# 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 vector space written as a finite union of proper subspaces

We raise here the following question: “can a vector space $$E$$ be written as a finite union of proper subspaces”?

Let’s consider the simplest case, i.e. writing $$E= V_1 \cup V_2$$ as a union of two proper subspaces. By hypothesis, one can find two non-zero vectors $$v_1,v_2$$ belonging respectively to $$V_1 \setminus V_2$$ and $$V_2 \setminus V_1$$. The relation $$v_1+v_2 \in V_1$$ leads to the contradiction $$v_2 = (v_1+v_2)-v_1 \in V_1$$ while supposing $$v_1+v_2 \in V_2$$ leads to the contradiction $$v_1 = (v_1+v_2)-v_2 \in V_2$$. Therefore, a vector space can never be written as a union of two proper subspaces.

We now analyze if a vector space can be written as a union of $$n \ge 3$$ proper subspaces. We’ll see that it is impossible when $$E$$ is a vector space over an infinite field. But we’ll describe a counterexample of a vector space over the finite field $$\mathbb{Z}_2$$ written as a union of three proper subspaces. Continue reading A vector space written as a finite union of proper subspaces

# A finitely generated soluble group isomorphic to a proper quotient group

Let $$\mathbb{Q}_2$$ be the ring of rational numbers of the form $$m2^n$$ with $$m, n \in \mathbb{Z}$$ and $$N = U(3, \mathbb{Q}_2)$$ the group of unitriangular matrices of dimension $$3$$ over $$\mathbb{Q}_2$$. Let $$t$$ be the diagonal matrix with diagonal entries: $$1, 2, 1$$ and put $$H = \langle t, N \rangle$$. We will prove that $$H$$ is finitely generated and that one of its quotient group $$G$$ is isomorphic to a proper quotient group of $$G$$. Continue reading A finitely generated soluble group isomorphic to a proper quotient group

# A (not finitely generated) group isomorphic to a proper quotient group

The basic question that we raise here is the following one: given a group $$G$$ and a proper subgroup $$H$$ (i.e. $$H \notin \{\{1\},G\}$$, can $$G/H$$ be isomorphic to $$G$$? A group $$G$$ is said to be hopfian (after Heinz Hopf) if it is not isomorphic with a proper quotient group.

All finite groups are hopfian as $$|G/H| = |G| \div |H|$$. Also, all simple groups are hopfian as a simple group doesn’t have proper subgroups.

So we need to turn ourselves to infinite groups to uncover non hopfian groups. Continue reading A (not finitely generated) group isomorphic to a proper quotient group

# Converse of Lagrange’s theorem does not hold

Lagrange’s theorem, states that for any finite group $$G$$, the order (number of elements) of every subgroup $$H$$ of $$G$$ divides the order of $$G$$ (denoted by $$\vert G \vert$$).

Lagrange’s theorem raises the converse question as to whether every divisor $$d$$ of the order of a group is the order of some subgroup. According to Cauchy’s theorem this is true when $$d$$ is a prime.

However, this does not hold in general: given a finite group $$G$$ and a divisor $$d$$ of $$\vert G \vert$$, there does not necessarily exist a subgroup of $$G$$ with order $$d$$. The alternating group $$G = A_4$$, which has $$12$$ elements has no subgroup of order $$6$$. We prove it below. Continue reading Converse of Lagrange’s theorem does not hold

# A module without a basis

Let’s start by recalling some background about modules.

Suppose that $$R$$ is a ring and $$1_R$$ is its multiplicative identity. A left $$R$$-module $$M$$ consists of an abelian group $$(M, +)$$ and an operation $$R \times M \rightarrow M$$ such that for all $$r, s \in R$$ and $$x, y \in M$$, we have:

1. $$r \cdot (x+y)= r \cdot x + r \cdot y$$ ($$\cdot$$ is left-distributive over $$+$$)
2. $$(r +s) \cdot x= r \cdot x + s \cdot x$$ ($$\cdot$$ is right-distributive over $$+$$)
3. $$(rs) \cdot x= r \cdot (s \cdot x)$$
4. $$1_R \cdot x= x$$

$$+$$ is the symbol for addition in both $$R$$ and $$M$$.
If $$K$$ is a field, $$M$$ is $$K$$-vector space. It is well known that a vector space $$V$$ is having a basis, i.e. a subset of linearly independent vectors that spans $$V$$.
Unlike for a vector space, a module doesn’t always have a basis. Continue reading A module without a basis

# A vector space not isomorphic to its double dual

In this page $$\mathbb{F}$$ refers to a field. Given any vector space $$V$$ over $$\mathbb{F}$$, the dual space $$V^*$$ is defined as the set of all linear functionals $$f: V \mapsto \mathbb{F}$$. The dual space $$V^*$$ itself becomes a vector space over $$\mathbb{F}$$ when equipped with the following addition and scalar multiplication:
$\left\{ \begin{array}{lll}(\varphi + \psi)(x) & = & \varphi(x) + \psi(x) \\ (a \varphi)(x) & = & a (\varphi(x)) \end{array} \right.$ for all $$\phi, \psi \in V^*$$, $$x \in V$$, and $$a \in \mathbb{F}$$.
There is a natural homomorphism $$\Phi$$ from $$V$$ into the double dual $$V^{**}$$, defined by $$(\Phi(v))(\phi) = \phi(v)$$ for all $$v \in V$$, $$\phi \in V^*$$. This map $$\Phi$$ is always injective. Continue reading A vector space not isomorphic to its double dual

# A field that can be ordered in two distinct ways

For a short reminder about ordered fields you can have a look to following post. We prove there that $$\mathbb{Q}$$ can be ordered in only one way.

That is also the case of $$\mathbb{R}$$ as $$\mathbb{R}$$ is a real-closed field. And one can prove that the only possible positive cone of a real-closed field is the subset of squares.

However $$\mathbb{Q}(\sqrt{2})$$ is a subfield of $$\mathbb{R}$$ that can be ordered in two distinct ways. Continue reading A field that can be ordered in two distinct ways