0
$\begingroup$

Disregarding the ordering of monomials, can a polynomial have two formal writings?

When studying symmetric polynomials my manual says that there's a unique representation of a symmetric polynomial in terms of elementary symmetric polynomials; But I think it can be generalized to all polynomials.

Let $f(x_1,x_2,x_3)$ be a polynomial $\in \mathbb{K}[x_1,x_2,x_3]$. Since $x_1 x_2,x_3$ are independent indeterminates, if there were a relation such as $x_1 = x_2^2\cdot x_1 + x_3\cdot x_1$ or the like, I could replace $x_1$ by the former relation and get a second formal writing. But if $x_1 x_2,x_3$ are independent it should not be possible. I'm not sure about this fact and I'd like some clarification.

$\endgroup$
3
  • 1
    $\begingroup$ If your polynomial is not symmetric, then you cannot write it as a polynomial expression of elementary symmetric polynomials. $\endgroup$
    – Qi Zhu
    Nov 10, 2020 at 9:42
  • $\begingroup$ it's not what I'm asking. In the case of symmetric polynomials: $\sum b_i* \sigma_1\cdot\ldots \cdot\sigma_n = \sum c_i* \sigma_1\cdot\ldots \cdot\sigma_n$ with $\sigma$s the symm. poly. then $b_i=c_i$ for all index i.e. there's a unique representaion of the polynomials. I'm asking if it is true for all polynomials that there's a unique representation. $\endgroup$ Nov 10, 2020 at 9:45
  • 2
    $\begingroup$ What you say is correct. Every polynomial in $K[x_0,x_1,\ldots,x_{n-1}]$ corresponds to exactly one "formal writing" of the form $\sum_{i\in\mathbb N^n}a_ix_0^{i_0}x_1^{i_1}\cdots x_{n-1}^{i_{n-1}}$, i.e. $a_i\in K$ for $i\in\mathbb N^n$ are uniquely determined. This is, however, not very related to the claim from your manual, and is usually taken either as a definition of polynomials in multiple variables, or it is proven (as one of the first statements) to be equivalent to whatever other definition is chosen... or is often just left out as intuitively obvious. $\endgroup$ Nov 10, 2020 at 9:51

2 Answers 2

1
$\begingroup$

That there are no relations between $x_1, x_2$, and $x_3$ is part of the definition of the polynomial ring.

That there are no relations between elementary symmetric polynomials is a non-trivial theorem.

One can easily construct rings of certain polynomials where there are relations, e.g. take the ring generated by $\{x^2,y^2,xy\}$ (i.e. all polynomials without linear terms). There is a relation between these generators, namely $x^2\cdot y^2 = (xy)^2$. Note that this is a property of a certain set of generators, not the ring itself.

$\endgroup$
0
$\begingroup$

The only intrinsic way to write polynomials in $n$ variables ${x}=(x_1,...,x_n)$ is as sum $$ P({x})=\sum_{I}a_Ix^I $$ where $I=(k_1,...,k_n)$ is a multindex with each $k_i\geq0$ and $x^I=x_1^{k_1}\cdots x_n^{k_n}$.

The polynomial identity principle says that two such polynomials are equal if and only if the coefficients $a_I$ of the two coincide.

For special purposes you can choose to write polynomials in some special form. For instance notice that $$ K[x_1,...,x_n]=\left(K[x_1,...,x_{n-1}]\right)[x_n], $$ but in the end it is just a reshuffling of the terms.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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