Image of End under reduction

Let K be a finite extension of F and let ϕ : A → K[τ] be a Drinfeld module over K whose characteristic morphism is the inclusion A ⊆ K. We prepare for the proof of Proposition 4.8.9, which will be an important tool for computing the endomorphism ring.

Letv be a place ofK and suppose thatϕis defined overO_v. Letp₀ be the characteristic ideal and h the height of ϕ_v. By Proposition 4.7.4, we have a canonical embedding End_K(ϕ)→End_kv(ϕ_v). Let K denote an algebraic closure of K and let O_v denote the integral closure of O_v in K. Fix a maximal ideal m_v of O_v lying over m_v. The residue field k_v := O_v/m_v is naturally an algebraic closure of k_v. We let R : O_v → k_v be the quotient map. This gives a commutative diagram

O_{v _} ^//

k_{v _}

O_v ^//k_v

In fact, this is a diagram of A-modules, where A acts via ϕ on the objects on the left half and viaϕ_v on those of the right half of the diagram.

By Proposition 4.7.4, for any nonzerof ∈End_K(ϕ) the zeros off inK are integral over O_v and therefore lie in O_v.

Lemma 4.8.1. The restriction of R to

Ker_Kf →Ker_kvfv

is surjective. In the case f =ϕ_a, with a 6∈p₀ it is an isomorphism.

Proof. The factorization inO_v[X]⊆K[X] into linear factors reduces to a factorization ink_v[X]. This shows surjectivity. Iff =ϕ_a anda6∈p₀, then ord_τϕ_v,a = 0, which means ϕ_v,a is separable. It follows that ϕ_a and ϕ_v,a have the same number of zeros, which implies that the restriction ofR is bijective, since we already know it is surjective.

Lemma 4.8.2. For any maximal ideal p of A and any n ≥0, the restriction ofR gives

Suppose first that for some b ∈ A we have pⁿ = (b). Then surjectivity follows from Lemma 4.8.1 for f = ϕ_b. In general, we can find N ≥ n, such that p^N = (b) for some

We have already shown that the top arrow is a surjection. The arrow on the right is surjective by Proposition 2.2.15, (b). It follows that the bottom arrow is also surjective, which is what we wanted to show.

Now (a) directly follows from the surjectivity since domain and range of the map in consideration are both isomorphic to (A/pⁿ)^⊕r by Proposition 2.2.15, (a) and therefore have the same finite cardinality.

It is left to show (b). Again by Proposition 2.2.15,the restriction of R is up to isomor-phisms a homomorphism of A/pⁿ₀-modules of the form

(A/pⁿ₀)^⊕r →(A/pⁿ₀)^⊕(r−h).

We already know it is surjective. Since the image is free, it is split surjective. The kernel is of the formLk

Comparing the submodules of both sides, which are annihilated byp shows that k =h.

Next we compare the lengths of both sides of the equality and find nr =n1+· · ·+nh+n(r−h)≤nh+n(r−h) = nr.

The inequality here cannot be strict. Since all n_i are bounded above by n, they must all be equal to n. This shows the kernel is of the desired form.

Now let g ∈ End_k(ϕ) be an endomorphism whose image g_v under reduction can be written as a product g_v = ϕ_v,ah_v = h_vϕ_v,a for some a ∈ A, and h_v ∈ End_kv(ϕ_v) (we donot assume that h_v is actually the reduction of an element of End_K(ϕ)). Let p be a maximal ideal ofA. For anyA-moduleM let_pM denote thep-power torsion submodule ofM, which is made up of those elements of M that are annihilated by a power of p.

Lemma 4.8.3. If p6=p₀, we have and Ker_Kg are A-submodules of Ker_Kϕ_b, and hence their p-power torsion submodules are contained in _p(Ker_Kϕ_b) = ϕ[pⁿ](K), where n = ord_p(b). By Lemma 4.8.2, the restriction ofR to ϕ[pⁿ](K) is injective. Therefore the inclusion (4.8.4) is equivalent to

R

p(Ker_Kϕ_a)

⊆ R

p(Ker_Kg)

. (4.8.6)

We claim this is exactly inclusion (4.8.5), which we know to be true. Showing this finishes the proof. It is generally true that for a torsionA-module M and anA-module homomorphism θ : M → N, we have _pθ(M) =θ(_pM). This follows from the fact that any torsion module decomposes as a direct sum of its prime power torsion modules for the different prime ideals of A. Now we combine this with Lemma 4.8.1 and find that for any endomorphism f ∈End_K(ϕ), we have

It remains to use this on both sides of (4.8.6).

Proposition 4.8.7.The torsion submodule of the cokernel of the embeddingEnd_K(ϕ),→ End_kv(ϕ_v) is annihilated by a power of p₀.

By the Chinese remainder theorem it is enough to check that the inclusion holds for all p-power torsion parts where p runs through the maximal ideals of A. For p 6= p₀, this follows directly from Lemma 4.8.3, sinceϕ_v,a divides g_v. For p₀, we have

p0(Ker_Kϕ_a) =ϕ[p^ord₀ ^p0(a)](K)⊆ϕ[p^N₀ ](K) = Ker_K(ϕ_b).

Since ϕ_a is separable, 4.8.8 implies that it divides gϕ_b from the right in K[τ], say gϕ_b = ˜hϕ_a ∈ End_K(ϕ). It follows that ϕ_a˜hϕ_a = ˜hϕ²_a. Right-division by ϕ_a shows that

˜hcommutes with ϕ_a, which implies ˜h∈End_K(ϕ), since we are in generic characteristic.

Now we pass to the reduction and find

˜h_vϕ_v,a =g_vϕ_v,b=ϕ_v,ah_vϕ_v,b=ϕ_v,bh_vϕ_v,a.

Right-division by ϕv,a in kv[τ] shows that ϕv,bhv lies in the image of EndK(ϕ). This shows that the image ofh_v in the cokernel is annihilated by a power of p₀.

Proposition 4.8.9. Let v and wbe places of K. Suppose ϕis defined over bothO_v and Ow and let pv be the maximal ideal of A lying under v and pw the one lying under w.

Suppose further thatp_v 6=p_w. Then the image of the combined reduction homomorphism End_K(ϕ)→End_kv(ϕ_v)⊕End_kw(ϕ_w)

g 7→(g_v, g_w)

is a saturated A-submodule of Endkv(ϕv)⊕Endkw(ϕw).

Proof. By Proposition 4.8.7, we are in the following situation: We have torsion-free A-modulesM, M₁ and M₂ with injective homomorphismsj₁ :M ,→M₁ andj₂ :M ,→M₂, such that the torsion submodule T₁ of M₁/j₁(M) is annihilated by a power of p_v and the torsion submodule T2 of M2/j2(M) by a power of pw. We want to show that the image ofj :M →M₁⊕M₂, m 7→(j₁(m), j₂(m)) is saturated in M₁⊕M₂.

Suppose p^k_v¹T₁ = 0 and p^k_w²T₂ = 0. By the Chinese remainder theorem, there is b ∈ A withb ∈p^k_v¹ and b6∈p_w. Sinceb is contained in only finitely many maximal ideals of A, there existsc∈Awithc∈p^k_w² and such thatbandcare relatively prime, sayxb+yc= 1 forx, y ∈A.

Now suppose we have m₁ ∈ M₁, m₂ ∈ M₂ and nonzero a ∈ A, such that a(m₁, m₂) = j(m) for some m ∈ M. It follows that m₁ +j₁(M) ∈ T₁ and, since by construction bT1 = 0, that bm1 =j1(n) for some n ∈M. Similarly, there is l ∈M with cm2 =j2(l).

Now we compute

j₁(an) = abm₁ =j₁(bm).

Since j₁ is injective, we have an=bm. This implies aj₂(n) = bj₂(m) =abm₂ and since M₂ is torsion free that j₂(n) = bm₂. Let ˜m=xn+yl. Then

j2( ˜m) =xbm2+ycm2 = (xb+yc)m2 =m2.

By symmetry of the argument we also have j₁( ˜m) = m₁. This shows that indeed the image ofj is saturated.

The following example shows that when we consider a single place v, the image of EndK(ϕ) in Endkv(ϕv) need not be saturated.

Example 4.8.10. Let A = Fq[t] with field of fractions F = Fq(t). Let K be the field extension obtained by first adjoining a 2(q−1)-st root s of t and then a solution ξ of the equationX^q−s^q−1X−s= 0, which is irreducible by the Eisenstein criterion. Let v be the valuation onF associated to the prime ideal (t) of Aand normalized byv(t) = 1.

Thisv extends uniquely to a valuation on K, which we also denote by v. We then have v(s) = 1

2(q−1), v(ξ) = 1 2q(q−1).

To see this, one considers the Newton polygons: The minimal polynomialX^2(q−1)−t of soverF has a Newton polygon with unique slope 1/(2(q−1)), so there is a unique way to extend v to F(s). Similarly the minimal polynomial X^q−s^q−1X−s of ξ over F(s) has the unique slope 1/(2q(q−1)), so there is also only one way to extend v further fromF(s) to all of K. The claimed values ofs and ξ can also be read off from this.

We define a Drinfeld A-module overK by

ϕ_t =τ²−(ξ^q−1+t^qξ^q(1−q))τ+t.

We claim thatϕhas good reduction atv but that End_K(ϕ) is not saturated in End_kv(ϕ_v):

The reduction of ϕ is given by ϕ_v,t = τ². Therefore ϕ has good, but not ordinary reduction at v. The endomorphism ring of the reduction ϕv : A→ F^q[τ] contains both τ and τ³ and inFq[τ] we have the relationτ³ =ϕ_v,tτ.

Claim 4.8.11. The endomorphism τ of ϕ_v does not lie in the image of End_K(ϕ).

Proof. Suppose τ is the image of some g ∈ End_K(ϕ). By injectivity of the reduction map we would then have g² =ϕ_t. By comparing highest and lowest coefficient in this equality,gmust be of the formg =τ+√

tfor some square root oftinK. Comparing the τ-coefficient ing² =ϕ_tthen gives√

t+√

t^q =−(ξ^q−1+t^qξ^q(1−q)). This is a contradiction, since the valuations of the respective sides of this equation are different.

Claim 4.8.12. The endomorphism τ³ of ϕ_v lies in the image of End_K(ϕ).

Proof. The additive polynomial

g = (τ −ξ^q−1)(τ −s^q−1)(τ−tξ^1−q)

is an endomorphism ofϕ and its reduction is equal to τ³. This can be seen by hand in the following way. Let π, ˜g and σ denote the first, second and third factor on the right hand side of this equality. Then one checks that πσ =ϕ_t and thatσπ = ˜g². From this it follows that

gϕ_t =π˜gσπσ =πg˜³σ =πσπgσ˜ =ϕ_tg.

The statement about the reduction follows from the fact that each factor reduces to τ.

5 Main algorithm

We now put the preceding results together. Throughout this sectionK denotes a finite extension ofF and ϕa DrinfeldA-module over K of rank r. By Theorem 4.7.8, exactly one of the following must be true: Either there is a place v of K at which ϕ has good and ordinary reduction, or End_K(ϕ) is inseparable over A.

If the endomorphism ring is separable over A, we can therefore use our results, in par-ticular Proposition 4.8.9 relating the endomorphism ring to its reductions, to compute End_K(ϕ), which is done in 5.1. If instead, there is an endomorphism of ϕ which is inseparable over A, we can use it to pass to a bigger coefficient ring and reduce to the computation of the endomorphism ring of a Drinfeld module over K of degree r/p whose coefficient ring is isomorphic toA. This is done in 5.2. Finally, we present a total algorithm in 5.3, which computes the endomorphism ring of ϕ by deciding which case we are in and calling the corresponding algorithm. In the inseparable case this gives a recursion, since the algorithm for the inseparable case calls the total algorithm. Each time this happens the rank of the Drinfeld module that is passed is strictly smaller than before, so the total algorithm still terminates.

5.1 The separable case

Algorithm

Algorithm 5.1.1 (Determining the endomorphism ring from the reduction at two places.). Given a Drinfeld module ϕ : A → K[τ] of rank r over a finite extension K ofF and places v₁, v₂ of K which fulfil the premise of Proposition 4.8.9 and at which ϕ has ordinary reduction, this algorithm computes a finite list of elements ofK[τ], which generate the endomorphism ring End_K(ϕ) as an A-algebra.

1. Compute the residue fields k_vi and the reductions ϕ_vi of ϕ for i= 1,2.

2. For i = 1,2 determine the minimal polynomial m_i of the Frobenius element τ^dvi over ϕ_vi(A) in k_vi[τ] with Algorithm 4.6.3 for i = 1,2. Let m^sep_i be the unique

separable irreducible polynomial over F for which m_i(X) = m^sep_i (X^pei) for some e_i ≥0.

3. Fori= 1,2 set E_i =F[X]/(m^sep_i ). Use Algorithm 3.1.2 to obtain a listIntFields of intermediate fields of the field extension E1/F. For each L in IntFields, compute a primitive element α_L over F and the minimal polynomial m_L of α_L overF.

4. Create a new list Embeddings. For L running through IntFields, determine all zeros of m_L in E₂ and for every such zero β ∈ E₂ add the pair (α_L, β) to Embeddings.

5. For every pair (α, β) in Embeddings determine a nonzero a ∈ A such that both aα∈A[X]/(m^s₁)⊆E₁ and aβ ∈A[X]/(m^s₂)⊆E₂. Replace (α, β) by (aα, aβ).

6. Iterating over (α, β) inEmbeddings, compute the minimal polynomialm ofα over F and the zeros ε₁, . . . , ε_k of m in K. Collect all the elements of K found in this way in the listConstants.

7. For every εin Constantsuse Algorithm 4.3.7 to find an endomorphism g_ε ∈K[τ]

with constant coefficientε, if it exists. Let S be the set of all those g_ε.

8. Using Algorithm 4.4.1, determine the saturation as anA-module of the subalgebra generated byS in the endomorphism ring. This returns generators for End_K(ϕ).

Correctness

To simplify the notation, we identify F with its image in any arising F-algebra. First we collect some observations about the algorithm, which are straightforward to check.

• By Proposition 4.2.3, the endomorphism ring of ϕ_vi is commutative for i= 1,2.

• It follows that End⁰_k

vi(ϕ_vi) = F(τ^dvi), and that it is a field of degree r overF.

• For i = 1,2 the field E_i maps bijectively onto the maximal separable subfield of End⁰_k

vi(ϕvi) over F by identifying the image of X with τ^dviei

. We identify it for the following considerations with this subfield.

• Under this identification A[X]/(m^sep_i ) is contained in the endomorphism ring of ϕ_vi.

• The list IntFields lists all intermediate fields of E₁/F, hence all intermediate fields of End⁰_kv

1(ϕv1)/F which are separable over F.

• The list Embeddings corresponds bijectively to the set of pairs (L, ι), where L is an intermediate field of E₁/F and ι : L → E₂ is an F-homomorphism. Hence it corresponds bijectively to the set of pairs (L, ι), where L is an intermediate field of End⁰_kv

1(ϕ_v1)/F which is separable over F and ι : L → End⁰_kv

2(ϕ_v2) is an F-homomorphism.

• In steps 6 and 7, exactly those endomorphisms of ϕ over K are computed which have the same minimal polynomial overF as someαwhich appears in a pair (α, β) of Embeddings.

Lemma 5.1.2. After or before step 5, the set of entries in Embeddings corresponds bijectively to the set of F-subalgebras of End⁰_kv

1(ϕ_v1)⊕End⁰_kv

2(ϕ_v2), which are separa-ble field extensions of F. A bijection is given by associating to a pair (α, β) the field generated by (α, β) over F in End⁰_kv

Proof. By the universal property of the direct product, the data of an F-subalgebra of End⁰_k

v1(ϕ_v1) ⊕End⁰_k

v2(ϕ_v2) which is a separable field extension of F is equivalent to the data of separable subfield L of End⁰_kv

1(ϕ_v1) containing F, together with an F -homomorphismL→End⁰_k

v2(ϕ_v2). The set of those corresponds bijectively toEmbeddings by the remarks above.

Abbreviate E := End⁰_K(ϕ). Algorithm 4.4.1 allows us to determine the full endomor-phism ring from any A-subalgebra S ⊆ End_K(ϕ) which has finite index in End_K(ϕ), or equivalently, which has also quotient field E. This happens in step 8 under the assumption that the set of endomorphismsS computed in step 7 generatesE overF. Therefore correctness of the algorithm follows from the

Claim 5.1.3. The set S contains some g0 ∈ EndK(ϕ), which is a primitive element for E/F.

By Proposition 4.7.4 the reduction homomorphisms End_K(ϕ) → End_kvi(ϕ_vi) extend to embeddings j_i : E → End⁰_k the same minimal polynomial over F as α and β. This implies that it will be found as an element ofS in step 7. Since j induces a field isomorphismE →j(E), it follows that g₀ generates E over F. This shows Claim 5.1.3.

Over the algebraic closure

In order to compute the Endomorphism ring of ϕ over the algebraic closure K one replaces step 6 in Algorithm 5.1.1 by the following variation:

6’. Iterating over (α, β) inEmbeddings, compute the minimal polynomialm ofα over F, extend K to a splitting field of m and compute the zeros ε₁, . . . , ε_k of m in the new K. Collect all the zeros found in this way in the listConstants.

To see why this works let K⁰ be a minimal finite separable extension of K, such that all endomorphisms of ϕover an algebraic closure of K are already defined over K⁰ and let v⁰_i be an extension of v_i to K⁰ for i = 1,2. Let E⁰ := End⁰_K0(ϕ), which is again a

i). In particular E⁰ embeds natu-rally as anF-algebra into

End⁰_kv

1(ϕ_v1)⊕End⁰_kv

2(ϕ_v2).

We can again apply Lemma 5.1.2 to see that after step 5 there is a pair (α, β) in Embeddings that generates the image of E⁰ over F in End⁰_kv

1(ϕ_v1)⊕End⁰_kv

2(ϕ_v2) and such thatα and β are actually endomorphisms. By saturatedness, it follows that there exists an endomorphism g of ϕ over K⁰ which maps to α under reduction. It follows that g has the same minimal polynomial m over F as α and that g generates E⁰ over F. Let ε⁰ ∈K⁰ denote the constant coefficient of g, which is also a zero of m. Then by Proposition 4.3.6,g is already defined overK[⁰]. It follows that the endomorphism ring overK⁰ is the same as over K[⁰] and hence that K⁰ =K[⁰] by minimality of K⁰. LetK⁰⁰ be the field obtained by repeatedly extendingK as in in step 6’. Since (α, β) is contained in Embeddings after step 5, in step 6’ it is arranged thatm has a zero inK⁰⁰ which is contained in the list Constants. It follows thatK⁰ embeds in K⁰⁰ and that the image ofg in K⁰⁰[τ] is added to S in step 7.

Im Dokument Computing the Endomorphism ring of a Drinfeld module in generic characteristic (Seite 34-42)