Monthly Archives: April 2017

Analysis

A monotonic function whose points of discontinuity form a dense set

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Consider a compact interval \([a,b] \subset \mathbb R\) with \(a \lt b\). Let’s build an increasing function \(f : [a,b] \to \mathbb R\) whose points of discontinuity is an arbitrary dense subset \(D = \{d_n \ ; \ n \in \mathbb N\}\) of \([a,b]\), for example \(D = \mathbb Q \cap [a,b]\).

Let \(\sum p_n\) be a convergent series of positive numbers whose sum is equal to \(p\) and define \(\displaystyle f(x) = \sum_{d_n \le x} p_n\).

\(f\) is strictly increasing

For \(a \le x \lt y \le b\) we have \[
f(y) – f(x) = \sum_{x \lt d_n \le y} p_n \gt 0\] as the \(p_n\) are positive and dense so it exists \(p_m \in (x, y]\).

\(f\) is right-continuous on \([a,b]\)

We pick-up \(x \in [a,b]\). For any \(\epsilon \gt 0\) is exists \(N \in \mathbb N\) such that \(0 \lt \sum_{n \gt N} p_n \lt \epsilon\). Let \(\delta > 0\) be so small that the interval \((x,x+\delta)\) doesn’t contain any point in the finite set \(\{p_1, \dots, p_N\}\). Then \[
0 \lt f(y) – f(x) \le \sum_{n \gt N} p_n \lt \epsilon,\] for any \(y \in (x,x+\delta)\) proving the right-continuity of \(f\) at \(x\). Continue reading A monotonic function whose points of discontinuity form a dense set →

Algebra

A normal extension of a normal extension may not be normal

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An algebraic field extension \(K \subset L\) is said to be normal if every irreducible polynomial, either has no root in \(L\) or splits into linear factors in \(L\).

One can prove that if \(L\) is a normal extension of \(K\) and if \(E\) is an intermediate extension (i.e., \(K \subset E \subset L\)), then \(L\) is a normal extension of \(E\).

However a normal extension of a normal extension may not be normal and the extensions \(\mathbb Q \subset \mathbb Q(\sqrt{2}) \subset \mathbb Q(\sqrt[4]{2})\) provide a counterexample. Let’s prove it.

As a short lemma, we prove that a quadratic extension \(k \subset K\) , i.e. an extension of degree two is normal. Suppose that \(P\) is an irreducible polynomial of \(k[x]\) with a root \(a \in K\). If \(a \in k\) then the degree of \(P\) is equal to \(1\) and we’re done. Otherwise \((1, a)\) is a basis of \(K\) over \(k\) and there exist \(\lambda, \mu \in k\) such that \(a^2 = \lambda a +\mu\). As \(a \notin k\), \(Q(x)= x^2 – \lambda x -\mu\) is the minimal polynomial of \(a\) over \(k\). As \(P\) is supposed to be irreducible, we get \(Q = P\). And we can conclude as \[
Q(x) = (x-a)(x- \lambda +a).\]

The entensions \(\mathbb Q \subset \mathbb Q(\sqrt{2})\) and \(\mathbb Q(\sqrt{2}) \subset \mathbb Q(\sqrt[4]{2})\) are quadratic, hence normal according to previous lemma and \(\sqrt[4]{2}\) is a root of the polynomial \(P(x)= x^4-2\) of \(\mathbb Q[x]\). According to Eisenstein’s criterion \(P\) is irreducible over \(\mathbb Q\). However \(\mathbb Q(\sqrt[4]{2}) \subset \mathbb R\) while the roots of \(P\) are \(\pm \sqrt[4]{2}, \pm i \sqrt[4]{2}\) and therefore not all real. We can conclude that \(\mathbb Q \subset \mathbb Q(\sqrt[4]{2})\) is not normal.

Algebra

The image of an ideal may not be an ideal

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If \(\phi : A \to B\) is a ring homomorphism then the image of a subring \(S \subset A\) is a subring \(\phi(A) \subset B\). Is the image of an ideal under a ring homomorphism also an ideal? The answer is negative. Let’s provide a simple counterexample.

Let’s take \(A=\mathbb Z\) the ring of the integers and for \(B\) the ring of the polynomials with integer coefficients \(\mathbb Z[x]\). The inclusion \(\phi : \mathbb Z \to \mathbb Z[x]\) is a ring homorphism. The subset \(2 \mathbb Z \subset \mathbb Z\) of even integers is an ideal. However \(2 \mathbb Z\) is not an ideal of \(\mathbb Z[x]\) as for example \(2x \notin 2\mathbb Z\).

Analysis

A function whose Maclaurin series converges only at zero

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Let’s describe a real function \(f\) whose Maclaurin series converges only at zero. For \(n \ge 0\) we denote \(f_n(x)= e^{-n} \cos n^2x\) and \[
f(x) = \sum_{n=0}^\infty f_n(x)=\sum_{n=0}^\infty e^{-n} \cos n^2 x.\] For \(k \ge 0\), the \(k\)th-derivative of \(f_n\) is \[
f_n^{(k)}(x) = e^{-n} n^{2k} \cos \left(n^2 x + \frac{k \pi}{2}\right)\] and \[
\left\vert f_n^{(k)}(x) \right\vert \le e^{-n} n^{2k}\] for all \(x \in \mathbb R\). Therefore \(\displaystyle \sum_{n=0}^\infty f_n^{(k)}(x)\) is normally convergent and \(f\) is an indefinitely differentiable function with \[
f^{(k)}(x) = \sum_{n=0}^\infty e^{-n} n^{2k} \cos \left(n^2 x + \frac{k \pi}{2}\right).\] Its Maclaurin series has only terms of even degree and the absolute value of the term of degree \(2k\) is \[
\left(\sum_{n=0}^\infty e^{-n} n^{4k}\right)\frac{x^{2k}}{(2k)!} > e^{-2k} (2k)^{4k}\frac{x^{2k}}{(2k)!} > \left(\frac{2kx}{e}\right)^{2k}.\] The right hand side of this inequality is greater than \(1\) for \(k \ge \frac{e}{2x}\). This means that for any nonzero \(x\) the Maclaurin series for \(f\) diverges.

Algebra

A group that is not a semi-direct product

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Given a group \(G\) with identity element \(e\), a subgroup \(H\), and a normal subgroup \(N \trianglelefteq G\); then we say that \(G\) is the semi-direct product of \(N\) and \(H\) (written \(G=N \rtimes H\)) if \(G\) is the product of subgroups, \(G = NH\) where the subgroups have trivial intersection \(N \cap H= \{e\}\).

Semi-direct products of groups provide examples of non abelian groups. For example the dihedral group \(D_{2n}\) with \(2n\) elements is isomorphic to a semidirect product of the cyclic groups \(\mathbb Z_n\) and \(\mathbb Z_2\). \(D_{2n}\) is the group of isometries preserving a regular polygon \(X\) with \(n\) edges.

Let’see that the converse is not true and present a group that is not a semi-direct product.

The Hamilton’s quaternions group is not a semi-direct product

The Hamilton’s quaternions group \(\mathbb H_8\) is the group consisting of the symbols \(\pm 1, \pm i, \pm j, \pm k\) where\[
-1 = i^2 =j^2 = k^2 \text{ and } ij = k = -ji,jk = i = -kj, ki = j = -ik.\] One can prove that \(\mathbb H_8\) endowed with the product operation above is indeed a group having \(8\) elements where \(1\) is the identity element.

\(\mathbb H_8\) is not abelian as \(ij = k \neq -k = ji\).

Let’s prove that \(\mathbb H_8\) is not the semi-direct product of two subgroups. If that was the case, there would exist a normal subgroup \(N\) and a subgroup \(H\) such that \(G=N \rtimes H\).

  • If \(\vert N \vert = 4\) then \(H = \{1,h\}\) where \(h\) is an element of order \(2\) in \(\mathbb H_8\). Therefore \(h=-1\) which is the only element of order \(2\). But \(-1 \in N\) as \(-1\) is the square of all elements in \(\mathbb H_8 \setminus \{\pm 1\}\). We get the contradiction \(N \cap H \neq \{1\}\).
  • If \(\vert N \vert = 2\) then \(\vert H \vert = 4\) and \(H\) is also normal in \(G\). Noting \(N=\{1,n\}\) we have for \(h \in H\) \(h^{-1}nh=n\) and therefore \(nh=hn\). This proves that the product \(G=NH\) is direct. Also \(N\) is abelian as a cyclic group of order \(2\). \(H\) is also cyclic as all groups of order \(p^2\) with \(p\) prime are abelian. Finally \(G\) would be abelian, again a contradiction.

We can conclude that \(G\) is not a semi-direct product.