Finite Fields and Cyclotomic Extensions: Examples and Counterexamples

$\pi$ has no root in ${\mathbb{F}}_2$: $\pi (0)=1$ and $\pi (1)=1+1+1=1$. By [L1] it is irreducible, so $K$ is a field by [L2].

$\pi$ is monic irreducible with $\pi (\alpha )=0$, so it is the minimal polynomial of $\alpha$ over ${\mathbb{F}}_2$ and $[K:{\mathbb{F}}_2]=3$ with power basis $1,\alpha ,{\alpha }^2$ by [L3]; hence $\mathrm{\mid }K\mathrm{\mid }=2^3=8$ by [L4].

By [L4] the extension $K\mathrm{/}{\mathbb{F}}_2$ is Galois with Galois group generated by $\sigma (x)=x^2$ and of order three.

The orbit of $\alpha$: ${\alpha }^2$ is ${\alpha }^2$; ${\alpha }^4=\alpha \cdot {\alpha }^3=\alpha (\alpha +1)={\alpha }^2+\alpha$; and $({\alpha }^2+\alpha {)}^2={\alpha }^4+{\alpha }^2=({\alpha }^2+\alpha )+{\alpha }^2=\alpha$, using that squaring is additive in characteristic two. So $\{\alpha ,{\alpha }^2,{\alpha }^2+\alpha \}$ is one orbit of size three.

The orbit of $\alpha +1$: $(\alpha +1{)}^2={\alpha }^2+1$, $({\alpha }^2+1{)}^2={\alpha }^4+1={\alpha }^2+\alpha +1$, and $({\alpha }^2+\alpha +1{)}^2={\alpha }^4+{\alpha }^2+1=\alpha +1$. So $\{\alpha +1,{\alpha }^2+1,{\alpha }^2+\alpha +1\}$ is the other orbit of size three, and together with $\{0\}$ and $\{1\}$ these account for all eight elements.

The positive divisors of three are $1$ and $3$, so by [L5] the intermediate fields are exactly two: ${\mathbb{F}}_2$ and $K$ itself. There is no field strictly between them.

The positive divisors of twelve are $1,2,3,4,6,12$, six in all, since a positive divisor $d$ of $12$ satisfies $d\le 12$ and direct inspection of $1,\dots ,12$ leaves exactly these.

By [L1] and [L2] the intermediate fields of $E\mathrm{/}{\mathbb{F}}_2$ are the ${\mathbb{F}}_{2^d}$ for those six $d$, one for each, with ${\mathbb{F}}_{2^d}\subseteq {\mathbb{F}}_{2^e}$ exactly when $d\mid e$.

Neither $4\mid 6$ nor $6\mid 4$, so by step 2.1 neither of ${\mathbb{F}}_{2^4}$ and ${\mathbb{F}}_{2^6}$ contains the other.

Their intersection is an intermediate field of $E\mathrm{/}{\mathbb{F}}_2$, being a subfield of $E$ containing ${\mathbb{F}}_2$, so it is ${\mathbb{F}}_{2^c}$ for a unique divisor $c$ of $12$ by step 2.1; from ${\mathbb{F}}_{2^c}\subseteq {\mathbb{F}}_{2^4}$ and ${\mathbb{F}}_{2^c}\subseteq {\mathbb{F}}_{2^6}$ one gets $c\mid 4$ and $c\mid 6$, so [L4] gives $c\mid \gcd (4,6)=2$; and $2\mid 4$ and $2\mid 6$ put ${\mathbb{F}}_{2^2}$ inside both, so $2\mid c$. Hence $c=2$ and the intersection is ${\mathbb{F}}_{2^2}$.

Their compositum is likewise an intermediate field ${\mathbb{F}}_{2^{c^{\mathrm{\prime }}}}$, and it contains both, so $4\mid c^{\mathrm{\prime }}$ and $6\mid c^{\mathrm{\prime }}$ by step 2.1. Thus [L4] gives $\mathrm{lcm}(4,6)=12\mid c^{\mathrm{\prime }}$; since $c^{\mathrm{\prime }}\mid 12$, $c^{\mathrm{\prime }}=12$ and the compositum is $E={\mathbb{F}}_{2^{12}}$.

The same six fields are what [L3] produces for $E$, whose order is $2^{12}$: its subfields are the $\{a:a^{2^e}=a\}$ for the divisors $e$ of $12$, and these are the sets named in [L2]. So the Galois indexing and the elementary one agree here.

The monic polynomials of degree one in ${\mathbb{F}}_2[t]$ are $t$ and $t+1$, and each is irreducible, having degree one; so $N_2(1)=2$.

A monic cubic over ${\mathbb{F}}_2$ is $f=t^3+a{t}^2+bt+c$ with $a,b,c\in {\mathbb{F}}_2$, so there are eight of them. Such an $f$ has no root in ${\mathbb{F}}_2$ exactly when $f(0)=c\mathrm{\ne }0$ and $f(1)=1+a+b+c\mathrm{\ne }0$, that is exactly when $c=1$ and $a+b=1$.

The positive divisors of $3$ are $1$ and $3$, so [L1] at $q=2$ and $n=3$ reads $N_2(1)+3N_2(3)=8$; with step 1.1 this gives $3N_2(3)=6$ and $N_2(3)=2$.

The pairs $(a,b)$ with $a+b=1$ in ${\mathbb{F}}_2$ are $(0,1)$ and $(1,0)$, so exactly two monic cubics have no root in ${\mathbb{F}}_2$, namely $t^3+t+1$ and $t^3+t^2+1$; by [L2] these two are irreducible and by [L2] and [L3] the other six are not, each having a root and hence a linear factor. This agrees with the count $N_2(3)=2$ of step 2.1.

$\pi$ has no root in ${\mathbb{F}}_2$, since $\pi (0)=1$ and $\pi (1)=1+1+1=1$; so by [L2] it has no factor of degree one.

The only monic irreducible quadratic in ${\mathbb{F}}_2[t]$ is $t^2+t+1$: the four monic quadratics are $t^2$, $t^2+1$, $t^2+t$ and $t^2+t+1$, and the first three have the root $0$, $1$ and $0$ respectively, so [L1] leaves only the last.

$\pi$ is irreducible. A factorisation of $\pi$ into two nonconstant factors has degrees summing to four by [L2], so it is either $1+3$, excluded by step 1.1, or $2+2$; and every monic quadratic factor would have to be irreducible, hence equal to $t^2+t+1$ by step 1.2, giving $\pi =(t^2+t+1{)}^2=t^4+t^2+1$, which is not $\pi$.

By [L3] the ring $K$ is a field; $\pi$ is monic irreducible with $\pi (\alpha )=0$, so $[K:{\mathbb{F}}_2]=4$ with power basis $1,\alpha ,{\alpha }^2,{\alpha }^3$ by [L4], and $\mathrm{\mid }K\mathrm{\mid }=2^4=16$ by [L5].

Compute the conjugates in that basis: ${\alpha }^4=\alpha +1$ by hypothesis, and ${\alpha }^8=({\alpha }^4{)}^2=(\alpha +1{)}^2={\alpha }^2+1$. So the four elements $\alpha ,{\alpha }^2,{\alpha }^4,{\alpha }^8$ have coordinate lists $(0,1,0,0)$, $(0,0,1,0)$, $(1,1,0,0)$ and $(1,0,1,0)$, which are pairwise different, so the four elements are pairwise distinct.

${\alpha }^{16}=({\alpha }^8{)}^2=({\alpha }^2+1{)}^2={\alpha }^4+1=(\alpha +1)+1=\alpha$, so the orbit closes after four steps.

By [L6] applied to $\pi$, of degree four, the elements $\alpha ,{\alpha }^2,{\alpha }^4,{\alpha }^8$ are exactly the roots of $\pi$ and $\pi =(t-\alpha )(t-{\alpha }^2)(t-{\alpha }^4)(t-{\alpha }^8)$ in $K[t]$; steps 4.1 and 5.1 verify the distinctness and the closing of the orbit directly.

By [L1] and [L2] the space $K$ is a three-dimensional ${\mathbb{F}}_2$-vector space with basis $1,\alpha ,{\alpha }^2$, and $\mathrm{Gal}(K\mathrm{/}{\mathbb{F}}_2)=\{\mathrm{i}\mathrm{d},\sigma ,{\sigma }^2\}$ acts by $x\mapsto x,x^2,x^4$.

The conjugate list of $\alpha$ is $(\alpha ,{\alpha }^2,{\alpha }^4)$ with ${\alpha }^4=\alpha \cdot {\alpha }^3=\alpha (\alpha +1)={\alpha }^2+\alpha$; hence $\alpha +{\alpha }^2+{\alpha }^4=\alpha +{\alpha }^2+{\alpha }^2+\alpha =0$. A vanishing combination with all coefficients $1$ is nontrivial, so this list is linearly dependent over ${\mathbb{F}}_2$ and is not a basis.

The conjugates of $\beta =\alpha +1$ are $\beta$, ${\beta }^2=(\alpha +1{)}^2={\alpha }^2+1$ and ${\beta }^4=({\beta }^2{)}^2=({\alpha }^2+1{)}^2={\alpha }^4+1={\alpha }^2+\alpha +1$.

The seven nonzero ${\mathbb{F}}_2$-combinations of $\beta ,{\beta }^2,{\beta }^4$ are nonzero: the three single terms are $\alpha +1$, ${\alpha }^2+1$ and ${\alpha }^2+\alpha +1$; the three pairwise sums are $\beta +{\beta }^2={\alpha }^2+\alpha$, $\beta +{\beta }^4={\alpha }^2$ and ${\beta }^2+{\beta }^4=\alpha$; and the total sum is $\beta +{\beta }^2+{\beta }^4=1$. None of these seven is $0$, as each has a nonzero coordinate list in the basis $1,\alpha ,{\alpha }^2$.

So the ${\mathbb{F}}_2$-linear map $T\text{\hspace{0.17em}⁣}:{\mathbb{F}}_2^3\to K$ sending $(c_1,c_2,c_3)$ to $c_1\beta +c_2{\beta }^2+c_3{\beta }^4$ has trivial kernel by step 4.1; both spaces have dimension three by step 1.1, so [L4] makes $T$ surjective as well, hence bijective, and [L3] makes $(\beta ,{\beta }^2,{\beta }^4)$ an ordered basis of $K$ over ${\mathbb{F}}_2$.

That list is the family of conjugates of $\beta$ under $\mathrm{Gal}(K\mathrm{/}{\mathbb{F}}_2)$ by step 3.1, so it is a normal basis, as [L5] guarantees exists for this cyclic extension.

By [L2] and [L3] the group $\mathrm{Aut}(\mathbb{C}\mathrm{/}\mathbb{R})$ has exactly the two elements $\mathrm{i}\mathrm{d}$ and conjugation, so its order is $2=[\mathbb{C}:\mathbb{R}]$ by [L1]; hence $\mathbb{C}\mathrm{/}\mathbb{R}$ is finite Galois with that Galois group by [L4].

The conjugate list of $1+i$ is $(1+i,1-i)$, whose members have coordinate lists $(1,1)$ and $(1,-1)$ in the ordered basis $(1,i)$ of [L1]. For $a,b\in \mathbb{R}$, $a(1+i)+b(1-i)=(a+b)+(a-b)i$ vanishes exactly when $a+b=0$ and $a-b=0$, hence when $2a=0$, that is $a=0$ and then $b=0$.

$(1,i)$ is a basis by [L1], but no $z\in \mathbb{C}$ has conjugate list $(z,\bar{z})$ with underlying set $\{1,i\}$: such a $z$ would lie in $\{1,i\}$, and the set for $z=1$ is $\{1\}$ while for $z=i$ it is $\{i,-i\}$, neither of which is $\{1,i\}$.

So the linear map ${\mathbb{R}}^2\to \mathbb{C}$ sending $(a,b)$ to $a(1+i)+b(1-i)$ has trivial kernel; both spaces have dimension two by [L1], so [L5] makes it bijective and $(1+i,1-i)$ an ordered $\mathbb{R}$-basis of $\mathbb{C}$. Being the conjugate list of $1+i$, it is a normal basis, in agreement with [L6].

$(i,-i)$ is the conjugate list of $i$, since $\bar{i}=-i$, and its two members are distinct; but $1\cdot i+1\cdot (-i)=0$ is a vanishing combination with nonzero coefficients, so the list is not independent and by [L5] is not a basis. With steps 3.1 and 2.2 this establishes all three claims.

$t^2+t+1$ has no root in ${\mathbb{F}}_2$, its values at $0$ and $1$ both being $1$, so it is irreducible and $L$ is a field by [L1]; it is the minimal polynomial of $\alpha$, so $[L:{\mathbb{F}}_2]=2$ with ordered basis $(1,\alpha )$ by [L2], and $L$ has four elements $0,1,\alpha ,\alpha +1$.

By [L3] the extension $L\mathrm{/}{\mathbb{F}}_2$ is Galois with $\mathrm{Gal}(L\mathrm{/}{\mathbb{F}}_2)=\{\mathrm{i}\mathrm{d},\sigma \}$, where $\sigma (x)=x^2$.

The conjugate lists of the four elements are $(0,0)$, $(1,1)$, $(\alpha ,\alpha +1)$ and $(\alpha +1,\alpha )$, using ${\alpha }^2=\alpha +1$ and $(\alpha +1{)}^2={\alpha }^2+1=\alpha$. Their underlying sets are $\{0\}$, $\{1\}$ and $\{\alpha ,\alpha +1\}$.

The list $(1,\alpha )$ is an ${\mathbb{F}}_2$-basis of $L$ by step 1.1, but its underlying set $\{1,\alpha \}$ is none of the three sets in step 3.1, so it is not the conjugate list of any element and hence is not a normal basis by [L4]. The false claim therefore fails already for $L\mathrm{/}{\mathbb{F}}_2$.

What is true is the existential statement: some element of $L$ generates a normal basis, and $\alpha$ does, since $(\alpha ,\alpha +1)$ is a list of two distinct elements whose only vanishing ${\mathbb{F}}_2$-combinations are trivial, as $\alpha \mathrm{\ne }0$, $\alpha +1\mathrm{\ne }0$ and $\alpha +(\alpha +1)=1\mathrm{\ne }0$. That is the content of [L5], which asserts existence and never universality.