# An irreducible integral polynomial reducible over all finite prime fields

A classical way to prove that an integral polynomial $$Q \in \mathbb{Z}[X]$$ is irreducible is to prove that $$Q$$ is irreducible over a finite prime field $$\mathbb{F}_p$$ where $$p$$ is a prime.

This raises the question whether an irreducible integral polynomial is irreducible over at least one finite prime field. The answer is negative and:
$P(X)=X^4+1$ is a counterexample. Continue reading An irreducible integral polynomial reducible over all finite prime fields

# A connected not locally connected space

In this article, I will describe a subset of the plane that is a connected space while not locally connected nor path connected.

Let’s consider the plane $$\mathbb{R}^2$$ and the two subspaces:
$A = \bigcup_{n \ge 1} [(0,0),(1,\frac{1}{n})] \text{ and } B = A \cup (\frac{1}{2},1]$ Where a segment noted $$|a,b|$$ stands for the plane segment $$|(a,0),(b,0)|$$. Continue reading A connected not locally connected space

# On polynomials having more roots than their degree

Let’s consider a polynomial of degree $$q \ge 1$$ over a field $$K$$. It is well known that the sum of the multiplicities of the roots of $$P$$ is less or equal to $$q$$.

The result remains for polynomials over an integral domain. What is happening for polynomials over a commutative ring? Continue reading On polynomials having more roots than their degree

# Counterexamples on real sequences (part 2)

In that article, I provide basic counterexamples on sequences convergence. I follow on here with some additional and more advanced examples.

#### If $$(u_n)$$ converges then $$(\vert u_n \vert )$$ converge?

This is true and the proof is based on the reverse triangle inequality: $$\bigl| \vert x \vert – \vert y \vert \bigr| \le \vert x – y \vert$$. However the converse doesn’t hold. For example, the sequence $$u_n=(-1)^n$$ is such that $$\lim \vert u_n \vert = 1$$ while $$(u_n)$$ diverges.

#### If for all $$p \in \mathbb{N}$$ $$\lim\limits_{n \to +\infty} (u_{n+p} – u_n)=0$$ then $$(u_n)$$ converges?

The assertion is wrong. A simple counterexample is $$u_n= \ln(n+1)$$. It is well known that $$(u_n)$$ diverges. However for any $$p \in \mathbb{N}$$ we have $$\lim\limits_{n \to +\infty} (u_{n+p} – u_n) =\ln(1+\frac{p}{n+1})=0$$.
The converse proposition is true. Assume that $$(u_n)$$ is a converging sequence with limit $$l$$ and $$p \ge 0$$ is any integer. We have $$\vert u_{n+p}-u_n \vert = \vert (u_{n+p}-l)-(u_n-l) \vert \le \vert u_{n+p}-l \vert – \vert u_n-l \vert$$ and both terms of the right hand side of the inequality are converging to zero. Continue reading Counterexamples on real sequences (part 2)