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.

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 \(q\).

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→

We consider a vector space \(E\) and a linear map \(T \in \mathcal{L}(E)\) having a left inverse \(S\) which means that \(S \circ T = S T =I\) where \(I\) is the identity map in \(E\).

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→

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.

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→

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:

\(r \cdot (x+y)= r \cdot x + r \cdot y\) (\( \cdot\) is left-distributive over \(+\))

\((r +s) \cdot x= r \cdot x + s \cdot x\) (\( \cdot\) is right-distributive over \(+\))

\((rs) \cdot x= r \cdot (s \cdot x)\)

\(1_R \cdot x= x \)

\(+\) 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→

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→

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.