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# 100th ring on the Database of Ring Theory

# Group homomorphism versus ring homomorphism

A ring homomorphism is a function between two rings which respects the structure. Let’s provide examples of functions between rings which respect the addition or the multiplication but not both.

### An additive group homomorphism that is not a ring homomorphism

We consider the ring \(\mathbb R[x]\) of real polynomials and the derivation \[

\begin{array}{l|rcl}

D : & \mathbb R[x] & \longrightarrow & \mathbb R[x] \\

& P & \longmapsto & P^\prime \end{array}\] \(D\) is an additive homomorphism as for all \(P,Q \in \mathbb R[x]\) we have \(D(P+Q) = D(P) + D(Q)\). However, \(D\) does not respect the multiplication as \[

D(x^2) = 2x \neq 1 = D(x) \cdot D(x).\] More generally, \(D\) satisfies the Leibniz rule \[

D(P \cdot Q) = P \cdot D(Q) + Q \cdot D(P).\]

### A multiplication group homomorphism that is not a ring homomorphism

The function \[

\begin{array}{l|rcl}

f : & \mathbb R & \longrightarrow & \mathbb R \\

& x & \longmapsto & x^2 \end{array}\] is a multiplicative group homomorphism of the group \((\mathbb R, \cdot)\). However \(f\) does not respect the addition.

# A nonzero continuous map orthogonal to all polynomials

Let’s consider the vector space \(\mathcal{C}^0([a,b],\mathbb R)\) of continuous real functions defined on a compact interval \([a,b]\). We can define an inner product on pairs of elements \(f,g\) of \(\mathcal{C}^0([a,b],\mathbb R)\) by \[

\langle f,g \rangle = \int_a^b f(x) g(x) \ dx.\]

It is known that \(f \in \mathcal{C}^0([a,b],\mathbb R)\) is the always vanishing function if we have \(\langle x^n,f \rangle = \int_a^b x^n f(x) \ dx = 0\) for all integers \(n \ge 0\). Let’s recall the proof. According to Stone-Weierstrass theorem, for all \(\epsilon >0\) it exists a polynomial \(P\) such that \(\Vert f – P \Vert_\infty \le \epsilon\). Then \[

\begin{aligned}

0 &\le \int_a^b f^2 = \int_a^b f(f-P) + \int_a^b fP\\

&= \int_a^b f(f-P) \le \Vert f \Vert_\infty \epsilon(b-a)

\end{aligned}\] As this is true for all \(\epsilon > 0\), we get \(\int_a^b f^2 = 0\) and \(f = 0\).

We now prove that the result becomes false if we change the interval \([a,b]\) into \([0, \infty)\), i.e. that one can find a continuous function \(f \in \mathcal{C}^0([0,\infty),\mathbb R)\) such that \(\int_0^\infty x^n f(x) \ dx\) for all integers \(n \ge 0\). In that direction, let’s consider the complex integral \[

I_n = \int_0^\infty x^n e^{-(1-i)x} \ dx.\] \(I_n\) is well defined as for \(x \in [0,\infty)\) we have \(\vert x^n e^{-(1-i)x} \vert = x^n e^{-x}\) and \(\int_0^\infty x^n e^{-x} \ dx\) converges. By integration by parts, one can prove that \[

I_n = \frac{n!}{(1-i)^{n+1}} = \frac{(1+i)^{n+1}}{2^{n+1}} n! = \frac{e^{i \frac{\pi}{4}(n+1)}}{2^{\frac{n+1}{2}}}n!.\] Consequently, \(I_{4p+3} \in \mathbb R\) for all \(p \ge 0\) which means \[

\int_0^\infty x^{4p+3} \sin(x) e^{-x} \ dx =0\] and finally \[

\int_0^\infty u^p \sin(u^{1/4}) e^{-u^{1/4}} \ dx =0\] for all integers \(p \ge 0\) using integration by substitution with \(x = u^{1/4}\). The function \(u \mapsto \sin(u^{1/4}) e^{-u^{1/4}}\) is one we were looking for.

# A group G isomorph to the product group G x G

Let’s provide an example of a nontrivial group \(G\) such that \(G \cong G \times G\). For a finite group \(G\) of order \(\vert G \vert =n > 1\), the order of \(G \times G\) is equal to \(n^2\). Hence we have to look at infinite groups in order to get the example we’re seeking for.

We take for \(G\) the infinite direct product \[

G = \prod_{n \in \mathbb N} \mathbb Z_2 = \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2 \dots,\] where \(\mathbb Z_2\) is endowed with the addition. Now let’s consider the map \[

\begin{array}{l|rcl}

\phi : & G & \longrightarrow & G \times G \\

& (g_1,g_2,g_3, \dots) & \longmapsto & ((g_1,g_3, \dots ),(g_2, g_4, \dots)) \end{array}\]

From the definition of the addition in \(G\) it follows that \(\phi\) is a group homomorphism. \(\phi\) is onto as for any element \(\overline{g}=((g_1, g_2, g_3, \dots),(g_1^\prime, g_2^\prime, g_3^\prime, \dots))\) in \(G \times G\), \(g = (g_1, g_1^\prime, g_2, g_2^\prime, \dots)\) is an inverse image of \(\overline{g}\) under \(\phi\). Also the identity element \(e=(\overline{0},\overline{0}, \dots)\) of \(G\) is the only element of the kernel of \(G\). Hence \(\phi\) is also one-to-one. Finally \(\phi\) is a group isomorphism between \(G\) and \(G \times G\).

# Counterexamples around series (part 1)

The purpose of this article is to provide some basic **counterexamples** on real series. Counterexamples are provided as answers to questions.

Unless otherwise stated, \((u_n)_{n \in \mathbb{N}}\) and \((v_n)_{n \in \mathbb{N}}\) are two real sequences.

### If \((u_n)\) is non-increasing and converges to zero then \(\sum u_n\) converges?

Is not true. A famous counterexample is the harmonic series \(\sum \frac{1}{n}\) which doesn’t converge as \[

\displaystyle \sum_{k=p+1}^{2p} \frac{1}{k} \ge \sum_{k=p+1}^{2p} \frac{1}{2p} = 1/2,\] for all \(p \in \mathbb N\).

### If \(u_n = o(1/n)\) then \(\sum u_n\) converges?

Does not hold as can be seen considering \(u_n=\frac{1}{n \ln n}\) for \(n \ge 2\). Indeed \(\int_2^x \frac{dt}{t \ln t} = \ln(\ln x) – \ln (\ln 2)\) and therefore \(\int_2^\infty \frac{dt}{t \ln t}\) diverges. We conclude that \(\sum \frac{1}{n \ln n}\) diverges using the integral test. However \(n u_n = \frac{1}{\ln n}\) converges to zero. Continue reading Counterexamples around series (part 1)

# Isomorphism of factors does not imply isomorphism of quotient groups

Let \(G\) be a group and \(H, K\) two isomorphic subgroups. We provide an example where the quotient groups \(G / H\) and \(G / K\) are not isomorphic.

Let \(G = \mathbb{Z}_4 \times \mathbb{Z}_2\), with \(H = \langle (\overline{2}, \overline{0}) \rangle\) and \(K = \langle (\overline{0}, \overline{1}) \rangle\). We have \[

H \cong K \cong \mathbb{Z}_2.\] The left cosets of \(H\) in \(G\) are \[

G / H=\{(\overline{0}, \overline{0}) + H, (\overline{1}, \overline{0}) + H, (\overline{0}, \overline{1}) + H, (\overline{1}, \overline{1}) + H\},\] a group having \(4\) elements and for all elements \(x \in G/H\), one can verify that \(2x = H\). Hence \(G / H \cong \mathbb{Z}_2 \times \mathbb{Z}_2\). The left cosets of \(K\) in \(G\) are \[

G / K=\{(\overline{0}, \overline{0}) + K, (\overline{1}, \overline{0}) + K, (\overline{2}, \overline{0}) + K, (\overline{3}, \overline{0}) + K\},\] which is a cyclic group of order \(4\) isomorphic to \(\mathbb{Z}_4\). We finally get the desired conclusion \[

G / H \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \ncong \mathbb{Z}_4 \cong G / K.\]

# An uncountable chain of subsets of the natural numbers

Consider the set \(\mathcal P(\mathbb N)\) of the subsets of the natural integers \(\mathbb N\). \(\mathcal P(\mathbb N)\) is endowed with the strict order \(\subset\). Let’s have a look to the chains of \((\mathcal P(\mathbb N),\subset)\), i.e. to the totally ordered subsets \(S \subset \mathcal P(\mathbb N)\).

### Some finite chains

It is easy to produce some finite chains like \(\{\{1\}, \{1,2\},\{1,2,3\}\}\) or one with a length of size \(n\) where \(n\) is any natural number like \[

\{\{1\}, \{1,2\}, \dots, \{1,2, \dots, n\}\}\] or \[

\{\{1\}, \{1,2^2\}, \dots, \{1,2^2, \dots, n^2\}\}\]

### Some infinite countable chains

It’s not much complicated to produce some countable infinite chains like \[

\{\{1 \},\{1,2 \},\{1,2,3\},…,\mathbb{N}\}\] or \[

\{\{5 \},\{5,6 \},\{5,6,7\},…,\mathbb N \setminus \{1,2,3,4\} \}\]

Let’s go further and define a one-to-one map from the real interval \([0,1)\) into the set of countable chains of \((\mathcal P(\mathbb N),\subset)\). For \(x \in [0,1)\) let \(\displaystyle x = \sum_{i=1}^\infty x_i 2^{-i}\) be its binary representation. For \(n \in \mathbb N\) we define \(S_n(x) = \{k \in \mathbb N \ ; \ k \le n \text{ and } x_k = 1\}\). It is easy to verify that \(\left(S_n(x))_{n \in \mathbb N}\right)\) is a countable chain of \((\mathcal P(\mathbb N),\subset)\) and that \(\left(S_n(x))\right) \neq \left(S_n(x^\prime))\right)\) for \(x \neq x^\prime\).

**What about defining an uncountable chain?** Continue reading An uncountable chain of subsets of the natural numbers

# Counterexamples on real sequences (part 3)

This article is a follow-up of Counterexamples on real sequences (part 2).

Let \((u_n)\) be a sequence of real numbers.

### If \(u_{2n}-u_n \le \frac{1}{n}\) then \((u_n)\) converges?

This is wrong. The sequence

\[u_n=\begin{cases} 0 & \text{for } n \notin \{2^k \ ; \ k \in \mathbb N\}\\

1- 2^{-k} & \text{for } n= 2^k\end{cases}\]

is a counterexample. For \(n \gt 2\) and \(n \notin \{2^k \ ; \ k \in \mathbb N\}\) we also have \(2n \notin \{2^k \ ; \ k \in \mathbb N\}\), hence \(u_{2n}-u_n=0\). For \(n = 2^k\) \[

0 \le u_{2^{k+1}}-u_{2^k}=2^{-k}-2^{-k-1} \le 2^{-k} = \frac{1}{n}\] and \(\lim\limits_{k \to \infty} u_{2^k} = 1\). \((u_n)\) does not converge as \(0\) and \(1\) are limit points.

### If \(\lim\limits_{n} \frac{u_{n+1}}{u_n} =1\) then \((u_n)\) has a finite or infinite limit?

This is not true. Let’s consider the sequence

\[u_n=2+\sin(\ln n)\] Using the inequality \(

\vert \sin p – \sin q \vert \le \vert p – q \vert\)

which is a consequence of the mean value theorem, we get \[

\vert u_{n+1} – u_n \vert = \vert \sin(\ln (n+1)) – \sin(\ln n) \vert \le \vert \ln(n+1) – \ln(n) \vert\] Therefore \(\lim\limits_n \left(u_{n+1}-u_n \right) =0\) as \(\lim\limits_n \left(\ln(n+1) – \ln(n)\right) = 0\). And \(\lim\limits_{n} \frac{u_{n+1}}{u_n} =1\) because \(u_n \ge 1\) for all \(n \in \mathbb N\).

I now assert that the interval \([1,3]\) is the set of limit points of \((u_n)\). For the proof, it is sufficient to prove that \([-1,1]\) is the set of limit points of the sequence \(v_n=\sin(\ln n)\). For \(y \in [-1,1]\), we can pickup \(x \in \mathbb R\) such that \(\sin x =y\). Let \(\epsilon > 0\) and \(M \in \mathbb N\) , we can find an integer \(N \ge M\) such that \(0 < \ln(n+1) - \ln(n) \lt \epsilon\) for \(n \ge N\). Select \(k \in \mathbb N\) with \(x +2k\pi \gt \ln N\) and \(N_\epsilon\) with \(\ln N_\epsilon \in (x +2k\pi, x +2k\pi + \epsilon)\). This is possible as \((\ln n)_{n \in \mathbb N}\) is an increasing sequence and the length of the interval \((x +2k\pi, x +2k\pi + \epsilon)\) is equal to \(\epsilon\). We finally get \[ \vert u_{N_\epsilon} - y \vert = \vert \sin \left(\ln N_\epsilon \right) - \sin \left(x + 2k \pi \right) \vert \le \left(\ln N_\epsilon - (x +2k\pi)\right) \le \epsilon\] proving that \(y\) is a limit point of \((u_n)\).

# A Commutative Ring with Infinitely Many Units

In a ring \(R\) a unit is any element \(u\) that has a multiplicative inverse \(v\), i.e. an element \(v\) such that \[

uv=vu=1,\] where \(1\) is the multiplicative identity.

The only units of the commutative ring \(\mathbb Z\) are \(-1\) and \(1\). For a field \(\mathbb F\) the units of the ring \(\mathrm M_n(\mathbb F)\) of the square matrices of dimension \(n \times n\) is the general linear group \(\mathrm{GL}_n(\mathbb F)\) of the invertible matrices. The group \(\mathrm{GL}_n(\mathbb F)\) is infinite if \(\mathbb F\) is infinite, but the ring \(\mathrm M_n(\mathbb F)\) is not commutative for \(n \ge 2\).

The commutative ring \(\mathbb Z[\sqrt{2}] = \{a + b\sqrt{2} \ ; \ (a,b) \in \mathbb Z^2\}\) is not a field. However it has infinitely many units.

### \(a + b\sqrt{2}\) is a unit if and only if \(a^2-2b^2 = \pm 1\)

For \(u = a + b\sqrt{2} \in \mathbb Z[\sqrt{2}]\) we denote \(\mathrm N(u) = a^2- 2b^2 \in \mathbb Z\). For any \(u,v \in \mathbb Z[\sqrt{2}]\) we have \(\mathrm N(uv) = \mathrm N(u) \mathrm N(v)\). Therefore for a unit \(u \in \mathbb Z[\sqrt{2}]\) with \(v\) as multiplicative inverse, we have \(\mathrm N(u) \mathrm N(v) = 1\) and \(\mathrm N(u) =a^2-2b^2 \in \{-1,1\}\).

### The elements \((1+\sqrt{2})^n\) for \(n \in \mathbb N\) are unit elements

The proof is simple as for \(n \in \mathbb N\) \[

(1+\sqrt{2})^n (-1 + \sqrt{2})^n = \left((1+\sqrt{2})(-1 + \sqrt{2})\right)^n=1\]

One can prove (by induction on \(b\)) that the elements \((1+\sqrt{2})^n\) are the only units \(u \in \mathbb Z[\sqrt{2}]\) for \(u \gt 1\).

# A strictly increasing continuous function that is differentiable at no point of a null set

We build in this article a strictly increasing continuous function \(f\) that is differentiable at no point of a null set \(E\). The null set \(E\) can be chosen arbitrarily. In particular it can have the cardinality of the continuum like the Cantor null set.

### A set of strictly increasing continuous functions

For \(p \lt q\) two real numbers, consider the function \[

f_{p,q}(x)=(q-p) \left[\frac{\pi}{2} + \arctan{\left(\frac{2x-p-q}{q-p}\right)}\right]\] \(f_{p,q}\) is positive and its derivative is \[

f_{p,q}^\prime(x) = \frac{2}{1+\left(\frac{2x-p-q}{q-p}\right)^2}\] which is always strictly positive. Hence \(f_{p,q}\) is strictly increasing. We also have \[

\lim\limits_{x \to -\infty} f_{p,q}(x) = 0 \text{ and } \lim\limits_{x \to \infty} f_{p,q}(x) = \pi(q-p).\] One can notice that for \(x \in (p,q)\), \(f_{p,q}^\prime(x) \gt 1\). Therefore for \(x, y \in (p,q)\) distinct we have according to the mean value theorem \(\frac{f_{p,q}(y)-f_{p,q}(x)}{y-x} \ge 1\).

### Covering \(E\) with an appropriate set of open intervals

As \(E\) is a null set, for each \(n \in \mathbb N\) one can find an open set \(O_n\) containing \(E\) and measuring less than \(2^{-n}\). \(O_n\) can be written as a countable union of disjoint open intervals as any open subset of the reals. Then \(I=\bigcup_{m \in \mathbb N} O_m\) is also a countable union of open intervals \(I_n\) with \(n \in \mathbb N\). The sum of the lengths of the \(I_n\) is less than \(1\). Continue reading A strictly increasing continuous function that is differentiable at no point of a null set