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11

This is a partial answer which highlights some of the subtleties of this question. I will use algebraic geometry language as this is the correct set up for such questions. First, this question is only really interesting for affine varieties as for projective varieties, every rational point is an integral point. I take the following set-up. Let $U$ be a ...

10

It is indeed true that among random polynomials only a very tiny fraction of them are counterexamples. Even after a few days my program found only very few examples among thousands of tested cases. In all findings it turned out that the critical point which is supposed to lie outside of $B$ actually is very close to the border of $B$. The following is a ...

6

Note that as soon as $p$ is decomposable, its Galois group ${\rm Gal}(p/\mathbb{Q})\leq S_d$ is imprimitive (in fact it is contained in a wreath product), in particular it is not $A_{d}$ or $S_{d}$. Many of the results for random polynomials that give irreducibility with high probability also give Galois group $S_d$ (or sometimes $S_d$ or $A_d$) with high ...

4

Expanding on some of the comments, I think that Vojta's conjecture will imply a strong result related to your question. First I'm going to reformulate the question in terms of Zariski density, which I think is more natural, since it can be applied inductively to reduce to the case of lower dimensional varieties. Set-Up: Let $K$ be a number field, let $R$ be ...

4

This follows from the work of Miller, Michael J., On Sendov’s conjecture for roots near the unit circle, J. Math. Anal. Appl. 175, No. 2, 632-639 (1993). ZBL0782.30007. and independently Vâjâitu, Viorel; Zaharescu, A., Ilyeff’s conjecture on a corona, Bull. Lond. Math. Soc. 25, No. 1, 49-54 (1993). ZBL0796.30004. who established Sendov's conjecture when the ...

4

This is an elaboration on the comment of Alexandre Eremenko. Algebraic multiplicity $n$ means that we have the equality of polynomials $$\det(t I_n -a_1A_1+\cdots+a_nA_n)=(t-\lambda)^n$$ for some $\lambda$. Comparing coefficients of $t^{n-1},t^{n-2},\ldots,1$, we find a system of $n$ equations $$\begin{cases} \mathrm{tr}(a_1A_1+\cdots+a_nA_n)=n\lambda,\\ ... 3 Indeed, \mathcal{P} is dense in \mathcal{F}. The Bernstein polynomials approximating f\in\mathcal{F} belong to \mathcal{P}, see Theorem 11.68 in: http://www.pitt.edu/~hajlasz/Teaching/Math1530Fall2018/selection.pdf 3 Yes, it is true. Every polynomial does have a decomposition like you ask for, with restrictions. To explain why this is the case, first suppose that p is homogeneous of degree d, and we seek a decomposition where each linear form L_i is a homogeneous linear form (no constant term) and each power k_i = d. The projective variety whose points are L^d ... 3 As I have commented, if f,g:[0,1]\rightarrow\mathbb{R} are C^{\infty} functions, and c\in(0,1) is a real number with f^{(n)}(c)=g^{(n)}(c) for each n, and f(x)\leq g(x) for each x\in[0,1], then there can be at most one polynomial p with f(x)\leq p(x)\leq g(x) for each x\in[0,1], and one can easily set up f,g so that there are no ... 3 The answers to the first two questions readily follow from Proposition 7.5.2 in [J. C. McConnell and J. C. Robson, "Noncommutative Noetherian rings", AMS, 1987]. Actually, they prove a much more general result which holds for skew polynomial rings and skew Laurent polynomial rings. The base ring R can be arbitrary, not necessarily commutative. 3 No. Let a:=\alpha and b:=\beta. If e.g. f(x)=(x+b)^{-a}, then a f(x) + (x + b) f'(x)=0 for all real x\ge0, so that the left-hand side (lhs) of your inequality is infinite, whereas the right-hand side (rhs) is 0. If you now insist that the lhs be a finite number, we can modify the above example as follows: let a=2, b=1, and$$f(x):=\frac{1 + (...

2

For a commutative ring $R$, write $P_{n,R}$ for the affine $(n+1)$-space of polynomials of degree $\leq n$ over $R$. In other words, its $S$-points for an $R$-algebra $S$ are given by $S[x]_{\leq n}$. Note that $P_{n,R} = P_{n,\mathbf Z} \times \operatorname{Spec} R$. Lemma. Let $R$ be a domain, and $g \in R[x]$ a monic polynomial of degree $d$. Then the map ...

2

Here is a purely real-analytic proof that is a bit simpler than my other answer and just relies on the Stone-Weierstrass theorem rather than complex analysis and Mergelyan's theorem. Theorem: Suppose that $f,g:[0,1]\rightarrow\mathbb{R}$ are functions such that $f\leq g,f(0)<g(0),f(1)<g(1)$, and there are finitely many points $c$ such that $f(c)=g(c)$....

2

Let me assume that $\newcommand\la{\langle}\newcommand\ra{\rangle}\newcommand\laa{\langle\!\langle}\newcommand\raa{\rangle\!\rangle} K$ is algebraically closed (this is not necessary, but it means I can work instead with the more symmetric polynomial $f_{n,a_i}=\displaystyle\sum_{r=1}^n(x_r)^{a_r}.$) If $K$ is of characteristic two then the ring $K\laa x_i\... 1 Claim 1:$f$has$\le3$positive roots if$t\ge1$and$p,q,r$are integers such that$0<p<q<r$. Claim 2:$f$has$\le3$positive roots if$t\ge2$and$p,q,r$are integers such that$p>q>r>0$. Let $$f_1(x):=\frac{f'(x)}{(p+x)^{t-1}},$$ $$f_2(x):=\frac{f_1'(x)}{(q+x)^{t-2}(q + p (t-1) + q t + 2 t x)},$$ $$f_3(x):=\frac{f_2'(x)}{t(r+x)^{t-3}}... 1 This is just a long comment, but it feels natural to look at the a_{ij} defined via$$ \sum_{i,j \geq 0} a_{i,j} x^i y^k = \frac{1}{P(x,y)} $$where P(x,y)=0 has infinitely many integer solutions. (Assuming (0,0) is not a solution). 1 I contacted some people privately and was suggested the following reference, which answers both the projective and injective dimension questions. The flat dimension question surely can be dealt with similarly. The paper is: S.Eilenberg, A.Rosenberg, and D.Zelinsky, "On the dimension of modules and algebras VIII. Dimension of tensor products", ... 1 I think that I can answer the first one of the three questions (which I guess is the simplest one), about the flat dimension. The idea is to reduce the problem to the case when R is a field, using the classical Hilbert's syzygy theorem as a black box. The argument proceeds by Noetherian induction in the ring R. So we assume that the assertion is true ... 1 Yes, it appears to be always possible. Let 0\neq v\in\mathbb{C}^n be a zero of F, v\in V(F). There are F_1,\dots,F_n\in R:=\mathbb{C}[x_1,\dots,x_n] of degree d-1 each, such that v is their only common zero, v\in V(F_1,\dots,F_n)\subset V(F). Hence F\in (F_1,\dots,F_n), which means that$$F=\sum_{k=1}^n \ell_k F_k, \quad\text{for$\ell_t\in R\$...

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