Section III.5.2

Finite fields

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Problem III.5.2.1

Prove that for any d∣n,d<nd \mid n, d<n, the relation Xn−1=(Xd−1)Φn(X)hd(X)X^{n}-1=\left(X^{d}-1\right) \Phi_{n}(X) h_{d}(X) holds, where hd∈Z[X]h_{d} \in \mathbb {Z}{}[X].

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Problem III.5.2.2

Let qq be a positive integer >1>1. By (10), Φn(q)∈Z\Phi_{n}(q) \in \mathbb {Z}. Show that Φn(q)∣(q−1)⟹n=1\Phi_{n}(q) \mid (q-1) \Longrightarrow n=1.

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Problem III.5.2.3

Verify that the cyclotomic polynomial

Φ15(X)=X8−X7+X5−X4+X3−X+1 \Phi _{15}(X)=X^{8}-X^{7}+X^{5}-X^{4}+X^{3}-X+1

considered over the field F2\mathbb {F}_{2}, is the product of two irreducible polynomials X4+X3+1X^{4}+X^{3}+1 and X4+X+1X^{4}+X+1. Using this fact, prove the irreducibility of Φ15(X)\Phi_{15}(X) over Q\mathbb {Q} (cf. Exercise 11 from [ВА I, Ch. 6, § 1]).

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Problem III.5.2.4

Verify the following properties of cyclotomic polynomials.

If pp is a prime number and p∣np \mid n, then Φpn(X)=Φn(Xp)\Phi_{p n}(X)=\Phi_{n}\left(X^{p}\right); if, however, p∤np \nmid n, then Φpn(X)=Φn(Xp)/Φn(X)\Phi_{p n}(X)=\Phi_{n}\left(X^{p}\right) / \Phi_{n}(X).

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Problem III.5.2.5

Starting from the chain of natural inclusions

GF(p)⊂GF(p2!)⊂GF(p3!)⊂…, G F(p) \subset G F\left(p^{2 !}\right) \subset G F\left(p^{3 !}\right) \subset \ldots ,

introduce the so-called limit field Ωp=GF(p∞)\Omega_{p}=G F\left(p^{\infty }\right), setting

α∈Ωp⟺{α∈GF(pn!) for sufficiently large n}.  \alpha \in \Omega _{p} \Longleftrightarrow \left\{ \alpha \in G F\left(p^{n !}\right) \text{ for sufficiently large } n\right\} \text{. }

Relying on the basic properties of finite fields, prove that Ωp\Omega_{p} is an algebraically closed field. In this way one obtains, together with the field of complex numbers C\mathbb {C}, examples of algebraically closed fields of any characteristic.

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Problem III.5.2.6

Let q=pnq=p^{n}. Show that for p=2p=2 all elements of the field Fq\mathbb {F}_{q} are squares, while for p>2p>2 the squares of the group Fq∗\mathbb {F}_{q}^{*} form in it a subgroup Fq∗2\mathbb {F}_{q}^{* 2} of index 2, with Fq∗2=Ker⁡(t↦t(q−1)/2)\mathbb {F}_{q}^{* 2}=\operatorname {Ker}\left(t \mapsto t^{(q-1) / 2}\right).

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Problem III.5.2.7

Let Fq\mathbb {F}_{q} be a finite field with an odd number q=pnq=p^{n} of elements. If qq is not equal to 3 or 5, then on the "circle" x2+y2=1x^{2}+y^{2}=1 there is a point with coordinates x,y∈Fq∗x, y \in \mathbb {F}_{q}^{*}. Prove this statement for p>5p>5.

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Problem III.5.2.8

Is every primitive element of the field Fq\mathbb {F}_{q} a generator of the multiplicative group Fq∗\mathbb {F}_{q}^{*}?

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Problem III.5.2.9

Let A(q)=Ass⁡F(X1,…,Xq)A(q)=\operatorname {Ass}_{F}\left(X_{1}, \ldots , X_{q}\right) be the free associative algebra over a field FF, generated by qq free generators — noncommuting variables X1,…,XqX_{1}, \ldots , X_{q}. Setting

Am(q)=⟨Xi1Xi2…Xim∣1⩽ij⩽q⟩F,dim⁡Am(q)=qm A_{m}(q)= \left\langle X_{i_{1}} X_{i_{2}} \ldots X_{i_{m}} \mid 1 \leqslant i_{j} \leqslant q\right\rangle _{F}, \quad \operatorname {dim} A_{m}(q)=q^{m}

we note that A(q)A(q) is a graded algebra:

A(q)=F⋅1⊕A1(q)⊕A2(q)⊕A3(q)⊕… A(q)=F \cdot 1 \oplus A_{1}(q) \oplus A_{2}(q) \oplus A_{3}(q) \oplus \ldots

A(q)A(q) contains the free Lie algebra L(q)=Lie⁡(X1,…,Xq)L(q)=\operatorname {Lie}\left(X_{1}, \ldots , X_{q}\right) with the same free generators and the commutator operation [UV]=UV−VU[U V]=U V-V U. The algebra L(q)L(q) is also graded:

L(q)=L1(q)⊕L2(q)⊕L3(q)⊕… L(q)=L_{1}(q) \oplus L_{2}(q) \oplus L_{3}(q) \oplus \ldots

where

L1(q)=⟨X1,…,Xq⟩F,L2(q)=⟨[Xi,Xj]∣i<j⟩F,… L_{1}(q)= \left\langle X_{1}, \ldots , X_{q}\right\rangle _{F}, \quad L_{2}(q)= \left\langle \left[X_{i}, X_{j}\right] \mid i<j\right\rangle _{F}, \quad \ldots

Using the Jacobi identity, verify that

L3(q)=⟨[[Xi,Xj],Xk]⟩,i<j,k⩽j;dim⁡L3(q)=13(q3−q) L_{3}(q)= \left\langle \left[\left[X_{i}, X_{j}\right], X_{k}\right]\right\rangle , \quad i<j, k \leqslant j ; \quad \operatorname {dim} L_{3}(q)=\frac{1}{3}\left(q^{3}-q\right)

In fact, the general Witt formula holds

dim⁡Lm(q)=Ψm(q)=1m∑d∣mμ(md)qd \operatorname {dim} L_{m}(q)=\Psi _{m}(q)=\frac{1}{m} \sum _{d \mid m} \mu \left(\frac{m}{d}\right) q^{d}

coinciding exactly with formula (12). The essential difference is only that in (12') qq is an arbitrary natural number, while in (12) it is a power of a prime number.

The number of necklaces from Problem 2 in the preamble to Ch. 3 is expressed by the same formula.

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