Exponent lifting for F -modules

Next, we shall show a series of simple propositions which will culminate (if this can be called a culmination) in a Carlitz analogue of the classical “lifting the exponent” theorem (see, e.g., [6, version with solutions (ancillary file), (12.68.8)]

for it).

Proposition 3.43. (a)The Fq-vector subspaceFF of F is a two-sided ideal of F.

(b)Let P ∈_Fq[T]. Then, CarlP ≡PmodFF.

Proof of Proposition 3.43. (a)First, we claim that

Fu∈ FF for everyu ∈ F. (78)

Proof of (78): Proposition 3.5(b)shows that theFq-module F is free with basis T^jFⁱ

i≥0, j≥0.

Let u ∈ F. We must prove the relation (78). Since this relation is Fq-linear in u (because FF is an Fq-vector subspace of F), we can WLOG assume that u belongs to the basis T^jFⁱ

i≥0, j≥0 of the Fq-module F. Assume this. Thus, u=T^jFⁱ for some i∈ _Nand j∈ N. Consider theseiand j. Now,

F u

|{z}

=T^jFⁱ

= FT^j

|{z}

=(^Tj)^qF

(by Proposition 3.6, applied toP=T^j)

Fⁱ =T^jq

FFⁱ

|{z}

=_Fⁱ⁺¹=_Fⁱ_F

=T^jq

Fⁱ

| {z }

∈F

F∈ FF.

This proves (78).

Now,

FF ={Fu | u∈ F } ⊆ FF (by (78)).

But it is clear thatFFis a left ideal ofF. Since we furthermore haveF F· F

| {z }

=FF ⊆FF

⊆ F F

|{z}⊆F

F ⊆ FF, we thus conclude that FF is a two-sided ideal of F. This proves Proposition 3.43(a).

(b) Proposition 3.43 (a) shows that FF is a two-sided ideal of F_{. Hence,} F/(FF) is a quotient ring of F, hence a quotient Fq-algebra of F. Let π de-note the canonical projection map F → F/(FF). Then, π is an Fq-algebra homomorphism (sinceF_/(FF) is a quotientFq-algebra ofF_).

But Carl(T) = F+T ≡ TmodFF (since F = 1

|{z}∈F

F ∈ FF). In other words, π(Carl(T)) = π(T) (since π is the canonical projection map F → F/(FF)).

Thus,

(π◦Carl) (T) = π(Carl(T)) = π



 T

|{z}

=_FincT(T)



=π(FincT(T))

= (π◦FincT) (T). (79)

But the three mapsπ, Carl and FincT areFq-algebra homomorphisms; hence, π◦Carl and π◦Finc_T are Fq-algebra homomorphisms as well. The two Fq -algebra homomorphismsπ◦Carl : Fq[T] → F/(FF) and π◦FincT : Fq[T] → F/(FF) are equal to each other on the generator T of the Fq-algebra Fq[T] (because of (79)). Therefore, these two homomorphisms must be identical. In other words,π◦Carl=π◦FincT.

Now,

π(CarlP) = (π◦Carl)

| {z }

=π◦Finc_T

(P) = (π◦Finc_T) (P) = π

















Finc_T(P)

| {z }

=P

(since we are regarding the map Finc_T as an inclusion)

















=_π(P).

In other words, CarlP ≡ PmodFF (since π is the canonical projection map F → F/(FF)). This proves Proposition 3.43(b).

Proposition 3.44. Let Abe an F-module. Let P∈ _Fq[T]. (a)We have FPA⊆P^qA.

(b)TheFq-vector subspace PAof Ais a leftF-submodule of A.

(c)Let kbe a positive integer. Then, FP^kA⊆P^k+1A.

(d)Let kbe a positive integer. Then, (CarlP)P^kA⊆P^k+1A.

Proof of Proposition 3.44. (a)Proposition 3.6 yieldsFP=P^qFinF. Hence, FP

|{z}

=P^qF

A = P^q FA

|{z}⊆A

⊆P^qA. Thus, Proposition 3.44 (a)is proven.

(b) Proposition 3.44 (a) yields FPA ⊆ P^q

Now, recall that theFq-algebraF is generated byFand T. From this, it is easy to derive the following fact: IfV is anFq-vector subspace of some leftF-module U satisfyingFV ⊆ V and TV ⊆ V, thenV is a leftF-submodule ofU. Applying this to U = A and V = PA, we conclude that PA is a left F-submodule of A (sinceFPA ⊆PAand TPA ⊆PA). Proposition 3.44(b)is thus shown.

(c)Proposition 3.44(a)(applied to P^k instead ofP) yields FP^kA⊆ P^kq (sincekis a positive

integer)

This establishes Proposition 3.44(c).

(d) Proposition 3.43 (b) yields CarlP ≡ PmodFF. In other words, CarlP− P ∈ FF. In other words, there exists some u ∈ F such that CarlP−P = uF.

Consider thisu.

Proposition 3.44 (b) (applied to P^k+1 instead of P) shows that the Fq-vector subspace P^k+1A of A is a left F-submodule of A. Hence, uP^k+1A ⊆ P^k+1A

This proves Proposition 3.44(d).

Proposition 3.45. Let A be an F-module. Let P ∈ _Fq[T]. Let k be a positive

assume thatP 6=0.

If degP =0, then the claim of Proposition 3.45(a)is true⁴⁰. Hence, we WLOG assume that degP6=0. Thus, degP≥_1.

Letd =degP. Then,d ≥1, so that F^d =FF^d−1.

But Proposition 3.44(b)(applied to P^k instead of P) shows that theFq-vector subspace P^kAof Ais a leftF-submodule of A. Hence,F ·P^kA ⊆P^kA.

Now, degP=d, so that

F^degPa−F^degPb =F^da−F^db = F^d

|{z}

=FF^d−1

(a−b)

| {z }

∈P^kA

∈ F F^d−1

| {z }

∈F

P^kA⊆ FF ·P^kA

| {z }

⊆P^kA

⊆FP^kA⊆P^k+1A (by Proposition 3.44(c)).

In other words,F^degPa≡ F^degPbmodP^k+1A. This proves Proposition 3.45 (a).

(b)We have

(CarlP)a−(CarlP)b = (_CarlP) (a−_b)

| {z }

∈P^kA

∈ (CarlP)P^kA⊆ _P^k+1_A

(by Proposition 3.44 (d)). In other words, (CarlP)a ≡ (CarlP)bmodP^k+1A.

This proves Proposition 3.45(b).

Corollary 3.46. Let A be an F-module. Let P ∈ _Fq[T]. Let k be a positive integer.

Let aand b be two elements of Asuch that a ≡_bmodP^k_A.

(a)We have F^deg(^{P`</sup>)<sub>a</sub> ≡F<sup>deg</sup>(<sup>P`})_bmodPk+`Afor every` ∈_N.

(b)We have Carl P^`

a≡ Carl P^`

bmodP<sup>k+`</sup>Afor every `∈_N.

Proof of Corollary 3.46. (a)We can prove Corollary 3.46 (a)by induction over`: Induction base: We have deg

P⁰

| {z }

=₁

= deg 1 = 0 and thus F^deg(^P0) = _F⁰ = 1.

Hence, F^deg(^P0)_a = 1a = a and similarly F^deg(^P0)_b = b. But a ≡ bmodP^kA.

Since k+₀ = k, this rewrites as a ≡ bmodP^k+0A. Now, F^deg(^P0)_a = a ≡ b = F^deg(^P0)_bmodPk+0A. In other words, Corollary 3.46 (a) holds for ` = 0. This completes the induction base.

Induction step: Let L∈ N. Assume that Corollary 3.46(a) holds for `= L. We must now prove that Corollary 3.46(a)holds for `=L+1.

We have assumed that Corollary 3.46(a) holds for` = L. In other words, we haveF^deg(^PL)_a ≡F^deg(^PL)_bmodPk+LA.

40Proof. Assume that degP = 0. Thus, the polynomialP is constant. SinceP 6= 0, this shows that the polynomial P is invertible in Fq[T]. Hence, P is invertible in F. Therefore, P^k+1 is also invertible in F. Hence, P^k+1A = A. But F^degPa ≡ F^degPbmodAis obviously true.

SinceP^k+1A= A, this rewrites asF^degPa≡ F^degPbmodP^k+1A. In other words, the claim of Proposition 3.45(a)is true; qed.

Butk is a positive integer, and hencek+Lis a positive integer. Hence, Propo-sition 3.45 (a) (applied to k+L, F^deg(^PL)_{a and F}^deg(^PL)_{b instead of k, a and b)} yields

F^degPF^deg(^PL)_a≡ F^degPF^deg(^PL)_bmodP^k+L+1_{A. (80)} Now, deg



P^L+1

| {z }

=PP^L



 = deg PP^L

= degP+deg P^L

. Hence, F^deg(^PL+1) = F^degP+deg(^PL) = F^degPF^deg(^PL). Therefore, (80) rewrites as follows:

F^deg(^PL+1)_a≡ F^deg(^PL+1)_bmodP^k+L+1_A.

In other words, Corollary 3.46 (a) holds for ` = L+1. This completes the induction step. The induction proof of Corollary 3.46(a)is thus finished.

(b)We can prove Corollary 3.46(b)by induction over `: Induction base: We have Carl

P⁰

| {z }

=1

= Carl 1 = 1 (since Carl is an Fq-algebra homomorphism). Hence, Carl P⁰

a=1a =aand similarly Carl P⁰ b=b.

But a ≡ bmodP^kA. Since k+0 = k, this rewrites as a ≡ bmodP^k+0A. Now, Carl P⁰

a = a ≡ b = Carl P⁰

bmodP^k+0A. In other words, Corollary 3.46(b)holds for `=0. This completes the induction base.

Induction step: Let L ∈N. Assume that Corollary 3.46(b)holds for `= L. We must now prove that Corollary 3.46(b)holds for `=L+1.

We have assumed that Corollary 3.46 (b)holds for ` = L. In other words, we have Carl P^L

a ≡ Carl P^L

bmodP^k+LA.

Butk is a positive integer, and hencek+Lis a positive integer. Hence, Propo-sition 3.45 (b) (applied to k+L, Carl P^L

a and Carl P^L

b instead of k, a andb) yields

(CarlP)Carl P^L

a≡(CarlP)Carl P^L

bmodP^k+L+1A. (81)

Now, Carl



P^L+1

| {z }

=PP^L



 =Carl PP^L

= (CarlP) Carl P^L

(since Carl is an Fq -algebra homomorphism). Thus, (81) rewrites as follows:

Carl

P^L+1

a≡Carl

P^L+1

bmodP^k+L+1A.

In other words, Corollary 3.46 (b) holds for ` = L+1. This completes the induction step. The induction proof of Corollary 3.46(b)is thus finished.

In order to state the last corollary in this section, we need a definition:

Definition 3.47. Let K be a field. Let π be a monic irreducible polynomial in K[T]. Let f be any polynomial in K[T]. Then, vπ(f) means the largest nonnegative integer m satisfying π^m | _f; this is set to be +∞ if f = 0. Thus, vπ(f) ∈_N∪ {+_∞}for each f.

We setP^+∞ =0 for eachP∈ K[T]. Thus,π^vπ(f) | f holds for each f ∈ K[T] (including the case when f =0).

Corollary 3.48. Let A be an F-module. Let N ∈ _Fq[T]. Let π be a monic irreducible polynomial inFq[T].

Let aand b be two elements of Asuch that a ≡bmodπA.

(a) We have F^degNa ≡ F^degNbmodπ^vπ(N)+1A. (Here, F^degN is understood to mean 0 when N =0.)

(b)We have(CarlN)a≡(CarlN)bmodπ^vπ(N)+1A.

Proof of Corollary 3.48. We have a ≡ bmodπA. In other words, a ≡ bmodπ¹A (sinceπ =π¹).

If N = 0, then Corollary 3.48 is easily seen to hold (since F^degN = 0 and Carl N

|{z}

=0

=Carl 0=0 in this case). Hence, we WLOG assume that N 6=0. Thus, vπ(N) ∈ _{N. Set} ` = vπ(N). Then, π<sup>`</sup> | N. In other words, there exists some polynomial M ∈_Fq[T] such that N =Mπ^`. Consider this M.

Proposition 3.44(b) (applied to P = π^{1+`</sup>) shows that theFq-vector subspace π<sup>1+`}Aof A is a leftF-submodule of A. Hence, F ·_π^{1+`</sup>A⊆<sub>π</sub><sup>1+`}A.

(a) From N = Mπ^{`</sup>, we obtain degN = deg Mπ<sup>`}

= degM+deg π^{`</sup> , so that F<sup>degN</sup> =F<sup>degM+deg</sup>(<sup>π`}) = F^degMF^deg(^π`)_.

Corollary 3.46(a)(applied to P=π and k=1) yields

F^deg(^{π`</sup>)<sub>a</sub>≡ F<sup>deg</sup>(<sup>π`})_bmodπ1+`A(sincea ≡bmodπ<sup>1</sup>A). In other words,F<sup>deg</sup>(<sup>π`</sup>)_a− F^deg(^{π`</sup>)<sub>b</sub> ∈ <sub>π</sub><sup>1+`}A. But

F^degNa−F^degNb

= F^degN

| {z }

=F^degMF^deg(^π`)

(a−b) = F^degM

| {z }

∈F

F^deg(^π`) (a−b)

| {z }

=F^deg(^{π`</sup>)<sub>a−F</sub><sup>deg</sup>(<sup>π`})_b∈

π^1+`A

∈ F ·π^{1+`</sup>A⊆π<sup>1+`}A.

In other words, F^degNa ≡ F^degNbmodπ<sup>1+`</sup>A. Since 1+ `

|{z}

=v_π(N)

=₁+v_π(N) = vπ(N) +1, this rewrites asF^degNa≡ F^degNbmodπ^vπ(N)+1A. This proves Corol-lary 3.48(a).

(b)From N = Mπ^{`</sup>, we obtain CarlN = Carl Mπ<sup>`}

= (CarlM) Carl π^` (since Carl is anFq-algebra homomorphism).

Corollary 3.46 (b) (applied to P = π and k = 1) yields Carl π^{`</sup> a ≡ Carl π<sup>`}

bmodπ^{1+`</sup>A(since a≡bmodπ<sup>1</sup>A). In other words, Carl π<sup>`} a−

Carl π^`

b ∈ _π^1+`A. But (CarlN)a−(CarlN)b

= (CarlN)

| {z }

=(CarlM)(^Carl(π^`))

(a−b) = (CarlM)

| {z }

∈F

Carl

π^`

(a−b)

| {z }

=(^Carl(^{π`</sup>))<sup>a−</sup>(<sup>Carl</sup>(<sup>π`}))^b∈π1+`A

∈ F ·π^{1+`</sup>A⊆π<sup>1+`}A.

In other words, (CarlN)a ≡ (CarlN)bmodπ<sup>1+`</sup>A. Since 1+ `

|{z}

=v_π(N)

= 1+ vπ(N) = vπ(N) +1, this rewrites as (CarlN)a ≡ (CarlN)bmodπ^vπ(N)+1A.

This proves Corollary 3.48(b).

Each of the two parts of Corollary 3.48 can be viewed as an analogue of the classical “exponent lifting lemma” [6, version with solutions (ancillary file), (12.68.8)].

Im Dokument Do the symmetric functions have a function-field analogue? (Seite 73-79)