# 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: $(1,0).(0,1)=(0,0)$