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 \(s_n=\cup_k \left[a_k,b_k\right]\), \[S_{n+1}=\bigcup_k \left(\left[a_k,\frac{a_k+b_k}{2}-\frac{1}{2^{2n+3}}\right] \cup \left[\frac{a_k+b_k}{2}+\frac{1}{2^{2n+3}},b_k\right]\right)\]

Continue reading The Smith Volterra Cantor Set

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
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 \[
0 & 1\\
0 & 0\end{pmatrix} \text{, }
0 & 0\\
0 & 1\end{pmatrix}\] which can be defined whatever the field we consider: \(\mathbb R, \mathbb C\) or even a field of finite characteristic.

One can verify that \[
0 & 1\\
0 & 0\end{pmatrix} \text{, }
0 & 0\\
0 & 0\end{pmatrix}\]

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 \[X^2=\mu_{AB} \neq \mu_{BA}=X.\]