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Over rational numbers, the set of all power functions up to a certain degree generate all symmetric polynomials in that degree.

My question is as follows. To be succinct, let's say we have four variables. Are all homogeneous symmetric functions of a given a degree (let's say four) in $x,y,z,w$ generated over rational numbers by the special power functions given by $$\begin{align*} &x^4,y^4,z^4,w^4,(x+y+z+w)^4, (x+y+z)^4, (x+y+w)^4, (x+w+z)^4, \\ &(w+y+z)^4, (x+y)^4, (z+y)^4, (w+y)^4, (x+w)^4, (w+z)^4,\text{ and }(x+z)^4\,? \end{align*}$$

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From the statement that the powers sums generate all symmetric polynomials it may be inferred that by "generate" you mean "generate as an $R$-algebra" for $R$ the ring of coefficients. Since the answers so far assumed otherwise, it may be useful to specify this. Note that apart from $(x+y+z+w)^4$ these are not symmetric polynomials. Your question is if the generated algebra contains all symmetric polynomials. – Marc van Leeuwen Dec 8 '11 at 10:07
@Marc van Leeuwen: I just meant as a formal linear combination over the rational numbers. – username Dec 8 '11 at 15:07
up vote 3 down vote accepted

I assume that by "generated over rational numbers" you mean "equal to a linear combination with rational coefficients". In this case the answer is no: for example, the symmetric function$$ p(x,y,z,w) = x^2y^2 + x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2 + z^2w^2 $$ is not a linear combination of the 15 polynomials you listed. To see this, note that if $p$ did equal a linear combination of those 15 power functions, then that identity would remain valid when the specific values $y=0$, $z=0$, $w=1$ are substituted in. However, $p(x,0,0,1) = x^2$, while each of those 15 power functions equals one of $0$, $1$, $x^4$, or $(x+1)^4$ when these values are substituted in; and it's easy to verify that $x^2$ cannot be written as a linear combination of $0$, $1$, $x^4$, and $(x+1)^4$.

(Some symmetric functions, such as $xyzw$, can be written as a linear combination of those 15 power functions; so I don't know what general principle underlies this.)

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Thanks, Greg for the answer and examples. Actually, I was interested in the product (in this case, $xyzw$) and worked that out for a few cases and thought this might work in general. Is there an argument that will work in general for products of variables when $n>4$? – username Dec 8 '11 at 15:04
Never mind, got it. – username Dec 8 '11 at 23:43

If you have $n$ variables and look at the dimension $H(d)$ of the vector space of homogeneous symmetric polynomials of degree $d$, $H$ grows like a polynomial in $d$ of degree $n-1$.

Meanwhile, the number of "special power functions" you have is always $2^n-1$, and doesn't grow when $d$ grows. Thus for $n>1$ and a high enough degree, you can't generate all the homogeneous symmetric polynomials of degree $d$ only with the special power functions.

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The answer is obviously no: since all polynomials you wrote down are homogeneous polynomials of degree $4$ with respect to the total degree, so they cannot be used to build anything with terms of total degree not divisible by $4$, like $xyzt+x+y+z+w$.

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Thanks, I meant to say homogeneous of a given degree. – username Dec 8 '11 at 15:05

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