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

A proper subspace without an orthogonal complement

We consider an inner product space $$V$$ over the field of real numbers $$\mathbb R$$. The inner product is denoted by $$\langle \cdot , \cdot \rangle$$.

When $$V$$ is a finite dimensional space, every proper subspace $$F \subset V$$ has an orthogonal complement $$F^\perp$$ with $$V = F \oplus F^\perp$$. This is no more true for infinite dimensional spaces and we present here an example.

Consider the space $$V=\mathcal C([0,1],\mathbb R)$$ of the continuous real functions defined on the segment $$[0,1]$$. The bilinear map
$\begin{array}{l|rcl} \langle \cdot , \cdot \rangle : & V \times V & \longrightarrow & \mathbb R \\ & (f,g) & \longmapsto & \langle f , g \rangle = \displaystyle \int_0^1 f(t)g(t) \, dt \end{array}$ is an inner product on $$V$$.

Let’s consider the proper subspace $$H = \{f \in V \, ; \, f(0)=0\}$$. $$H$$ is an hyperplane of $$V$$ as $$H$$ is the kernel of the linear form $$\varphi : f \mapsto f(0)$$ defined on $$V$$. $$H$$ is a proper subspace as $$\varphi$$ is not always vanishing. Let’s prove that $$H^\perp = \{0\}$$.

Take $$g \in H^\perp$$. By definition of $$H^\perp$$ we have $$\int_0^1 f(t) g(t) \, dt = 0$$ for all $$f \in H$$. In particular the function $$h : t \mapsto t g(t)$$ belongs to $$H$$. Hence
$0 = \langle h , g \rangle = \displaystyle \int_0^1 t g(t)g(t) \, dt$ The map $$t \mapsto t g^2(t)$$ is continuous, non-negative on $$[0,1]$$ and its integral on this segment vanishes. Hence $$t g^2(t)$$ is always vanishing on $$[0,1]$$, and $$g$$ is always vanishing on $$(0,1]$$. As $$g$$ is continuous, we finally get that $$g = 0$$.

$$H$$ doesn’t have an orthogonal complement.

Moreover we have
$(H^\perp)^\perp = \{0\}^\perp = V \neq H$

Intersection and sum of vector subspaces

Let’s consider a vector space $$E$$ over a field $$K$$. We’ll look at relations involving basic set operations and sum of subspaces. We denote by $$F, G$$ and $$H$$ subspaces of $$E$$.

The relation $$(F \cap G) + H \subset (F+H) \cap (G + H)$$

This relation holds. The proof is quite simple. For any $$x \in (F \cap G) + H$$ there exists $$y \in F \cap G$$ and $$h \in H$$ such that $$x=y+h$$. As $$y \in F$$, $$x \in F+H$$ and by a similar argument $$x \in F+H$$. Therefore $$x \in (F+H) \cap (G + H)$$.

Is the inclusion $$(F \cap G) + H \subset (F+H) \cap (G + H)$$ always an equality? The answer is negative. Take for the space $$E$$ the real 3 dimensional space $$\mathbb R^3$$. And for the subspaces:

• $$H$$ the plane of equation $$z=0$$,
• $$F$$ the line of equations $$y = 0, \, x=z$$,
• $$G$$ the line of equations $$y = 0, \, x=-z$$,

A non Archimedean ordered field

Let’s recall that an ordered field $$K$$ is said to be Archimedean if for any $$a,b \in K$$ such that $$0 \lt a \lt b$$ it exists a natural number $$n$$ such that $$na > b$$.

The ordered fields $$\mathbb Q$$ or $$\mathbb R$$ are Archimedean. We introduce here the example of an ordered field which is not Archimedean. Let’s consider the field of rational functions
$\mathbb R(x) = \left\{\frac{S(x)}{T(x)} \ | \ S, T \in \mathbb R[x] \right\}$ For $$f(x)=\frac{S(x)}{T(x)} \in \mathbb R(x)$$ we can suppose that the polynomials have a constant polynomial greatest common divisor.

Now we define $$P$$ as the set of elements $$f(x)=\frac{S(x)}{T(x)} \in \mathbb R(x)$$ in which the leading coefficients of $$S$$ and $$T$$ have the same sign.

One can verify that the subset $$P \subset \mathbb R(x)$$ satisfies following two conditions:

ORD 1
Given $$f(x) \in \mathbb R(x)$$, we have either $$f(x) \in P$$, or $$f(x)=0$$, or $$-f(x) \in P$$, and these three possibilities are mutually exclusive. In other words, $$\mathbb R(x)$$ is the disjoint union of $$P$$, $$\{0\}$$ and $$-P$$.
ORD 2
For $$f(x),g(x) \in P$$, $$f(x)+g(x)$$ and $$f(x)g(x)$$ belong to $$P$$.

This means that $$P$$ is a positive cone of $$\mathbb R(x)$$. Hence, $$\mathbb R(x)$$ is ordered by the relation
$f(x) > 0 \Leftrightarrow f(x) \in P.$

Now let’s consider the rational fraction $$h(x)=\frac{x}{1} \in \mathbb R(x)$$. $$h(x)$$ is a positive element, i.e. belongs to $$P$$ as $$h-1 = \frac{x-1}{1}$$. For any $$n \in \mathbb N$$, we have
$h – n 1=\frac{x-n}{1} \in P$ as the leading coefficients of $$x-n$$ and $$1$$ are both equal to $$1$$. Therefore, we have $$h \gt n 1$$ for all $$n \in \mathbb N$$, proving that $$\mathbb R(x)$$ is not Archimedean.

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 $$(1,1)$$.
• $$R$$ 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: $(1,0).(0,1)=(0,0)$

Two matrices A and B for which AB and BA have different minimal polynomials

We consider here the algebra of matrices $$\mathcal{M}_n(\mathbb F)$$ of dimension $$n \ge 1$$ over a field $$\mathbb F$$.

It is well known that for $$A,B \in \mathcal{M}_n(\mathbb F)$$, the characteristic polynomial $$p_{AB}$$ of the product $$AB$$ is equal to the one (namely $$p_{BA}$$) of the product of $$BA$$. What about the minimal polynomial?

Unlikely for the characteristic polynomials, the minimal polynomial $$\mu_{AB}$$ of $$AB$$ maybe different to the one of $$BA$$.

Consider the two matrices $A=\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix} \text{, } B=\begin{pmatrix} 0 & 0\\ 0 & 1\end{pmatrix}$ which can be defined whatever the field we consider: $$\mathbb R, \mathbb C$$ or even a field of finite characteristic.

One can verify that $AB=A=\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix} \text{, } BA=\begin{pmatrix} 0 & 0\\ 0 & 0\end{pmatrix}$

As $$BA$$ is the zero matrix, its minimal polynomial is $$\mu_{BA}=X$$. Regarding the one of $$AB$$, we have $$(AB)^2=A^2=0$$ hence $$\mu_{AB}$$ divides $$X^2$$. Moreover $$\mu_{AB}$$ cannot be equal to $$X$$ as $$AB \neq 0$$. Finally $$\mu_{AB}=X^2$$ and we verify that $X^2=\mu_{AB} \neq \mu_{BA}=X.$

A finite extension that contains infinitely many subfields

Let’s consider $$K/k$$ a finite field extension of degree $$n$$. The following theorem holds.

Theorem: the following conditions are equivalent:

1. The extension contains a primitive element.
2. The number of intermediate fields between $$k$$ and $$K$$ is finite.

Our aim here is to describe a finite field extension having infinitely many subfields. Considering the theorem above, we have to look at an extension without a primitive element.

The extension $$\mathbb F_p(X,Y) / \mathbb F_p(X^p,Y^p)$$ is finite

For $$p$$ prime, $$\mathbb F_p$$ denotes the finite field with $$p$$ elements. $$\mathbb F_p(X,Y)$$ is the algebraic fraction field of two variables over the field $$\mathbb F_p$$. $$\mathbb F_p(X^p,Y^p)$$ is the subfield of $$\mathbb F_p(X,Y)$$ generated by the elements $$X^p,Y^p$$. Continue reading A finite extension that contains infinitely many subfields

A group isomorphic to its automorphism group

We consider a group $$G$$ and we look at its automorphism group $$\text{Aut}(G)$$. Can $$G$$ be isomorphic to
$$\text{Aut}(G)$$?
The answer is positive and we’ll prove that it is the case for the symmetric group $$S_3$$.

Consider the morphism $\begin{array}{l|rcl} \Phi : & S_3 & \longrightarrow & \text{Aut}(S_3) \\ & a & \longmapsto & \varphi_a \end{array}$
where $$\varphi_a$$ is the inner automorphism $$\varphi_a : x \mapsto a^{-1}xa$$. It is easy to verify that $$\Phi$$ is indeed a group morphism. The kernel of $$\Phi$$ is the center of $$S_3$$ which is having the identity for only element. Hence $$\Phi$$ is one-to-one and $$S_3 \simeq \Phi(S_3)$$. Therefore it is sufficient to prove that $$\Phi$$ is onto. As $$|S_3|=6$$, we’ll be finished if we prove that $$|\text{Aut}(S_3)|=6$$.

Generally, for $$G_1,G_2$$ groups and $$f : G_1 \to G_2$$ a one-to-one group morphism, the image of an element $$x$$ of order $$k$$ is an element $$f(x)$$ having the same order $$k$$. So for $$\varphi \in \text{Aut}(S_3)$$ the image of a transposition is a transposition. As the transpositions $$\{(1 \ 2), (1 \ 3), (2 \ 3)\}$$ generate $$(S_3)$$, $$\varphi$$ is completely defined by $$\{\varphi((1 \ 2)), \varphi((1 \ 3)), \varphi((2 \ 3))\}$$. We have 3 choices to define the image of $$(1 \ 2)$$ under $$\varphi$$ and then 2 choices for the image of $$(1 \ 3)$$ under $$\varphi$$. The image of $$(2 \ 3)$$ under $$\varphi$$ is the remaining transposition.

Finally, we have proven that $$|\text{Aut}(S_3)|=6$$ as desired and $$S_3 \simeq \text{Aut}(S_3)$$.