Tag Archives: algebra

Subset of elements of finite order of a group

February 12, 2017 Jean-Pierre Merx Leave a comment

Consider a group $G$ and have a look at the question: is the subset $S$ of elements of finite order a subgroup of $G$ ?

The answer is positive when any two elements of $S$ commute. For the proof, consider $x,y \in S$ of order $m,n$ respectively. Then $\left(xy\right)^{mn} = x^{mn} y^{mn} = (x^m)^n (y^n)^m = e$ where $e$ is the identity element. Hence $xy$ is of finite order (less or equal to $mn$ ) and belong to $S$ .

Example of a non abelian group

In that cas, $S$ might not be subgroup of $G$ . Let’s take for $G$ the general linear group over $\mathbb Q$ (the set of rational numbers) of $2 \times 2$ invertible matrices named $\text{GL}_2(\mathbb Q)$ . The matrices $A = \begin{pmatrix}0&1\\1&0\end{pmatrix},\ B=\begin{pmatrix}0 & 2\\\frac{1}{2}& 0\end{pmatrix}$ are of order $2$ . They don’t commute as $AB = \begin{pmatrix}\frac{1}{2}&0\\0&2\end{pmatrix} \neq \begin{pmatrix}2&0\\0&\frac{1}{2}\end{pmatrix}=BA.$ Finally, $AB$ is of infinite order and therefore doesn’t belong to $S$ proving that $S$ is not a subgroup of $G$ .

Algebra

Field not algebraic over an intersection but algebraic over each initial field

January 15, 2017 Jean-Pierre Merx Leave a comment

Let’s describe an example of a field $K$ which is of degree $2$ over two distinct subfields $M$ and $N$ respectively, but not algebraic over $M \cap N$ .

Let $K=F(x)$ be the rational function field over a field $F$ of characteristic $0$ , $M=F(x^2)$ and $N=F(x^2+x)$ . I claim that those fields provide the example we’re looking for.

$K$ is of degree $2$ over $M$ and $N$

The polynomial $\mu_M(t)=t^2-x^2$ belongs to $M[t]$ and $x \in K$ is a root of $\mu_M$ . Also, $\mu_M$ is irreducible over $M=F(x^2)$ . If that wasn’t the case, $\mu_M$ would have a root in $F(x^2)$ and there would exist two polynomials $p,q \in F[t]$ such that $p^2(x^2) = x^2 q^2(x^2)$ which cannot be, as can be seen considering the degrees of the polynomials of left and right hand sides. This proves that $[K:M]=2$ . Considering the polynomial $\mu_N(t)=t^2-t-(x^2+x)$ , one can prove that we also have $[K:N]=2$ .

We have $M \cap N=F$

The mapping $\sigma_M : x \mapsto -x$ extends uniquely to an $F$ -automorphism of $K$ and the elements of $M$ are fixed under $\sigma_M$ . Similarly, the mapping $\sigma_N : x \mapsto -x-1$ extends uniquely to an $F$ -automorphism of $K$ and the elements of $N$ are fixed under $\sigma_N$ . Also $(\sigma_N\circ\sigma_M)(x)=\sigma_N(\sigma_M(x))=\sigma_N(-x)=-(-x-1)=x+1.$ An element $z=p(x)/q(x) \in M \cap N$ where $p(x),q(x)$ are coprime polynomials of $K=F(x)$ is fixed under $\sigma_M \circ \sigma_N$ . Therefore following equality holds $\frac{p(x)}{q(x)}=z=(\sigma_2\circ\sigma_1)(z)=\frac{p(x+1)}{q(x+1)},$ which is equivalent to $p(x)q(x+1)=p(x+1)q(x).$ By induction, we get for $n \in \mathbb Z$ $p(x)q(x+n)=p(x+n)q(x).$ Assume $p(x)$ is not a constant polynomial. Then it has a root $\alpha$ in some finite extension $E$ of $F$ . As $p(x),q(x)$ are coprime polynomials, $q(\alpha) \neq 0$ . Consequently $p(\alpha+n)=0$ for all $n \in \mathbb Z$ and the elements $\alpha +n$ are all distinct as the characteristic of $F$ is supposed to be non zero. This implies that $p(x)$ is the zero polynomial, in contradiction with our assumption. Therefore $p(x)$ is a constant polynomial and $q(x)$ also according to a similar proof. Hence $z$ is constant as was supposed to be proven.

Finally, $K=F(x)$ is not algebraic over $F=M \cap N$ as $(1,x, x^2, \dots, x^n, \dots)$ is independent over the field $F$ which concludes our claims on $K, M$ and $N$ .

Algebra

Complex matrix without a square root

December 26, 2016 Jean-Pierre Merx Leave a comment

Consider for $n \ge 2$ the linear space $\mathcal M_n(\mathbb C)$ of complex matrices of dimension $n \times n$ . Is a matrix $T \in \mathcal M_n(\mathbb C)$ always having a square root $S \in \mathcal M_n(\mathbb C)$ , i.e. a matrix such that $S^2=T$ ? is the question we deal with.

First, one can note that if $T$ is similar to $V$ with $T = P^{-1} V P$ and $V$ has a square root $U$ then $T$ also has a square root as $V=U^2$ implies $T=\left(P^{-1} U P\right)^2$ .

Diagonalizable matrices

Suppose that $T$ is similar to a diagonal matrix $D=\begin{bmatrix} d_1 & 0 & \dots & 0 \\ 0 & d_2 & \dots & 0 \\ \vdots & \vdots & \ddots & 0 \\ 0 & 0 & \dots & d_n \end{bmatrix}$ Any complex number has two square roots, except $0$ which has only one. Therefore, each $d_i$ has at least one square root $d_i^\prime$ and the matrix $D^\prime=\begin{bmatrix} d_1^\prime & 0 & \dots & 0 \\ 0 & d_2^\prime & \dots & 0 \\ \vdots & \vdots & \ddots & 0 \\ 0 & 0 & \dots & d_n^\prime \end{bmatrix}$ is a square root of $D$ . Continue reading Complex matrix without a square root →

Algebra

Additive subgroups of vector spaces

December 10, 2016 Jean-Pierre Merx Leave a comment

Consider a vector space $V$ over a field $F$ . A subspace $W \subseteq V$ is an additive subgroup of $(V,+)$ . The converse might not be true.

If the characteristic of the field is zero, then a subgroup $W$ of $V$ might not be an additive subgroup. For example $\mathbb R$ is a vector space over $\mathbb R$ itself. $\mathbb Q$ is an additive subgroup of $\mathbb R$ . However $\sqrt{2}= \sqrt{2}.1 \notin \mathbb Q$ proving that $\mathbb Q$ is not a subspace of $\mathbb R$ .

Another example is $\mathbb Q$ which is a vector space over itself. $\mathbb Z$ is an additive subgroup of $\mathbb Q$ , which is not a subspace as $\frac{1}{2} \notin \mathbb Z$ .

Yet, an additive subgroup of a vector space over a prime field $\mathbb F_p$ with $p$ prime is a subspace. To prove it, consider an additive subgroup $W$ of $(V,+)$ and $x \in W$ . For $\lambda \in F$ , we can write $\lambda = \underbrace{1 + \dots + 1}_{\lambda \text{ times}}$ . Consequently $\lambda \cdot x = (1 + \dots + 1) \cdot x= \underbrace{x + \dots + x}_{\lambda \text{ times}} \in W.$

Finally an additive subgroup of a vector space over any finite field is not always a subspace. For a counterexample, take the non-prime finite field $\mathbb F_{p^2}$ (also named $\text{GF}(p^2)$ ). $\mathbb F_{p^2}$ is also a vector space over itself. The prime finite field $\mathbb F_p \subset \mathbb F_{p^2}$ is an additive subgroup that is not a subspace of $\mathbb F_{p^2}$ .

Algebra

Non commutative rings

October 16, 2016 Jean-Pierre Merx Leave a comment

Let’s recall that a set $R$ equipped with two operations $(R,+,\cdot)$ is a ring if and only if $(R,+)$ is an abelian group, multiplication $\cdot$ is associative and has a multiplicative identity $1$ and multiplication is left and right distributive with respect to addition.

$(\mathbb Z, +, \cdot)$ is a well known infinite ring which is commutative. The rational, real and complex numbers are other infinite commutative rings. Those are in fact fields as every non-zero element have a multiplicative inverse.

For a field $F$ (finite or infinite), the polynomial ring $F[X]$ is another example of infinite commutative ring.

Also for $n$ integer, the integers modulo n is a finite ring that is commutative. Finally, according to Wedderburn theorem every finite division ring is commutative.

So what are examples of non commutative rings? Let’s provide a couple. Continue reading Non commutative rings →

Algebra

A linear map without any minimal polynomial

August 8, 2016 Jean-Pierre Merx Leave a comment

Given an endomorphism $T$ on a finite-dimensional vector space $V$ over a field $\mathbb F$ , the minimal polynomial $\mu_T$ of $T$ is well defined as the generator (unique up to units in $\mathbb F$ ) of the ideal: $I_T= \{p \in \mathbb F[t]\ ; \ p(T)=0\}.$

For infinite-dimensional vector spaces, the minimal polynomial might not be defined. Let’s provide an example.

We take the real polynomials $V = \mathbb R [t]$ as a real vector space and consider the derivative map $D : P \mapsto P^\prime$ . Let’s prove that $D$ doesn’t have any minimal polynomial. By contradiction, suppose that $\mu_D(t) = a_0 + a_1 t + \dots + a_n t^n \text{ with } a_n \neq 0$ is the minimal polynomial of $D$ , which means that for all $P \in \mathbb R[t]$ we have $a_0 + a_1 P^\prime + \dots + a_n P^{(n)} = 0.$ Taking for $P$ the polynomial $t^n$ we get $a_0 t^n + n a_1 t^{n-1} + \dots + n! a_n = 0,$ which doesn’t make sense as $n! a_n \neq 0$ , hence $a_0 t^n + n a_1 t^{n-1} + \dots + n! a_n$ cannot be the zero polynomial.

We conclude that $D$ doesn’t have any minimal polynomial.

Algebra

Non linear map preserving Euclidean norm

August 3, 2016 Jean-Pierre Merx Leave a comment

Let $V$ be a real vector space endowed with an Euclidean norm $\Vert \cdot \Vert$ .

A bijective map $T : V \to V$ that preserves inner product $\langle \cdot, \cdot \rangle$ is linear. Also, Mazur-Ulam theorem states that an onto map $T : V \to V$ which is an isometry ( $\Vert T(x)-T(y) \Vert = \Vert x-y \Vert$ for all $x,y \in V$ ) and fixes the origin ( $T(0) = 0$ ) is linear.

What about an application that preserves the norm ( $\Vert T(x) \Vert = \Vert x \Vert$ for all $x \in V$ )? $T$ might not be linear as we show with following example: $\begin{array}{l|rcll} T : & V & \longrightarrow & V \\ & x & \longmapsto & x & \text{if } \Vert x \Vert \neq 1\\ & x & \longmapsto & -x & \text{if } \Vert x \Vert = 1\end{array}$

It is clear that $T$ preserves the norm. However $T$ is not linear as soon as $V$ is not the zero vector space. In that case, consider $x_0$ such that $\Vert x_0 \Vert = 1$ . We have: $\begin{cases} T(2 x_0) &= 2 x_0 \text{ as } \Vert 2 x_0 \Vert = 2\\ \text{while}\\ T(x_0) + T(x_0) = -x_0 + (-x_0) &= – 2 x_0 \end{cases}$

Algebra

Non linear map preserving orthogonality

July 24, 2016 Jean-Pierre Merx Leave a comment

Let $V$ be a real vector space endowed with an inner product $\langle \cdot, \cdot \rangle$ .

It is known that a bijective map $T : V \to V$ that preserves the inner product $\langle \cdot, \cdot \rangle$ is linear.

That might not be the case if $T$ is supposed to only preserve orthogonality. Let’s consider for $V$ the real plane $\mathbb R^2$ and the map $\begin{array}{l|rcll} T : & \mathbb R^2 & \longrightarrow & \mathbb R^2 \\ & (x,y) & \longmapsto & (x,y) & \text{for } xy \neq 0\\ & (x,0) & \longmapsto & (0,x)\\ & (0,y) & \longmapsto & (y,0) \end{array}$

The restriction of $T$ to the plane less the x-axis and the y-axis is the identity and therefore is bijective on this set. Moreover $T$ is a bijection from the x-axis onto the y-axis, and a bijection from the y-axis onto the x-axis. This proves that $T$ is bijective on the real plane.

$T$ preserves the orthogonality on the plane less x-axis and y-axis as it is the identity there. As $T$ swaps the x-axis and the y-axis, it also preserves orthogonality of the coordinate axes. However, $T$ is not linear as for non zero $x \neq y$ we have: $\begin{cases} T[(x,0) + (0,y)] = T[(x,y)] &= (x,y)\\ \text{while}\\ T[(x,0)] + T[(0,y)] = (0,x) + (y,0) &= (y,x) \end{cases}$

Algebra

A simple ring which is not a division ring

May 29, 2016 Jean-Pierre Merx Leave a comment

Let’s recall that a simple ring is a non-zero ring that has no two-sided ideal besides the zero ideal and itself. A division ring is a simple ring. Is the converse true? The answer is negative and we provide here a counterexample of a simple ring which is not a division ring.

We prove that for $n \ge 1$ the matrix ring $M_n(F)$ of $n \times n$ matrices over a field $F$ is simple. $M_n(F)$ is obviously not a division ring as the matrix with $1$ at position $(1,1)$ and $0$ elsewhere is not invertible.

Let’s prove first following lemma. Continue reading A simple ring which is not a division ring →

Algebra

Unique factorization domain that is not a Principal ideal domain

April 24, 2016 Jean-Pierre Merx Leave a comment

In this article, we provide an example of a unique factorization domain – UFD that is not a principal ideal domain – PID. However, it is known that a PID is a UFD.

We take a field $F$ , for example $\mathbb Q$ , $\mathbb R$ , $\mathbb F_p$ (where $p$ is a prime) or whatever more exotic.

The polynomial ring $F[X]$ is a UFD. This follows from the fact that $F[X]$ is a Euclidean domain. It is also known that for a UFD $R$ , $R[X]$ is also a UFD. Therefore the polynomial ring $F[X_1,X_2]$ in two variables is a UFD as $F[X_1,X_2] = F[X_1][X_2]$ . However the ideal $I=(X_1,X_2)$ is not principal. Let’s prove it by contradiction.

Suppose that $(X_1,X_2) = (P)$ with $P \in F[X_1,X_2]$ . Then there exist two polynomials $Q_1,Q_2 \in F[X_1,X_2]$ such that $X_1=PQ_1$ and $X_2=PQ_2$ . As a polynomial in variable $X_2$ , the polynomial $X_1$ is having degree $0$ . Therefore, the degree of $P$ as a polynomial in variable $X_2$ is also equal to $0$ . By symmetry, we get that the degree of $P$ as a polynomial in variable $X_1$ is equal to $0$ too. Which implies that $P$ is an element of the field $F$ and consequently that $(X_1,X_2) = F[X_1,X_2]$ .

But the equality $(X_1,X_2) = F[X_1,X_2]$ is absurd. Indeed, the degree of a polynomial $X_1 T_1 + X_2 T_2$ cannot be equal to $0$ for any $T_1,T_2 \in F[X_1,X_2]$ . And therefore $1 \notin F[X_1,X_2]$ .

Math Counterexamples

Tag Archives: algebra

Subset of elements of finite order of a group

Example of a non abelian group

Field not algebraic over an intersection but algebraic over each initial field

$K$ is of degree $2$ over $M$ and $N$

We have $M \cap N=F$

Complex matrix without a square root

Diagonalizable matrices

Additive subgroups of vector spaces

Non commutative rings

A linear map without any minimal polynomial

Non linear map preserving Euclidean norm

Non linear map preserving orthogonality

A simple ring which is not a division ring

Unique factorization domain that is not a Principal ideal domain

Mathematical exceptions to the rules or intuition

Example of a non abelian group

KK is of degree 22 over MM and NN

We have M∩N=FM \cap N=F

Diagonalizable matrices

Mathematical exceptions to the rules or intuition

$K$ is of degree $2$ over $M$ and $N$

We have $M \cap N=F$