# A Commutative Ring with Infinitely Many Units

In a ring $$R$$ a unit is any element $$u$$ that has a multiplicative inverse $$v$$, i.e. an element $$v$$ such that $uv=vu=1,$ where $$1$$ is the multiplicative identity.

The only units of the commutative ring $$\mathbb Z$$ are $$-1$$ and $$1$$. For a field $$\mathbb F$$ the units of the ring $$\mathrm M_n(\mathbb F)$$ of the square matrices of dimension $$n \times n$$ is the general linear group $$\mathrm{GL}_n(\mathbb F)$$ of the invertible matrices. The group $$\mathrm{GL}_n(\mathbb F)$$ is infinite if $$\mathbb F$$ is infinite, but the ring $$\mathrm M_n(\mathbb F)$$ is not commutative for $$n \ge 2$$.

The commutative ring $$\mathbb Z[\sqrt{2}] = \{a + b\sqrt{2} \ ; \ (a,b) \in \mathbb Z^2\}$$ is not a field. However it has infinitely many units.

### $$a + b\sqrt{2}$$ is a unit if and only if $$a^2-2b^2 = \pm 1$$

For $$u = a + b\sqrt{2} \in \mathbb Z[\sqrt{2}]$$ we denote $$\mathrm N(u) = a^2- 2b^2 \in \mathbb Z$$. For any $$u,v \in \mathbb Z[\sqrt{2}]$$ we have $$\mathrm N(uv) = \mathrm N(u) \mathrm N(v)$$. Therefore for a unit $$u \in \mathbb Z[\sqrt{2}]$$ with $$v$$ as multiplicative inverse, we have $$\mathrm N(u) \mathrm N(v) = 1$$ and $$\mathrm N(u) =a^2-2b^2 \in \{-1,1\}$$.

### The elements $$(1+\sqrt{2})^n$$ for $$n \in \mathbb N$$ are unit elements

The proof is simple as for $$n \in \mathbb N$$ $(1+\sqrt{2})^n (-1 + \sqrt{2})^n = \left((1+\sqrt{2})(-1 + \sqrt{2})\right)^n=1$

One can prove (by induction on $$b$$) that the elements $$(1+\sqrt{2})^n$$ are the only units $$u \in \mathbb Z[\sqrt{2}]$$ for $$u \gt 1$$.

# The image of an ideal may not be an ideal

If $$\phi : A \to B$$ is a ring homomorphism then the image of a subring $$S \subset A$$ is a subring $$\phi(A) \subset B$$. Is the image of an ideal under a ring homomorphism also an ideal? The answer is negative. Let’s provide a simple counterexample.

Let’s take $$A=\mathbb Z$$ the ring of the integers and for $$B$$ the ring of the polynomials with integer coefficients $$\mathbb Z[x]$$. The inclusion $$\phi : \mathbb Z \to \mathbb Z[x]$$ is a ring homorphism. The subset $$2 \mathbb Z \subset \mathbb Z$$ of even integers is an ideal. However $$2 \mathbb Z$$ is not an ideal of $$\mathbb Z[x]$$ as for example $$2x \notin 2\mathbb Z$$.

# Non commutative rings

Let’s recall that a set $$R$$ equipped with two operations $$(R,+,\cdot)$$ is a ring if and only if $$(R,+)$$ is an abelian group, multiplication $$\cdot$$ is associative and has a multiplicative identity $$1$$ and multiplication is left and right distributive with respect to addition.

$$(\mathbb Z, +, \cdot)$$ is a well known infinite ring which is commutative. The rational, real and complex numbers are other infinite commutative rings. Those are in fact fields as every non-zero element have a multiplicative inverse.

For a field $$F$$ (finite or infinite), the polynomial ring $$F[X]$$ is another example of infinite commutative ring.

Also for $$n$$ integer, the integers modulo n is a finite ring that is commutative. Finally, according to Wedderburn theorem every finite division ring is commutative.

So what are examples of non commutative rings? Let’s provide a couple. Continue reading Non commutative rings

# Database of Ring Theory

You want to find rings having some properties but not having other properties? Go there: Database of Ring Theory! A great repository of rings, their properties, and more ring theory stuff.

# A simple ring which is not a division ring

Let’s recall that a simple ring is a non-zero ring that has no two-sided ideal besides the zero ideal and itself. A division ring is a simple ring. Is the converse true? The answer is negative and we provide here a counterexample of a simple ring which is not a division ring.

We prove that for $$n \ge 1$$ the matrix ring $$M_n(F)$$ of $$n \times n$$ matrices over a field $$F$$ is simple. $$M_n(F)$$ is obviously not a division ring as the matrix with $$1$$ at position $$(1,1)$$ and $$0$$ elsewhere is not invertible.

Let’s prove first following lemma. Continue reading A simple ring which is not a division ring

# Unique factorization domain that is not a Principal ideal domain

In this article, we provide an example of a unique factorization domain – UFD that is not a principal ideal domain – PID. However, it is known that a PID is a UFD.

We take a field $$F$$, for example $$\mathbb Q$$, $$\mathbb R$$, $$\mathbb F_p$$ (where $$p$$ is a prime) or whatever more exotic.

The polynomial ring $$F[X]$$ is a UFD. This follows from the fact that $$F[X]$$ is a Euclidean domain. It is also known that for a UFD $$R$$, $$R[X]$$ is also a UFD. Therefore the polynomial ring $$F[X_1,X_2]$$ in two variables is a UFD as $$F[X_1,X_2] = F[X_1][X_2]$$. However the ideal $$I=(X_1,X_2)$$ is not principal. Let’s prove it by contradiction.

Suppose that $$(X_1,X_2) = (P)$$ with $$P \in F[X_1,X_2]$$. Then there exist two polynomials $$Q_1,Q_2 \in F[X_1,X_2]$$ such that $$X_1=PQ_1$$ and $$X_2=PQ_2$$. As a polynomial in variable $$X_2$$, the polynomial $$X_1$$ is having degree $$0$$. Therefore, the degree of $$P$$ as a polynomial in variable $$X_2$$ is also equal to $$0$$. By symmetry, we get that the degree of $$P$$ as a polynomial in variable $$X_1$$ is equal to $$0$$ too. Which implies that $$P$$ is an element of the field $$F$$ and consequently that $$(X_1,X_2) = F[X_1,X_2]$$.

But the equality $$(X_1,X_2) = F[X_1,X_2]$$ is absurd. Indeed, the degree of a polynomial $$X_1 T_1 + X_2 T_2$$ cannot be equal to $$0$$ for any $$T_1,T_2 \in F[X_1,X_2]$$. And therefore $$1 \notin F[X_1,X_2]$$.

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