# Unique factorization domain that is not a Principal ideal domain

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 $$1 \notin F[X_1,X_2]$$.

# A non-measurable set

We describe here a non-measurable subset of the segment $$I=[0,1] \subset \mathbb R$$.

Let’s define on $$I$$ an equivalence relation by $$x \sim y$$ if and only if $$x-y \in \mathbb Q$$. The equivalence relation $$\sim$$ induces equivalence classes on $$I$$. For $$x \in I$$, it’s equivalence class $$[x]$$ is $$[x] = \{y \in I \ : \ y-x \in \mathbb Q\}$$. By the Axiom of Choice, we can form a set $$A$$ by selecting a single point from each equivalence class.

We claim that the set $$A$$ is not Lebesgue measurable.

For all $$q \in \mathbb Q$$ we denote $$A_q = \{q+x \ : x \in A\}$$. Let’s take $$p,q \in \mathbb Q$$. If it exists $$z \in A_p \cap A_q$$, it means that there exist $$u,v \in A$$ such that
$z= p+u=q+v$ hence $$u-v=q-p=0$$ as $$u,v$$ are supposed to be unique representatives of the classes of the equivalence relation $$\sim$$. Finally if $$p,q$$ are distincts, $$A_p \cap A_q = \emptyset$$.

As Lebesgue measure $$\mu$$ is translation invariant, we have for $$q \in \mathbb Q \cap [0,1]$$ : $$\mu(A) = \mu(A_q)$$ and also $$A_q \subset [0,2]$$. Hence if we denote
$B = \bigcup_{q \in \mathbb Q \cap [0,1]} A_q$ we have $$B \subset [0,2]$$. If we suppose that $$A$$ is measurable, we get
$\mu(B) = \sum_{q \in \mathbb Q \cap [0,1]} \mu(A_q) = \sum_{q \in \mathbb Q \cap [0,1]} \mu(A) \le 2$ by countable additivity of Lebesgue measure (the set $$\mathbb Q \cap [0,1]$$ being countable infinite). This implies $$\mu(A) = 0$$.

Let’s prove now that
$[0,1] \subset \bigcup_{q \in \mathbb Q \cap [-1,1]} A_q$ For $$z \in [0,1]$$, there exists $$u \in A$$ such that $$z \in [u]$$. As $$A \subset [0,1]$$, we have $$q = z-u \in \mathbb Q$$ and $$-1 \le q \le 1$$. And $$z=q+u$$ means that $$z \in A_q$$. This proves the inclusion. However the inclusion implies the contradiction
$1 = \mu([0,1]) \le \sum_{q \in \mathbb Q \cap [-1,1]} \mu(A_q) = \sum_{q \in \mathbb Q \cap [-1,1]} \mu(A) =0$

Finally $$A$$ is not Lebesgue measurable.

# Isometric versus affine

Throughout this article we let $$E$$ and $$F$$ denote real normed vector spaces. A map $$f : E \rightarrow F$$ is an isometry if $$\Vert f(x) – f(y) \Vert = \Vert x – y \Vert$$ for all $$x, y \in E$$, and $$f$$ is affine if $f((1-t) a + t b ) = (1-t) f(a) + t f(b)$ for all $$a,b \in E$$ and $$t \in [0,1]$$. Equivalently, $$f$$ is affine if the map $$T : E \rightarrow F$$, defined by $$T(x)=f(x)-f(0)$$ is linear.

First note that an isometry $$f$$ is always one-to-one as $$f(x) = f(y)$$ implies $0 = \Vert f(x) – f(y) \Vert = \Vert x- y \Vert$ hence $$x=y$$.

There are two important cases when every isometry is affine:

1. $$f$$ is bijective (equivalently surjective). This is Mazur-Ulam theorem, which was proven in 1932.
2. $$F$$ is a strictly convex space. Recall that a normed vector space $$(S, \Vert \cdot \Vert)$$ is strictly convex if and only if for all distinct $$x,y \in S$$, $$\Vert x \Vert = \Vert y \Vert =1$$ implies $$\Vert \frac{x+y}{2} \Vert <1$$. For example, an inner product space is strictly convex. The sequence spaces $$\ell_p$$ for $$1 < p < \infty$$ are also strictly convex.

# Counterexample around Morera’s theorem

Let’s recall Morera’s theorem.

Morera’s theorem
Suppose that $$f$$ is a continuous complex-valued function in a connected open set $$\Omega \subset \mathbb C$$ such that
$\int_{\partial \Delta} f(z) \ dz = 0$ for every closed triangle $$\Delta \subset \Omega \setminus \{p\}$$ where $$p \in \Omega$$. Then $$f$$ is holomorphic in $$\Omega$$.

Does the conclusion of Morera’s theorem still hold if $$f$$ is supposed to be continuous only in $$\Omega \setminus \{p\}$$? The answer is negative and we provide a counterexample.

Let $$\Omega$$ be the entire complex plane, $$f$$ defined as follows
$f(z)=\begin{cases} \frac{1}{z^2} & \text{if } z \neq 0\\ 0 & \text{otherwise} \end{cases}$ and $$p$$ the origin.

For $$a,b \in \Omega \setminus \{0\}$$ we have
\begin{aligned} \int_{[a,b]} f(z) \ dz &= \int_{[a,b]} \frac{dz}{z^2}\\ &= \int_0^1 \frac{b-a}{[a+t(b-a)]^2} \ dt\\ &=\left[ -\frac{1}{a+t(b-a)} \right]_0^1 = \frac{1}{a} – \frac{1}{b} \end{aligned}

Hence for a triangle $$\Delta$$ with vertices at $$a,b ,c \in \Omega \setminus \{0\}$$:
$\int_{\partial \Delta} f(z) \ dz = \left( \frac{1}{a} – \frac{1}{b} \right) + \left( \frac{1}{b} – \frac{1}{c} \right) + \left( \frac{1}{c} – \frac{1}{a} \right)=0$

However, $$f$$ is not holomorphic in $$\Omega$$ as it is even not continuous at $$0$$.

# Counterexamples to theorems on Clifford algebras

This page provides counterexamples on Clifford algebras, spinors, spin groups and the exterior algebra. You’ll find a list of counterexamples references at the end of the article.