Tag Archives: rings

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

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 \((0,0)\).
  • \(R\) 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: \[(1,0).(0,1)=(0,0)\]

On polynomials having more roots than their degree

Let’s consider a polynomial of degree \(q \ge 1\) over a field \(K\). It is well known that the sum of the multiplicities of the roots of \(P\) is less or equal to \(q\).

The result remains for polynomials over an integral domain. What is happening for polynomials over a commutative ring? Continue reading On polynomials having more roots than their degree