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
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
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 \(E\).
When \(E\) is of finite dimension, \(S\) is invertible. Continue reading A linear map having a left inverse which is not a right inverse
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.
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
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:
\(+\) 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.
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
