Infinite rings and fields with positive characteristic

Familiar to us are infinite fields whose characteristic is equal to zero like \(\mathbb Z, \mathbb Q, \mathbb R\) or the field of constructible numbers.

We’re also familiar with rings having infinite number of elements and zero for characteristic like:

  • The rings of polynomials \(\mathbb Z[X], \mathbb Q[X], \mathbb R[X]\).
  • The rings of matrices \(\mathcal{M}_2(\mathbb R)\).
  • Or the ring of real continuous functions defined on \(\mathbb R\).

We also know rings or fields like integers modulo \(n\) (with \(n \ge 2\)) \(\mathbb Z_n\) or the finite field \(\mathbb F_q\) with \(q=p^r\) elements where \(p\) is a prime.

We provide below examples of infinite rings or fields with positive characteristic.

Infinite rings with positive characteristic

Consider the ring \(\mathbb Z_n[X]\) of polynomials in one variable \(X\) with coefficients in \(\mathbb Z_n\) for \(n \ge 2\) integer. It is an infinite ring since \(\mathbb X^m \in \mathbb{Z}_n[X]\) for all positive integers \(m\), and \(X^r \neq X^s\) for \(r \neq s\). But the characteristic of \(\mathbb Z_n[X]\) is clearly \(n\).

Another example is based on product of rings. If \(I\) is an index set and \((R_i)_{i \in I}\) a family of rings, one can define the product ring \(\displaystyle \prod_{i \in I} R_i\). The operations are defined the natural way with \((a_i)_{i \in I} + (b_i)_{i \in I} = (a_i+b_i)_{i \in I}\) and \((a_i)_{i \in I} \cdot (b_i)_{i \in I} = (a_i \cdot b_i)_{i \in I}\). Fixing \(n \ge 2\) integer and taking \(I = \mathbb N\), \(R_i = \mathbb Z_n\) for all \(i \in I\) we get the ring \(\displaystyle R = \prod_{k \in \mathbb N} \mathbb Z_n\). \(R\) multiplicative identity is the sequence with all terms equal to \(1\). The characteristic of \(R\) is \(n\) and \(R\) is obviously infinite. Continue reading Infinite rings and fields with positive characteristic

Playing with images and inverse images

We’re looking here to relational expressions involving image and inverse image.

We consider a function \(f : X \to Y\) from the set \(X\) to the set \(Y\). If \(x\) is a member of \(X\), \(f(x)\) is the image of \(x\) under \(f\).
The image of a subset \(A \subset X\) under \(f\) is the subset (of \(Y\)) \[f(A)\stackrel{def}{=} \{f(x) : x \in A\}.\]

The inverse image of a subset \(B \subset Y\) is the subset of \(A\) \[f^{-1}(B)\stackrel{def}{=} \{x \in X : f(x) \in B\}.\] Important to understand is that here, \(f^{-1}\) is not the inverse function of \(f\).

We now look at relational expressions involving the image and the inverse image under \(f\).

Inverse images with unions and intersections

Following relations hold:
f^{-1}(B_1 \cup B_2) = f^{-1}(B_1) \cup f^{-1}(B_2)\\
f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)
\end{array}\] Let’s prove the first equality as an example. We have \(x \in f^{-1}(B_1 \cup B_2)\) if and only if \(f(x) \in B_1 \cup B_2\) if and only if \(f(x) \in B_1\) or \(f(x) \in B_2\) which means exactly \(x \in f^{-1}(B_1) \cup f^{-1}(B_2)\). Continue reading Playing with images and inverse images

Counterexamples around balls in metric spaces

Let’s play with balls in a metric space \((M,d)\). We denote by

  • \(B_r(p) = \{x \in M : d(x,p) < r\}\) the open ball.
  • \(B_r[p] = \{x \in M : d(x,p) \le r\}\) the closed ball.

A ball of radius \(r\) included in a ball of radius \(r^\prime < r\)

We take for \(M\) the space \(\{0\} \cup [2, \infty)\) equipped with the standard metric distance \(d(x,y)=\vert x – y \vert\).

We have \(B_4(0) = \{0\} \cup [2, 4)\) while \(B_3(2) = \{0\} \cup [2, 5)\). Despite having a strictly smaller radius, the ball \(B_3(2)\) strictly contains the ball \(B_4(0)\).

The phenomenon cannot happen in a normed vector space \((M, \Vert \cdot \Vert)\). For the proof, take two open balls \(B_r(p),B_{r^\prime}(p^\prime) \subset M\), \(0 < r^\prime < r\) and suppose that \(p \in B_{r^\prime}(p^\prime)\). If \(p=p^\prime\) and \(q \in B_{r^\prime}(p^\prime) \setminus \{p^\prime\}\) then \(p + \frac{\frac{r+r^\prime}{2} }{\Vert p q \Vert} p q \in B_r(p) \setminus B_{r^\prime}(p^\prime) \). And if \(p \neq p^\prime\), \(p \in B_{r^\prime}(p^\prime)\) then \(p^\prime + \frac{\frac{r+r^\prime}{2} }{\Vert p^\prime p \Vert} p^\prime p \in B_r(p) \setminus B_{r^\prime}(p^\prime) \).

An open ball \(B_r(p)\) whose closure is not equal to the closed ball \(B_r[p]\)

Here we take for \(M\) a subspace of \(\mathbb R^2\) which is the union of the origin \(\{0\}\) with the unit circle \(S^1\). For the distance, we use the Euclidean norm.
The open unit ball centered at the origin \(B_1(0)\) is reduced to the origin: \(B_1(0) = \{0\}\). Its closure \(\overline{B_1(0)}\) is itself. However the closed ball \(B_1[0]\) is the all space \(\{0\} \cup S^1\).

Again one can prove that for a normed vector space this cannot happen. The closure of an open ball is the closed ball for a normed vector space.

Counterexamples around Dini’s theorem

In this article we look at counterexamples around Dini’s theorem. Let’s recall:

Dini’s theorem: If \(K\) is a compact topological space, and \((f_n)_{n \in \mathbb N}\) is a monotonically decreasing sequence (meaning \(f_{n+1}(x) \le f_n(x)\) for all \(n \in \mathbb N\) and \(x \in K\)) of continuous real-valued functions on \(K\) which converges pointwise to a continuous function \(f\), then the convergence is uniform.

We look at what happens to the conclusion if we drop some of the hypothesis.

Cases if \(K\) is not compact

We take \(K=(0,1)\), which is not closed equipped with the common distance. The sequence \(f_n(x)=x^n\) of continuous functions decreases pointwise to the always vanishing function. But the convergence is not uniform because for all \(n \in \mathbb N\) \[\sup\limits_{x \in (0,1)} x^n = 1\]

The set \(K=\mathbb R\) is closed but unbounded, hence also not compact. The sequence defined by \[f_n(x)=\begin{cases}
0 & \text{for } x < n\\ \frac{x-n}{n} & \text{for } n \le x < 2n\\ 1 & \text{for } x \ge 2n \end{cases}\] is continuous and monotonically decreasing. It converges to \(0\). However, the convergence is not uniform as for all \(n \in \mathbb N\): \(\sup\{f_n(x) : x \in \mathbb R\} =1\). Continue reading Counterexamples around Dini’s theorem