# A normal extension of a normal extension may not be normal

An algebraic field extension $$K \subset L$$ is said to be normal if every irreducible polynomial, either has no root in $$L$$ or splits into linear factors in $$L$$.

One can prove that if $$L$$ is a normal extension of $$K$$ and if $$E$$ is an intermediate extension (i.e., $$K \subset E \subset L$$), then $$L$$ is a normal extension of $$E$$.

However a normal extension of a normal extension may not be normal and the extensions $$\mathbb Q \subset \mathbb Q(\sqrt{2}) \subset \mathbb Q(\sqrt{2})$$ provide a counterexample. Let’s prove it.

As a short lemma, we prove that a quadratic extension $$k \subset K$$ , i.e. an extension of degree two is normal. Suppose that $$P$$ is an irreducible polynomial of $$k[x]$$ with a root $$a \in K$$. If $$a \in k$$ then the degree of $$P$$ is equal to $$1$$ and we’re done. Otherwise $$(1, a)$$ is a basis of $$K$$ over $$k$$ and there exist $$\lambda, \mu \in k$$ such that $$a^2 = \lambda a +\mu$$. As $$a \notin k$$, $$Q(x)= x^2 – \lambda x -\mu$$ is the minimal polynomial of $$a$$ over $$k$$. As $$P$$ is supposed to be irreducible, we get $$Q = P$$. And we can conclude as $Q(x) = (x-a)(x- \lambda +a).$

The entensions $$\mathbb Q \subset \mathbb Q(\sqrt{2})$$ and $$\mathbb Q(\sqrt{2}) \subset \mathbb Q(\sqrt{2})$$ are quadratic, hence normal according to previous lemma and $$\sqrt{2}$$ is a root of the polynomial $$P(x)= x^4-2$$ of $$\mathbb Q[x]$$. According to Eisenstein’s criterion $$P$$ is irreducible over $$\mathbb Q$$. However $$\mathbb Q(\sqrt{2}) \subset \mathbb R$$ while the roots of $$P$$ are $$\pm \sqrt{2}, \pm i \sqrt{2}$$ and therefore not all real. We can conclude that $$\mathbb Q \subset \mathbb Q(\sqrt{2})$$ is not normal.

# Field not algebraic over an intersection but algebraic over each initial field

Let’s describe an example of a field $$K$$ which is of degree $$2$$ over two distinct subfields $$M$$ and $$N$$ respectively, but not algebraic over $$M \cap N$$.

Let $$K=F(x)$$ be the rational function field over a field $$F$$ of characteristic $$0$$, $$M=F(x^2)$$ and $$N=F(x^2+x)$$. I claim that those fields provide the example we’re looking for.

### $$K$$ is of degree $$2$$ over $$M$$ and $$N$$

The polynomial $$\mu_M(t)=t^2-x^2$$ belongs to $$M[t]$$ and $$x \in K$$ is a root of $$\mu_M$$. Also, $$\mu_M$$ is irreducible over $$M=F(x^2)$$. If that wasn’t the case, $$\mu_M$$ would have a root in $$F(x^2)$$ and there would exist two polynomials $$p,q \in F[t]$$ such that $p^2(x^2) = x^2 q^2(x^2)$ which cannot be, as can be seen considering the degrees of the polynomials of left and right hand sides. This proves that $$[K:M]=2$$. Considering the polynomial $$\mu_N(t)=t^2-t-(x^2+x)$$, one can prove that we also have $$[K:N]=2$$.

### We have $$M \cap N=F$$

The mapping $$\sigma_M : x \mapsto -x$$ extends uniquely to an $$F$$-automorphism of $$K$$ and the elements of $$M$$ are fixed under $$\sigma_M$$. Similarly, the mapping $$\sigma_N : x \mapsto -x-1$$ extends uniquely to an $$F$$-automorphism of $$K$$ and the elements of $$N$$ are fixed under $$\sigma_N$$. Also $(\sigma_N\circ\sigma_M)(x)=\sigma_N(\sigma_M(x))=\sigma_N(-x)=-(-x-1)=x+1.$ An element $$z=p(x)/q(x) \in M \cap N$$ where $$p(x),q(x)$$ are coprime polynomials of $$K=F(x)$$ is fixed under $$\sigma_M \circ \sigma_N$$. Therefore following equality holds $\frac{p(x)}{q(x)}=z=(\sigma_2\circ\sigma_1)(z)=\frac{p(x+1)}{q(x+1)},$ which is equivalent to $p(x)q(x+1)=p(x+1)q(x).$ By induction, we get for $$n \in \mathbb Z$$ $p(x)q(x+n)=p(x+n)q(x).$ Assume $$p(x)$$ is not a constant polynomial. Then it has a root $$\alpha$$ in some finite extension $$E$$ of $$F$$. As $$p(x),q(x)$$ are coprime polynomials, $$q(\alpha) \neq 0$$. Consequently $$p(\alpha+n)=0$$ for all $$n \in \mathbb Z$$ and the elements $$\alpha +n$$ are all distinct as the characteristic of $$F$$ is supposed to be non zero. This implies that $$p(x)$$ is the zero polynomial, in contradiction with our assumption. Therefore $$p(x)$$ is a constant polynomial and $$q(x)$$ also according to a similar proof. Hence $$z$$ is constant as was supposed to be proven.

Finally, $$K=F(x)$$ is not algebraic over $$F=M \cap N$$ as $$(1,x, x^2, \dots, x^n, \dots)$$ is independent over the field $$F$$ which concludes our claims on $$K, M$$ and $$N$$.

# Additive subgroups of vector spaces

Consider a vector space $$V$$ over a field $$F$$. A subspace $$W \subseteq V$$ is an additive subgroup of $$(V,+)$$. The converse might not be true.

If the characteristic of the field is zero, then a subgroup $$W$$ of $$V$$ might not be an additive subgroup. For example $$\mathbb R$$ is a vector space over $$\mathbb R$$ itself. $$\mathbb Q$$ is an additive subgroup of $$\mathbb R$$. However $$\sqrt{2}= \sqrt{2}.1 \notin \mathbb Q$$ proving that $$\mathbb Q$$ is not a subspace of $$\mathbb R$$.

Another example is $$\mathbb Q$$ which is a vector space over itself. $$\mathbb Z$$ is an additive subgroup of $$\mathbb Q$$, which is not a subspace as $$\frac{1}{2} \notin \mathbb Z$$.

Yet, an additive subgroup of a vector space over a prime field $$\mathbb F_p$$ with $$p$$ prime is a subspace. To prove it, consider an additive subgroup $$W$$ of $$(V,+)$$ and $$x \in W$$. For $$\lambda \in F$$, we can write $$\lambda = \underbrace{1 + \dots + 1}_{\lambda \text{ times}}$$. Consequently $\lambda \cdot x = (1 + \dots + 1) \cdot x= \underbrace{x + \dots + x}_{\lambda \text{ times}} \in W.$

Finally an additive subgroup of a vector space over any finite field is not always a subspace. For a counterexample, take the non-prime finite field $$\mathbb F_{p^2}$$ (also named $$\text{GF}(p^2)$$). $$\mathbb F_{p^2}$$ is also a vector space over itself. The prime finite field $$\mathbb F_p \subset \mathbb F_{p^2}$$ is an additive subgroup that is not a subspace of $$\mathbb F_{p^2}$$.

# 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 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

# The skew field of Hamilton’s quaternions

We give here an example of a division ring which is not commutative. According to Wedderburn theorem every finite division ring is commutative. So we must turn to infinite division rings to find a non-commutative one, i.e. a skew field.

Let’s introduce the skew field of the Hamilton’s quaternions $\mathbb H = \left\{\begin{pmatrix} u & -\overline{v} \\ v & \overline{u} \end{pmatrix} \ | \ u,v \in \mathbb C\right\}$ Continue reading The skew field of Hamilton’s quaternions

# An irreducible integral polynomial reducible over all finite prime fields

A classical way to prove that an integral polynomial $$Q \in \mathbb{Z}[X]$$ is irreducible is to prove that $$Q$$ is irreducible over a finite prime field $$\mathbb{F}_p$$ where $$p$$ is a prime.

This raises the question whether an irreducible integral polynomial is irreducible over at least one finite prime field. The answer is negative and:
$P(X)=X^4+1$ is a counterexample. Continue reading An irreducible integral polynomial reducible over all finite prime fields

# 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

# A vector space written as a finite union of proper subspaces

We raise here the following question: “can a vector space $$E$$ be written as a finite union of proper subspaces”?

Let’s consider the simplest case, i.e. writing $$E= V_1 \cup V_2$$ as a union of two proper subspaces. By hypothesis, one can find two non-zero vectors $$v_1,v_2$$ belonging respectively to $$V_1 \setminus V_2$$ and $$V_2 \setminus V_1$$. The relation $$v_1+v_2 \in V_1$$ leads to the contradiction $$v_2 = (v_1+v_2)-v_1 \in V_1$$ while supposing $$v_1+v_2 \in V_2$$ leads to the contradiction $$v_1 = (v_1+v_2)-v_2 \in V_2$$. Therefore, a vector space can never be written as a union of two proper subspaces.

We now analyze if a vector space can be written as a union of $$n \ge 3$$ proper subspaces. We’ll see that it is impossible when $$E$$ is a vector space over an infinite field. But we’ll describe a counterexample of a vector space over the finite field $$\mathbb{Z}_2$$ written as a union of three proper subspaces. Continue reading A vector space written as a finite union of proper subspaces