Tag Archives: rings

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

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

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