Would you like to be the contributor for the 100th ring on the Database of Ring Theory? Go here!

# Tag Archives: rings

# A prime ideal that is not a maximal ideal

Every maximal ideal is a prime ideal. The converse is true in a principal ideal domain – PID, i.e. every nonzero prime ideal is maximal in a PID, but this is not true in general. Let’s produce a counterexample.

\(R= \mathbb Z[x]\) is a ring. \(R\) is not a PID as can be shown considering the ideal \(I\) generated by the set \(\{2,x\}\). \(I\) cannot be generated by a single element \(p\). If it was, \(p\) would divide \(2\), i.e. \(p=1\) or \(p=2\). We can’t have \(p=1\) as it means \(R = I\) but \(3 \notin I\). We can’t have either \(p=2\) as it implies the contradiction \(x \notin I\). The ideal \(J = (x)\) is a prime ideal as \(R/J \cong \mathbb Z\) is an integral domain. Since \(\mathbb Z\) is not a field, \(J\) is not a maximal ideal.

# Four elements rings

A group with four elements is isomorphic to either the cyclic group \(\mathbb Z_4\) or to the Klein four-group \(\mathbb Z_2 \times \mathbb Z_2\). Those groups are commutative. Endowed with the usual additive and multiplicative operations, \(\mathbb Z_4\) and \(\mathbb Z_2 \times \mathbb Z_2\) are commutative rings.

Are all four elements rings also isomorphic to either \(\mathbb Z_4\) or \(\mathbb Z_2 \times \mathbb Z_2\)? The answer is negative. Let’s provide two additional examples of commutative rings with four elements not isomorphic to \(\mathbb Z_4\) or \(\mathbb Z_2 \times \mathbb Z_2\).

The first one is the field \(\mathbb F_4\). \(\mathbb F_4\) is a commutative ring with four elements. It is not isomorphic to \(\mathbb Z_4\) or \(\mathbb Z_2 \times \mathbb Z_2\) as both of those rings have zero divisor. Indeed we have \(2 \cdot 2 = 0\) in \(\mathbb Z_4\) and \((1,0) \cdot (0,1)=(0,0)\) in \(\mathbb Z_2 \times \mathbb Z_2\).

A second one is the ring \(R\) of the matrices \(\begin{pmatrix}

x & 0\\

y & x\end{pmatrix}\) where \(x,y \in \mathbb Z_2\). One can easily verify that \(R\) is a commutative subring of the ring \(M_2(\mathbb Z_2)\). It is not isomorphic to \(\mathbb Z_4\) as its characteristic is \(2\). This is not isomorphic to \(\mathbb Z_2 \times \mathbb Z_2\) either as \(\begin{pmatrix}

0 & 0\\

1 & 0\end{pmatrix}\) is a non-zero matrix solution of the equation \(X^2=0\). \((0,0)\) is the only solution of that equation in \(\mathbb Z_2 \times \mathbb Z_2\).

One can prove that the four rings mentioned above are the only commutative rings with four elements up to isomorphism.

# Group homomorphism versus ring homomorphism

A ring homomorphism is a function between two rings which respects the structure. Let’s provide examples of functions between rings which respect the addition or the multiplication but not both.

### An additive group homomorphism that is not a ring homomorphism

We consider the ring \(\mathbb R[x]\) of real polynomials and the derivation \[

\begin{array}{l|rcl}

D : & \mathbb R[x] & \longrightarrow & \mathbb R[x] \\

& P & \longmapsto & P^\prime \end{array}\] \(D\) is an additive homomorphism as for all \(P,Q \in \mathbb R[x]\) we have \(D(P+Q) = D(P) + D(Q)\). However, \(D\) does not respect the multiplication as \[

D(x^2) = 2x \neq 1 = D(x) \cdot D(x).\] More generally, \(D\) satisfies the Leibniz rule \[

D(P \cdot Q) = P \cdot D(Q) + Q \cdot D(P).\]

### A multiplication group homomorphism that is not a ring homomorphism

The function \[

\begin{array}{l|rcl}

f : & \mathbb R & \longrightarrow & \mathbb R \\

& x & \longmapsto & x^2 \end{array}\] is a multiplicative group homomorphism of the group \((\mathbb R, \cdot)\). However \(f\) does not respect the addition.

# 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 \(1 \notin F[X_1,X_2]\).