S}_n$. The standard symmetric polynomial verify this, so a linear combination of them verifies this too. It's then an exercise to show the converse is true.

In your example, if you exchange the variables $x_2$ and $x_3$ (the permutation $(2 \; 3) \in \mathfrak{S}_5$), then the polynomial changes because of the signs in front of the monomials $x_1^2 x_2^2$ and $x_1^2 x_3^2$. So the polynomial is not symmetric.

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