# A ring whose characteristic is a prime having a zero divisor

Consider a ring $$R$$ whose characteristic is a composite number $$p=ab$$ with $$a,b$$ integers greater than $$1$$. Then $$R$$ has a zero divisor as we have $0=p.1=(a.b).1=(a.1).(b.1).$

What can we say of a ring $$R$$ having zero divisors? It is known that the rings $$\mathbb{Z}/p.\mathbb{Z}$$ where $$p$$ is a prime are fields and therefore do not have zero divisors. Is this a general fact? That is, does a ring whose characteristic is a prime do not have zero divisors?

The answer is negative and we give below a counterexample.

Let’s consider the field $$\mathbb{F}_p = \mathbb{Z}/p.\mathbb{Z}$$ where $$p$$ is a prime and the product of rings $$R=\mathbb{F}_p \times \mathbb{F}_p$$. One can verify following facts:

• $$R$$ additive identity is equal to $$(0,0)$$.
• $$R$$ multiplicative identity is equal to $$(1,1)$$.
• $$R$$ is commutative.
• The characteristic of $$R$$ is equal to $$p$$ as for $$n$$ integer, we have $$n.(1,1)=(n.1,n.1)$$ which is equal to $$(0,0)$$ if and only if $$p$$ divides $$n$$.

However, $$R$$ does have zero divisors as following identity holds: $(1,0).(0,1)=(0,0)$

# The Smith Volterra Cantor Set

In Cantor set article, I presented the Cantor set which is a null set having the cardinality of the continuum. I present here a modification of the Cantor set named the Smith-Volterra-Cantor set.

### Construction of the Smith-Volterra-Cantor set

The Smith-Volterra-Cantor set (also named SVC set below) $$S$$ is a subset of the real segment $$I=[0,1]$$. It is built by induction:

• Starting with $$S_0=I$$
• $$S_1=[0,\frac{3}{8}] \cup [\frac{5}{8},1]$$
• If $$S_n$$ is a finite disjoint union of segments $$s_n=\cup_k \left[a_k,b_k\right]$$, $S_{n+1}=\bigcup_k \left(\left[a_k,\frac{a_k+b_k}{2}-\frac{1}{2^{2n+3}}\right] \cup \left[\frac{a_k+b_k}{2}+\frac{1}{2^{2n+3}},b_k\right]\right)$

# No minimum at the origin but a minimum along all lines

We look here at an example, from the Italian mathematician Giuseppe Peano of a real function $$f$$ defined on $$\mathbb{R}^2$$. $$f$$ is having a local minimum at the origin along all lines passing through the origin, however $$f$$ does not have a local minimum at the origin as a function of two variables.

The function $$f$$ is defined as follows
$\begin{array}{l|rcl} f : & \mathbb{R}^2 & \longrightarrow & \mathbb{R} \\ & (x,y) & \longmapsto & f(x,y)=3x^4-4x^2y+y^2 \end{array}$ One can notice that $$f(x, y) = (y-3x^2)(y-x^2)$$. In particular, $$f$$ is strictly negative on the open set $$U=\{(x,y) \in \mathbb{R}^2 \ : \ x^2 < y < 3x^2\}$$, vanishes on the parabolas $$y=x^2$$ and $$y=3 x^2$$ and is strictly positive elsewhere. Consider a line $$D$$ passing through the origin. If $$D$$ is different from the coordinate axes, the equation of $$D$$ is $$y = \lambda x$$ with $$\lambda > 0$$. We have $f(x, \lambda x)= x^2(\lambda-3x)(\lambda -x).$ For $$x \in (-\infty,\frac{\lambda}{3}) \setminus \{0\}$$, $$f(x, \lambda x) > 0$$ while $$f(0,0)=0$$ which proves that $$f$$ has a local minimum at the origin along the line $$D \equiv y – \lambda x=0$$. Along the $$x$$-axis, we have $$f(x,0)=3 x^ 4$$ which has a minimum at the origin. And finally, $$f$$ also has a minimum at the origin along the $$y$$-axis as $$f(0,y)=y^2$$.

However, along the parabola $$\mathcal{P} \equiv y = 2 x^2$$ we have $$f(x,2 x^2)=-x^4$$ which is strictly negative for $$x \neq 0$$. As $$\mathcal{P}$$ is passing through the origin, $$f$$ assumes both positive and negative values in all neighborhood of the origin.

This proves that $$f$$ does not have a minimum at $$(0,0)$$.

# Two matrices A and B for which AB and BA have different minimal polynomials

We consider here the algebra of matrices $$\mathcal{M}_n(\mathbb F)$$ of dimension $$n \ge 1$$ over a field $$\mathbb F$$.

It is well known that for $$A,B \in \mathcal{M}_n(\mathbb F)$$, the characteristic polynomial $$p_{AB}$$ of the product $$AB$$ is equal to the one (namely $$p_{BA}$$) of the product of $$BA$$. What about the minimal polynomial?

Unlikely for the characteristic polynomials, the minimal polynomial $$\mu_{AB}$$ of $$AB$$ maybe different to the one of $$BA$$.

Consider the two matrices $A=\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix} \text{, } B=\begin{pmatrix} 0 & 0\\ 0 & 1\end{pmatrix}$ which can be defined whatever the field we consider: $$\mathbb R, \mathbb C$$ or even a field of finite characteristic.

One can verify that $AB=A=\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix} \text{, } BA=\begin{pmatrix} 0 & 0\\ 0 & 0\end{pmatrix}$

As $$BA$$ is the zero matrix, its minimal polynomial is $$\mu_{BA}=X$$. Regarding the one of $$AB$$, we have $$(AB)^2=A^2=0$$ hence $$\mu_{AB}$$ divides $$X^2$$. Moreover $$\mu_{AB}$$ cannot be equal to $$X$$ as $$AB \neq 0$$. Finally $$\mu_{AB}=X^2$$ and we verify that $X^2=\mu_{AB} \neq \mu_{BA}=X.$