3. Proofs 20
3.4. q-polynomials from subspaces
We shall now see a classical way to constructq-polynomials.
Definition 3.16. Let Abe a commutativeFq-algebra. For every finite subsetV of A, let fV be the polynomial ∏
v∈V
(X+v) ∈ A[X].
The following result is a consequence of [15, (7.7)] (and also appears in [3, Theorem A.1 2)] in the particular case when Ais an integral domain):
Theorem 3.17. Let Abe a commutativeFq-algebra. LetV be a finiteFq-vector subspace of A. Then, fV is aq-polynomial.
We shall prove Theorem 3.17 following an idea that appears in [15, proof of (7.15)]; but first, let us slightly generalize it:
22Proof of (34): Letb ∈ B. From f = ∑
n∈NanXqn, we obtain f(b) = ∑
n∈Nanbqn. Comparing this with
bf(b) =
∑
n∈N
anFrobn(b)
| {z }
=bqn (by (33))
since bf =
∑
n∈N
anFrobn
!
=
∑
n∈N
anbqn, this yields f(b) = bf(b), qed.
Definition 3.18. Let Abe a commutativeFq-algebra. For every finite setVand every map ϕ: V → A, we let fV,ϕ be the polynomial ∏
v∈V
(X+ϕ(v))∈ A[X]. Theorem 3.19. Let Abe a commutativeFq-algebra. LetV be a finiteFq-vector space, and let ϕ: V → Abe an Fq-linear map. Then, fV,ϕ is aq-polynomial.
Theorem 3.19 is not significantly more general than Theorem 3.17 (it is easily derived from the latter), but this little generality helps in proving it. The proof will need the following lemmas:
Lemma 3.20. Let A be a commutative Fq-algebra. LetV and W be two finite Fq-vector spaces. Let ϕ : V → A and ψ : W → A be two Fq-linear maps.
Assume that fW,ψ is a q-polynomial. Leth : A → Abe an Fq-linear map such that every a∈ Asatisfies
h(a) = fW,ψ(a). (35) Let χ : V⊕W → A be the Fq-linear map which sends every (v,w) ∈ V⊕W to ϕ(v) +ψ(w) ∈ A. Then,
fV⊕W,χ = fV,h◦ϕ◦ fW,ψ in A[X]. Proof of Lemma 3.20. The definition of fW,ψ yields
fW,ψ =
∏
v∈W
(X+ψ(v)) =
∏
w∈W
(X+ψ(w)) (36) (here, we renamed the summation indexvasw).
Fix somev∈ V. If we substituteX+ϕ(v) forXon both sides of (36), then we obtain
fW,ψ(X+ϕ(v)) =
∏
w∈W
(X+ϕ(v) +ψ(w)). (37) We have assumed that fW,ψis aq-polynomial. In other words, fW,ψ ∈ A[X]q−lin. Hence, Proposition 3.14 (applied toB= A[X]and f = fW,ψ) shows that the map A[X] → A[X], b 7→ fW,ψ(b) is Fq-linear. Hence, fW,ψ(x1+x2) = fW,ψ(x1) + fW,ψ(x2) for every x1,x2 ∈ A[X]. Applying this to x1 = X and x2 = ϕ(v), we obtain
fW,ψ(X+ϕ(v)) = fW,ψ(X)
| {z }
=fW,ψ
+ fW,ψ(ϕ(v))
| {z }
=h(ϕ(v))
(because (35) (applied toa=ϕ(v)) yieldsh(ϕ(v))=fW,ψ(ϕ(v)))
= fW,ψ+h(ϕ(v))
| {z }
=(h◦ϕ)(v)
= fW,ψ+ (h◦ϕ) (v).
Comparing this with (37), we obtain
w
∏
∈W(X+ϕ(v) +ψ(w)) = fW,ψ+ (h◦ϕ) (v). (38) Let us now forget that we fixed v. We thus have shown proven the equality (38) for allv∈ V.
The definition of fV,h◦ϕ yields fV,h◦ϕ =
∏
v∈V
(X+ (h◦ϕ) (v)).
Substituting fW,ψ forX on both sides of this equality, we obtain fV,h◦ϕ fW,ψ
=
∏
v∈V
fW,ψ+ (h◦ϕ) (v). (39) The definition of fV⊕W,χ yields
fV⊕W,χ =
∏
v∈V⊕W
(X+χ(v)) =
∏
(v,w)∈V⊕W
| {z }
=∏
v∈V ∏
w∈W
X+ χ(v,w)
| {z }
=ϕ(v)+ψ(w) (by the definition ofχ)
(here, we renamed the indexv as (v,w) in the product)
=
∏
v∈V
∏
w∈W
(X+ϕ(v) +ψ(w))
| {z }
=fW,ψ+(h◦ϕ)(v) (by (38))
=
∏
v∈V
fW,ψ+ (h◦ϕ) (v)
= fV,h◦ϕ fW,ψ
(by (39))
= fV,h◦ϕ◦ fW,ψ. This proves Lemma 3.20.
Lemma 3.21. We have
∏
λ∈Fq
(X−λY) = Xq−XYq−1 (40) in the polynomial ringFq[X,Y].
Proof of Lemma 3.21. It is well-known that
∏
λ∈Fq
(X−λ) = Xq−X (41)
in the polynomial ringFq[X] 23.
Now, consider the element X/Y in the quotient field Fq(X,Y) of the ring Fq[X,Y]. Substituting this elementX/Yfor X in (41), we obtain
∏
λ∈Fq
(X/Y−λ) = (X/Y)q−X/Y.
Multiplying this equality byYq, we obtain Yq
∏
23Let us give aproof of (41)for the sake of completeness:
The polynomial ∏
λ∈Fq
(X−λ)is a product of Fq
=qmonic polynomials of degree 1. Thus, it is a monic polynomial of degree q. Hence, both polynomials ∏
λ∈Fq
(X−λ) and Xq−X are monic polynomials of degreeq. Their difference ∏
λ∈Fq
(X−λ)−(Xq−X) therefore is a polynomial of degree<q(since the subtraction causes their leading terms to cancel).
On the other hand, everyµ∈Fqsatisfies
λ∈
∏
Fq(µ−λ)
| {z }
(since one of the factors of=0 this product isµ−µ=0)
ButFqis a field. Hence, any polynomial inFq[X]whose degree is smaller than its number of roots must be the zero polynomial. The polynomial ∏
λ∈Fq
(X−λ)−(Xq−X) is such a polynomial (since its degree is < q, but it has at least q roots), and thus must be the zero polynomial. In other words, ∏
λ∈Fq
(X−λ) = (Xq−X). This proves (41).
Lemma 3.22. Let Abe a commutativeFq-algebra. LetV be a one-dimensional Fq-vector space. Let ϕ : V → A be an Fq-linear map. Let e be a nonzero element ofV. Then, fV,ϕ = Xq−(ϕ(e))q−1X.
Proof of Lemma 3.22. The element −e ofV is nonzero (sincee is nonzero).
The Fq-vector space V is one-dimensional, and thus any nonzero element of Vforms a basis ofV. Thus, −eforms a basis ofV(since−eis a nonzero element of V). In other words, the map Fq → V, λ 7→ λ(−e) is a bijection. Now, the definition of fV,ϕ yields
fV,ϕ =
∏
Proof of Theorem 3.19. We shall prove Theorem 3.19 by induction over dimV:
Induction base: Theorem 3.19 holds in the case when dimV = 0 24. This completes the induction base.
Induction step:Let N ∈N. Assume (as the induction hypothesis) that Theorem 3.19 holds in the case when dimV = N. We need to show that Theorem 3.19 holds in the case when dimV = N+1.
Consider the setting of Theorem 3.19, and assume that dimV = N+1. Thus, dimV = N+1 > 0. Hence, V contains a nonzero element e. Consider this e.
Let U be the Fq-vector subspaceFqe of V; thus, dimU =1 (since eis nonzero).
24Proof.Consider the setting of Theorem 3.19, and assume that dimV=0. From dimV=0, we obtainV=0. The definition of fV,ϕyields
fV,ϕ=
∏
Thus, fV,ϕis aq-polynomial (sinceXis aq-polynomial). Thus, Theorem 3.19 is proven in the case when dimV=0.
Pick any complementW to the subspace U of V (such a complement exists by one of the basic theorems of linear algebra). Then, W is an Fq-vector subspace of V satisfying U⊕W = V. We shall identify V with the external direct sum ofU and W (that is, we shall identify each element v ofV with the unique pair (u,w) ∈ U×W satisfying v = u+w). Thus, the Fq-linear map ϕ : V → A can be regarded as anFq-linear map ϕ: U⊕W → A.
Define two Fq-linear maps γ : U → A and ψ : W → A by γ = ϕ |U and ψ= ϕ|W. Then, theFq-linear map ϕ: U⊕W → Asends every (v,w) ∈ U⊕W toγ(v) +ψ(w) 25.
From V = U⊕W, we obtain dimV = dimU +dimW, so that dimW = dimV
| {z }
=N+1
−dimU
| {z }
=1
= N+1−1= N. Thus, (according to the induction hypothesis) Theorem 3.19 can be applied toW andψinstead of V and ϕ. As a consequence, we obtain that fW,ψ is a q-polynomial. In other words, fW,ψ ∈ A[X]q−lin. Thus, Proposition 3.14 (applied to f = fW,ψ and B = A) shows that the map A → A, b 7→ fW,ψ(b) is Fq-linear. Let us denote this map by h. Thus, h is the map A→ A, b 7→ fW,ψ(b), and isFq-linear. Every a∈ Asatisfies h(a) = fW,ψ(a)(by the definition of h).
Now, Lemma 3.20 (applied to U, γ and ϕ instead of V, ϕ and χ) shows that fU⊕W,ϕ = fU,h◦γ◦ fW,ψ in A[X].
But theFq-vector space U is one-dimensional (since dimU = 1) and contains the nonzero vectore (since U = Fqe ⊇ e). Thus, Lemma 3.22 (applied to U and h◦γ instead of V and ϕ) shows that fU,h◦γ = Xq−((h◦γ) (e))q−1X. This is clearly a q-polynomial (since ((h◦γ) (e))q−1 is just a coefficient in A). In other words, fU,h◦γ ∈ A[X]q−lin.
Proposition 3.10 (b) shows that A[X]q−lin is a submonoid of the monoid (A[X],◦). Hence, A[X]q−lin is closed under the binary operation ◦. There-fore, fU,h◦γ◦ fW,ψ ∈ A[X]q−lin (since fU,h◦γ ∈ A[X]q−lin and fW,ψ ∈ A[X]q−lin).
But V = U⊕W, so that fV,ϕ = fU⊕W,ϕ = fU,h◦γ ◦ fW,ψ ∈ A[X]q−lin. In other words, fV,ϕ is a q-polynomial. Thus, Theorem 3.19 is proven in the case when dimV =N+1. This completes the induction step.
The proof of Theorem 3.19 is thus complete.
As a consequence of Theorem 3.19, we can remove one unneeded assumption from Lemma 3.20:
25Proof.Let(v,w)∈U⊕W. We must show that ϕ(v,w) =γ(v) +ψ(w).
We havev∈U, and thusγ(v) =ϕ(v)(sinceγ= ϕ|U). We havew∈W, and thusψ(w) = ϕ(w) (since ψ = ϕ |W). The map ϕ is Fq-linear, and thus ϕ(v+w) = ϕ(v)
| {z }
=γ(v)
+ϕ(w)
| {z }
=ψ(w)
= γ(v) +ψ(w). But recall that we are identifying (v,w) ∈ U⊕W with v+w ∈ V. Thus, ϕ(v,w) =ϕ(v+w) =γ(v) +ψ(w), qed.
Corollary 3.23. Let Abe a commutative Fq-algebra. Let V and W be two Fq -vector spaces. Let ϕ : V → A and ψ : W → A be two Fq-linear maps. Let h : A→ Abe anFq-linear map such that everya ∈ Asatisfiesh(a) = fW,ψ(a). Let χ : V⊕W → A be the Fq-linear map which sends every (v,w) ∈ V⊕W to ϕ(v) +ψ(w) ∈ A. Then,
fV⊕W,χ = fV,h◦ϕ◦ fW,ψ in A[X].
Proof of Corollary 3.23. Theorem 3.19 (applied to W and ψ instead of V and ϕ) shows that fW,ψ is a q-polynomial. Thus, Lemma 3.20 shows that fV⊕W,χ =
fV,h◦ϕ◦ fW,ψ in A[X]. This proves Corollary 3.23.
Let us finally derive Theorem 3.17 from Theorem 3.19:
Proof of Theorem 3.17. Let ι be the canonical inclusion map V → A. Thus, ι is an Fq-linear map. Hence, Theorem 3.19 (applied to ϕ = ι) shows that fV,ι is a q-polynomial. But the definition of fV,ι shows that
fV,ι=
∏
v∈V
X+ ι(v)
|{z}=v (sinceιis an inclusion map)
=
∏
v∈V
(X+v) = fV
(since this is how fV is defined). Thus, fV is a q-polynomial (since fV,ι is a q-polynomial). This proves Theorem 3.17.