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Questions in which polynomials (single or several variables) play a key role. It is typically important that this tag is combined with other tags; polynomials appear in very different contexts. Please, use at least one of the top-level tags, such as nt.number-theory, co.combinatorics, ac.commutative-algebra, in addition to it. Also, note the more specific tags for some special types of polynomials, e.g., orthogonal-polynomials, symmetric-polynomials.

6 votes
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If some powers of polynomials are linearly independent, does it imply higher powers are also...

Let $P_1,\dotsc,P_k$ be polynomials. Assume they are pairwise non-proportional (i.e., any two of them are linearly independent). … The question Are large powers of polynomials linearly independent? concerns showing that there exists an $N$ giving independence. …
Zach Teitler's user avatar
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5 votes
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Ideals: If $\langle f_1,f_2 \rangle = \langle g_1,g_2 \rangle$, then $\langle f_1-\lambda,f_...

The answer to your questions is no. The ideals $\langle x, y(1-xy) \rangle$ and $\langle x, y \rangle$ are equal, and maximal; but $$ \langle x-\lambda, y(1-xy)\rangle \neq \langle x-\delta, y-\epsilo …
Zach Teitler's user avatar
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3 votes
Accepted

Modulo $x^2 + y^2 - 1$, is every homogeneous polynomial that is a square of a polynomial, ne...

These are the same polynomials that Will Sawin's answer ends with. …
Zach Teitler's user avatar
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4 votes

Dimension of a general partial derivative of a linear subspace of polynomials

Change coordinates, or act by a linear transformation, so that $U$ is a general subspace and we are differentiating by $\partial = \frac{\partial}{\partial x_1}$. Since $U$ is general, it has a basis …
Zach Teitler's user avatar
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1 vote
Accepted

Conjectures inspired in the context of Casas-Alvero conjecture, via the logarithmic derivati...

I'm probably making a stupid mistake, but is Conjecture 1 possibly rather easy? At $\ell=0$ we're assuming $$ p(x) = \frac{a_n}{n!} \left( \frac{n}{\frac{d}{dx} \log p(x)} \right)^n $$ Write $\frac{d} …
Zach Teitler's user avatar
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1 vote

The relationship between the symmetric tensor product and sum of squares

z^2\})$, although of course they map to the same polynomials. … But if $V = \operatorname{Sym}^d(\operatorname{span}\{x_1,\dotsc,x_n\})$ is already the space of degree $d$ polynomials in the $x_i$, then $\operatorname{Sym}^2(V)$ is not the space of degree $2d$ polynomials
Zach Teitler's user avatar
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4 votes
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Polynomial as a sum of powers of linear forms (with restrictions)

So, this argument shows that you can impose any finite number of your nonvanishing restrictions and the $L^d$'s that remain are still sufficient to span all the degree $d$ homogeneous polynomials. … A reference for this is: Białynicki-Birula, Schinzel, Representations of multivariate polynomials by sums of univariate polynomials in linear forms, Colloq. Math. 112 (2008), no. 2, 201–233. …
Zach Teitler's user avatar
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8 votes
Accepted

Linear spaces secant to Veronese varieties

Here is an answer in terms of power sum decompositions of polynomials. A point $p \in \mathbb{P}^9$ corresponds to a homogeneous polynomial $P$ of degree $3$ in $3$ variable, defining a plane cubic. … For more on homogeneous polynomials with infinitely many rank decompositions, look for keywords like "identifiable polynomials" (or non-identifiable, really). Good luck! …
Zach Teitler's user avatar
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4 votes
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System of polynomial equations with a known root

If $p = (a_1,\dotsc,a_n)$ is a known solution and your system of equations is given by the ideal $I$, then a system of equations for all the other solutions is given as follows. Let $m_p = (x_1-a_1,\d …
Zach Teitler's user avatar
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7 votes
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Can a positive polynomial on sphere be represented as the sum of squares of spherical harmonics

It is not true that positivity on the unit sphere implies representation as a sum of squares of polynomials. … In either case, it does not imply that $p$ is a sum of squares of polynomials. …
Zach Teitler's user avatar
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2 votes
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Lüroth theorem for $k \subset k(f,g) \subseteq k(x)$

This answer is just repeating comments, above, about a negative answer to conjecture (2). I'm unable to say anything for question (1). The conjecture (2) is not correct. As you note, $k(x) = k(x^3,x^ …
Zach Teitler's user avatar
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3 votes
Accepted

Combinatorial formula for Betti numbers of a $k[x,y]$-module

It seems unlikely that the Betti numbers can be determined by the data you list (bigraded Hilbert function and ranks of multiplication maps). Consider the following possibility: $\dim M_{1,0} = \dim M …
Zach Teitler's user avatar
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4 votes

Complexity of a polynomial

Let $k_1,\dotsc,k_r$ be the even integers with even $2$-adic valuation in the range you want. For any polynomial $p$ meeting your condition, $p(x)-2^{n-1}$ has a root between $k_i-2$ and $k_i$, and a …
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1 vote

Linear homogenous polynomials that generates one quadratic polynomial

Each ideal $\langle P_i, Q_j \rangle$ is a radical ideal. So in fact $x_0 \in \langle P_i, Q_j \rangle$. That means that $x_0 \in \operatorname{span}(P_i,Q_j)$. This holds for each $i,j$. Case 1. Eve …
Zach Teitler's user avatar
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1 vote

About local maxima of multivariable polynomials

The set of critical points (in the domain) of a polynomial is the solution set of a system of polynomial equations viz the vanishing of the first derivatives. So it has finitely many irreducible compo …
Zach Teitler's user avatar
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