# 100th ring on the Database of Ring Theory

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# 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)

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)

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