Basic concepts: Field extensions, primitive polynomials.
The Frobenius automorphism, the Galois group, the trace.
A general construction idea for good codes over some field Fq is to first construct a code over a larger field Fqr and then to go down in some way from the large field to the small field. The most fruitful methods of “going down” are the trace codes and the subfield codes. Before we can discuss these methods, we need to know more about finite fields, in particula,r about pairs of finite fields contained in one another.
We saw that, for every primepand every natural numberr,there is a field Fpr of orderpr. It can be constructed as an extension of the prime fieldFp, using an irreducible polynomial of degreerwith coefficients inFp.In fact, we chose not to give all the details, as this would have led us too deeply into field theory. We chose to accept without proof that all the required irreducible polynomials exist over finite fields (not over all fields: there is, for example, no irreducible polynomial of odd degree over the reals) and also that, for each orderpr, there is only one field.
Consider fieldsFpsandFpr of the same characteristicp,wheres≤r.When is Fps ⊂ Fpr? If this is the case, then the larger field Fpr is in particular a vector space over Fps (the field axioms are stronger than the vector space axioms), of some dimensiond.Thenpr= (ps)d=psd.It follows thatsdivides r, and the dimension isd=r/s.If, on the other hand,s|r,we can choose an irreducible polynomial of degreed=r/swith coefficients inFps and construct Fpr as an extension field of Fps,by the method from Section 3.2 (we believe that these irreducible polynomials always exist).
135
This shows the following:
12.1 Proposition. We haveFps⊂Fpr if and only if sdivides r.
In fact, there are good and not so good irreducible polynomials.
12.2 Definition. Letf(X)∈Fq[X]be irreducible of degreer,andǫ∈Fqr the element corresponding to X.We say that f(X)is a primitive polynomial if 1 = ǫ0, ǫ, . . . , ǫqr−2 are different, equivalently if the powers of ǫ are all nonzero elements of Fqr. An element whose powers run through all nonzero field elements is called aprimitive element.
It can be shown that primitive polynomials always exist. They are also known as maximum exponent polynomials. In practice one always works with primitive polynomials. Should we come across an irreducible polynomial, which is not primitive, we will throw it away and replace it with a primitive polynomial.
12.3 Example. f(X) =X2+ 1 certainly is an irreducible polynomial over F3. In fact, as f(0) = 1, f(1) = f(2) = 2, it has no roots and therefore no linear factor. However, we will not use the polynomial to describe F9. Let ǫ=X mod f(X). Then ǫ2=−1 = 2 and therefore ǫ4 = 1. This shows that f(X)is not primitive.
The existence of primitive elements has important consequences. Letǫ∈Fq be primitive. Then ǫq−1 = 1. Each 06=x∈Fq has the formx=ǫi for some i.It follows thatxq−1=ǫi(q−1)= 1i= 1 andxq =xfor allx∈Fq (including x= 0).
12.4 Theorem. Each nonzero element x ∈ Fq satisfies xq−1 = 1. Each element x∈Fq satisfies xq =x. An element x∈Fqr is inFq if and only if xq=x.
PROOF Only the last statement still needs a proof. We know already that xq=xfor allx∈Fq.Consider the polynomialXq−X as a polynomial with coefficients inFqr.It has degreeqand we knowqof its roots, the elements of Fq.A polynomial of degreeqcannot have more thanqroots.
LetF be a finite field of characteristicp.The mappingx7→φ(x) =xphas surprising properties. It is known as the Frobenius automorphism. The multiplicative structure ofF is respected: φ(xy) =φ(x)φ(y). What happens ifφis applied to a sum? The binomial formula
(x+y)p= Xp
i=0
p i
xiyp−i
is of course valid in all fields. The decisive observation is that the integers pi are multiples of pexcept for i= 0 andi=p.In our field of characteristic p,
these coefficients are therefore equal to 0. This means that all but two terms of the sum vanish, and we obtain the simple formula
(x+y)p=xp+yp in fields of characteristicp.
Some call this the freshman’s dream. It means that φ respects both the multiplicative and the additive field structure. It is clear thatφis a one-to-one mapping.
12.5 Proposition. For each finite fieldF of characteristicp,theFrobenius automorphism φ(x) =xp is a field automorphism. In other words, φ is a one-to-one mapping satisfying
φ(x+y) =φ(x) +φ(y) andφ(xy) =φ(x)φ(y) for allx, y ∈F.
It follows from Theorem 12.4 that φ acts trivially on Fp : it maps each element of the prime field to itself.
We continue to consider a field extensionFq ⊂Fqr =F, whereq=pf.As φis a field automorphism ofF,the same is true of thef-th power ofφ.This is the mapping x7→ xq. By Theorem 12.4 it acts trivially on Fq. Repeated application yields automorphisms x 7→ xqi. Theorem 12.4 shows that, for i=r, the identity automorphism ofF is obtained.
12.6 Definition. Let gi be the field automorphism of F =Fqr defined by gi(x) =xqi, i= 0,1, . . . r−1.
Call G={g0, g1, . . . , gr−1} theGalois group ofFqr overFq.
In fact,g0is the identity,giis thei-th power ofg1(in the sense of applying g1 i times) and gr =g0. In particular, Gis closed under multiplication. It consists of the powers of g1. The inverse of g1 is gr−1, and in general the inverse ofgi is gr−i. Groups which consist of the powers of one element are known as cyclic groups. The Galois group is cyclic of orderr.The term that Gis the Galois groupoverFq reflects the fact that each element ofGmaps each element of the ground fieldFq to itself.
The Galois group allows us to go down from the extension field F to the ground field Fq, as follows. Let x∈ F. Consider the images of xunder the elements of the Galois group and add up:
y =g0(x) +g1(x) +· · ·+gr−1(x).
Applyg1 toy:
g1(y) =yq =g1(x) +g2(x) +· · ·+gr(x).
However, gr = g0, and so we are adding up the same terms as in the sum definingy.This showsyq =y.By Theorem 12.4 we havey∈Fq.The mapping x7→y is known as thetrace fromF =Fqr toFq.
12.7 Definition. Let Fq ⊂ Fqr = F be an extension of finite fields. The trace
tr=trqr|q :F →Fq is defined by
tr(x) =x+g1(x) +· · ·+gr−1(x) =x+xq+xq2+· · ·+xqr−1. It follows from the definition of the trace that it is additive (tr(x1+x2) = tr(x1) +tr(x2)) and Fq-linear (tr(λx) = λtr(x) for λ ∈ Fq). In particular, tr is a linear mapping from the r-dimensional vector space F to the one- dimensional vector spaceFq (linear overFq).
Let us check the trace in our favorite finite extension fields.
12.8 Example. Let tr=tr4|2:F4→F2.We have tr(0) =tr(1) = 0and tr(ω) =ω+ω2= 1, likewise tr(ω2) = 1.
12.9 Example. Let tr =tr8|2 : F8→ F2. As always tr(0) = 0, but tr(1) = 3 = 1. Also
tr(ǫ) =ǫ+ǫ2+ǫ4=ǫ(1 +ǫ+ǫ3) =ǫ(ǫ5+ǫ3) =ǫ4(1 +ǫ2) = 1 (miraculously). The elements of trace 0 are 0, ǫ3, ǫ5, ǫ6; the remaining four elements have trace1.
1. Fps ⊂Fpr if and only if sdividesr.
2. An elementǫ∈F isprimitiveif every nonzero element ofF is a power ofǫ.
3. Every finite field has primitive elements.
4. An element ofFqr is inFq if and only if it is a root of the polynomialXq−X.
5. The Frobenius automorphismφ:Fpr →Fpr is defined by φ(x) =xp.It is a field automorphism and fixes precisely the elements ofFp.
6. TheGalois groupGofFqr overFq is cyclic of orderr.
It consists of the powers ofg1,whereg1(x) =xq is a power of the Frobenius automorphism.
7. Thetrace tr:Fqr →Fq is defined by tr(x) =P
g∈Gg(x) (Gis the Galois group).
8. The trace is anFq-linear mapping.
Exercises 12.1
12.1.1. What is the dimension of F212 as a vector space overF8?
12.1.2. Show that every irreducible polynomial of degree 5 over F2 must be primitive.
12.1.3. Letǫbe a primitive element of Fq, whereq is odd. How can you read off from the exponent j if ǫj is a square in Fq?
How many nonzero squares are there in Fq? 12.1.4. The normN fromFqr toFq is defined by
N(x) =x·g1(x)· · · · ·gr−1(x) = Y
g∈G
g(x).
Prove that indeedN(x)∈Fq for every x∈Fqr.
12.1.5. Let N : Fq2 → Fq be the norm. How many elements x ∈Fq2 have N(x) = 1?
12.1.6. Show that there are exactly qr−1 elements in Fqr whose trace is 0.
Continue to show that thiskernelker(tr)of the trace is an(r−1)-dimensional subspace of F whenF is considered as a vector space overFq.
12.1.7. Determine the kernelker(tr)in the cases tr:F4→F2 and tr:F8→F2.
12.1.8. Let tr : Fqr → Fq. Under which conditions on q, r is it true that Fq ⊆ker(tr)? When do we haveker(tr) =Fq?
12.1.9. Determine all values tr(x)for tr:F9→F3.
12.1.10. Let α ∈ Fqr. We studied the minimal polynomial in a series of exercises in Section 3.2.
Prove that αandαq have the same minimal polynomial.
12.1.11. Provetr(αq) =tr(α), wheretr:Fqr →Fq.
12.1.12. Let α∈Fqr.Show that Fq(α) =Fqr if and only if the imagesg(α) under the elementsg∈Gare pairwise different.
12.1.13. Why is the trace called the trace? The reason is that tr(α) is the trace of the Fq-linear mapping : Fqr →Fqr defined by multiplication withα.
We want to prove this in a special case.
Let F =Fqr =Fq(α). Prove that the Fq-linear mapping : F →F defined by x7→αx has tracetr(α).
12.1.14. Show thatX3−X−1 is an irreducible polynomial overF3. Decide if it is primitive.
12.1.15. Show thatX3−X2+ 1is an irreducible polynomial overF3.Use it to constructF27.Determine ker(tr), wheretr is the trace down toF3. 12.1.16. Prove thetransitivity of the trace:
if Fq⊂L=Fqs ⊂F =Fqr, then
trF|Fq(x) =trL|Fq(trF|L(x))for allx∈F.