# 100th ring on the Database of Ring Theory

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# 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 $$1 \notin F[X_1,X_2]$$.