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3 votes
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Is every finite collection of points in $\mathbb{C}^n$ the solution to a compatible algebraic system?

Consider a set of 6 points in $\mathbb{C}^2$ which do not all lie on a common conic -- for example, $\{ (0,0), (1,0), (2,0), (0,1), (1,1), (2,2) \}$. Then $6 = 2 \cdot 3$, but clearly the set cannot ...
Daniel Schepler's user avatar
3 votes

$P(x)$ with rational coefficients, but $P(P(x))$ with integer coefficients

(Not a solution. Too long to be a comment. I pointed out that Srini's solution is in error, so here is my attempt at fixing it.) Degree 2 case: $$f(x) = ax^2 + bx + c \\ f(f(x)) = a^3x^4 + 2a^2bx^3 + ...
Calvin Lin's user avatar
  • 72.2k
2 votes

Proof of this derivative definition for polynomials in a finite field polynomial ring? $f(x)=q(x)(x-\alpha)^2+f'(\alpha)(x-\alpha)+\beta$

Ok, first of all, I don't like how they define it because it is kinda confusing. First note that the set $B_{\alpha}=\{(x-\alpha)^n\mid n\in\mathbb{N}_0\}$ is a basis of $\mathbb{F}_q[x]$ for all $\...
schiepy's user avatar
  • 208
2 votes
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Studying the irreducibility of polynomials

For $g$ you write it as a cubic polynomial in $y$: $$g = x y^3 + (x + 1) y^2 + (x^4 + x^3) y + x^5 \in K[x]\bigl[y\bigr].$$ Since this is of degree $3$ and $K[x]$ is factorial, it suffices to show ...
Martin Brandenburg's user avatar
2 votes
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Proving Simplicity of Zeros of a Polynomial.

Take $f$ to be the polynomial $$f(z)=1+z+z^2+\cdots+z^n.$$ Its roots are the $n$-th roots of unity different from $1$. For each of those roots, take a small disk $D_k$ around it, such that the disks $...
Simon Pitte's user avatar
2 votes
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computing local cohomlogy of a polynomial ideal

Let me address the first question(s). This depends on how you define the local cohomology modules. A very popular reference is the book Local Cohomology by Brodmann and Sharp. If you use their ...
Jose's user avatar
  • 749
2 votes

Necessary and sufficient condition to at least two cubic polynomial roots have absolute value lower than 1

Let's call your polynomial $f(\lambda)$, and let $C$ be the positively oriented unit circle. Assuming none of the roots are on $C$, the number of roots (counted by multiplicity) inside the unit ...
Robert Israel's user avatar
2 votes
Accepted

If $L/K$ is an extension, then how to get an irreducible factor of $f(x) \in K[x]$, when an irreducible factor of it is given in $L[x]$?

As stated, this is wrong. Consider $K=\mathbb Q$, $L=\mathbb Q(\sqrt[3]2)$, $f=X^3-2$ and $g=X-\sqrt[3]2$. Then $\operatorname{Aut}_K(L)=\{\operatorname{id}\}$, so the linked definition of the norm ...
anankElpis's user avatar
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1 vote

If $(x-\lambda)^{2n-2}$ divides $\sum_{i,j}p_{i,j}(x)$, does it divide each $p_{i,j}(x)$?

Double indices are unnecessary here. Anyway, here is a counterexample: $$p_{1,1} =(x-\lambda)^{2n-2} +1, p_{1,2} =(x-\lambda)^{2n-2} -1$$ where $i \in \{1\}$ and $j \in \{1,2\}$.
Kritiker der Elche's user avatar
1 vote
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A homogeneous polynomial divides a polynomial if and if it divides each homogeneous component?

That‘s correct. First suppose that for all i $G|f_i$, then clearly $G|f$. On the other hand if $G|f$ let $f = G * h$ with $h = h_d + .. + h_0$ with homogenous terms $h_i$. Then $f = G*h_d + .. + G*h_0$...
Andreas Könen's user avatar
1 vote

Polynomial $f(x)$ has positive coefficients and all roots as real. How many polynomials formed from terms of $f(x)$ also have only real roots?

Some results from this answer are already in Blue's answer. Following it, we shall call a polynomial real-rooted, if each its root is real. We assume that the coefficients $a_0,\dots,a_n$ of the real-...
Alex Ravsky's user avatar
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1 vote
Accepted

$X$, $Y$, $U$, $V$ are independent on $\{0,1,\ldots,n-1\}$ with $X$, $Y$ uniform and $X+Y \sim U+V$

For $n=12$, we'll construct a counterexample . . . Claim:$\;$There exist independent random variables $X,Y,U,V$, valued on $\{0,...,11\}$, such that $X,Y$ are uniform.$\\[4pt]$ $U+V\sim X+Y$.$\\[4pt]...
quasi's user avatar
  • 59.5k
1 vote

Show a polynomial in two variables is irreducible

To answer your last question first: It's fine to write $f=q(x,y)y+r(x)$, but not the most helpful for the problem. However, it's not possible to write $f(x,y)=q(x)y^2+r(x)$, because the polynomial on ...
anankElpis's user avatar
  • 1,890
1 vote
Accepted

Infinite number of decompositions into sum of four cubes

If there is at least a single integer solution to, $$a^3+b^3+c^3+d^3 = N$$ such that $(a+b)(c+d)<0$ and $(a+b)(c+d) \neq -\square$, then we can find an infinite more using a Pell equation. We use ...
Tito Piezas III's user avatar
1 vote

Proof of this derivative definition for polynomials in a finite field polynomial ring? $f(x)=q(x)(x-\alpha)^2+f'(\alpha)(x-\alpha)+\beta$

By the binomial theorems one gets the Horner-Ruffini formula (that they used, not invented) for long division of a polynomial by a linear factor $$ f(X)=f(\alpha)+g(X,\alpha)(X-\alpha) $$ $g$ is a ...
Lutz Lehmann's user avatar
1 vote

Necessary and sufficient condition to at least two cubic polynomial roots have absolute value lower than 1

In the book titled, "Periodicities in Nonlinear Difference Equations", E. A. Grove and G. Ladas (2005, CRC Press) give many important results for the stability of polynomials that arise in ...
Sundar's user avatar
  • 2,772

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