5
$\begingroup$

This theorem stays that any symmetric polynomial can be expressed as a polynomial of elementary polynomials. So let's suppose I have a symmetric polynomial $f(x_1,x_2,...,x_n)$ in $R[\mathbb{X}]$. I can find a unique polynomial $g(s_1,s_2,...,s_n)$ in $R[\mathbb{X}]$ so I can express f in term of g, where $s_1,s_2,...,s_n$ are elementary polynomials. My question to you is: Is there a algorithm/set of stepts to determine g? Thanks.

$\endgroup$

2 Answers 2

2
$\begingroup$

Let $c x_1^{a_1} x_2^{a_2} \dots x_n^{a_n}$ be the lexicographically largest monomial of $f$, that is there are no monomials with strictly larger exponent of $x_1$, and no monomials with $x_1$ exponent ${a_1}$ that have a higher power of $x_2$, and so on. We'll think of this as being the leading term of the polynomial.

Now the key thing to notice is that $e_n^{a_n}e_{n-1}^{(a_{n-1}-{a_n})} \dots e_1^{(a_1-a_2)}$ contains the monomial $x_1^{a_1} x_2^{a_2} \dots x_n^{a_n}$ with coefficient $1$ and all other monomials it contains are smaller lexicographically.

Now the point is you can consider the leading term of $f - c e_n^{a_n}e_{n-1}^{(a_{n-1}-{a_n})} \dots e_1^{(a_1-a_2)}$, and so on. Each step reduces the leading term (in lexicographic order), so this process must eventually terminate, at which point you have written $f$ in terms of the elementary symmetric functions.

$\endgroup$
2
  • $\begingroup$ Ok, you kinda lost me at the last paragraph. Can you please show me how this work on a example. Let's say I have the following polynomial: $$ P(x_1,x_2,x_3) = x_1^4+x_2^4+x_3^4-2x_1^2x_2^2-2x_2^2x_3^2-2x_1^2x_3^2 $$. Thanks a lot for your help. $\endgroup$ Mar 9, 2016 at 22:35
  • $\begingroup$ The lexicographically largest term is $x_1^4$. So according to the algorithm described you should subtract off $e_1^4 = (x_1+x_2+x_3)^4$. The point being that what remains, while having more terms, will have a smaller leading term namely $-4x_1^3x_2$. Then we apply the same operation again to this polynomial $P - e_1^4$, this time adding $4e_2e_1^2$ to cancel the leading term. Next we would cancel the leading term $P - e_1^4 + 4e_2e_1^2$, and so on. Eventually we are left with $P - g(e_1,e_2,e_3)$ having no leading term and hence $P - g(e_1,e_2,e_3) = 0$ as desired. $\endgroup$
    – Nate
    Mar 10, 2016 at 15:14
1
$\begingroup$

You can use divided difference operators to express the polynomial in terms of Schur polynomials, then you can apply the second Jacobi-Trudi formula, which expresses a Schur polynomial as the determinant of a matrix whose entries are elementary symmetric polynomials. Computationally this is rather inefficient, but it is nonrecursive, which is sometimes advantageous.

$\endgroup$
2
  • $\begingroup$ Hey, thank you for your answare. Can you please show me how this would work on a real example. See the polynomial in the comment section for @Nate 's answare. $\endgroup$ Mar 10, 2016 at 15:10
  • $\begingroup$ @Raducu I don't have a computer algebra system here, but my hand calculations resulted in $\partial^{4123}(p)=3$, $\partial^{5423}(p)=-1$, $\partial^{3423}(p)=-2$, so $p=3s_{(31)}-s_{(211)}-2s_{(22)}$, and you can go to the Wikipedia article on the Jacobi-Trudi formula and substitute what it gives you. $\endgroup$ Mar 10, 2016 at 16:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .