Proofs of the first two bounding lemmas – preliminary version
Zoé Chatzidakis, Piotr Kowalski and Françoise Point
March 26, 2013
Needed
φq : the Frobenius map x 7→ xq in a ring of characteristic p.
Mq : functor defined on difference rings. Let R be a difference ring, Jq (R) the ideal generated by pR
and by σ(r) − rq for r ∈ R. Then Mq (R) = R/Jq (R). If X = Specσ R, then Mq (X) = Spec Mq (R).
If X is a difference scheme, one uses gluing to define the algebraic scheme Mq (X).
Xq,y : fiber above y in the algebraic scheme Mq (X); = Mq (X)y .
Remark 0.1. If the (reduced) total dimension of the well-mixed R is finite, then its transformal
dimension is 0. The converse holds when R is contained in the localisation of a finitely generated
over k difference ring. See 2.6 in [Box 3].
Needed
{H-4.35}
0.2. In a difference ring R which is finitely generated over some Noetherian difference ring (or is
contained in a localisation of such a difference ring), every perfect σ-ideal is the intersection of
finitely many transformally prime ideals. More precisely, the topological space Specσ R is Noetherian. [Hr-4.35] and [Co], Theorems 3.5.II, 3.6.III and 3.8.V.
1
Proof of the First bounding lemma - estimate of the number of points without multiplicities
{H-9.8}
Proposition 1.1. Let X be a difference scheme of finite type over a Noetherian difference scheme
Y.
(1) Assume that X has transformal dimension d over Y . Then for all prime powers q and all
points y ∈ Mq (Y )(L), L a field, the fiber Xq,y has dimension at most d as a scheme over Ly .
(2) If X has reduced total dimension ≤ e over Y , then there exists b ∈ N such that for all large
enough prime power q and all y ∈ Mq (Y )(L), L a perfect field, the zero-dimensional scheme
Xq,y over Ly has at most bq e points.
Proof. (The proof in [H] is fairly clear - but my proof of the claim is a little different: I work in
quotients instead of localizations). The statement is local, so we may assume that X = Specσ R,
and Y = Specσ D, where D is a difference domain (**Not necessary**), and R a finitely generated
1
difference D-algebra. We assume that the transformal dimension of X is d [resp. red.t.dim R = e].
Let S be a difference subring of R containing D, let Z = Specσ S. We will use Noetherian induction
on Y and then on Z.
Claim. Assume that the lemma holds for Specσ RU 0 → U 0 and for Specσ SU → U whenever U is
an open subset (for the σ-topology) of X, and U 0 is an open subset U 0 of Z. Then the lemma holds
for X → Y .
Proof. We give the proof for (2), it is similar for (1). Note that we are trying to bound the size of
certain quotients R/P , with P ∈ Specσ R. Let P ∈ Specσ R be minimal. Then P ∩ S is minimal
in S, by 1.14 (exp3). Let K, L, M , be the fields of fractions of D, S/P ∩ S and R/P respectively.
If e1 = tr.deg (M/L) and e2 = tr.deg (L/K), then e1 + e2 ≤ e. By 2.10 in exp3 there is an open
subset U of Y containing P ∩ D = (0), and an open subset U 0 of Z containing P ∩ S such that for
any difference field K containing k, over any point z ∈ U 0 (K), the relative reduced dimension of
XK,z over z is ≤ e1 , and over any point y ∈ U (K), the relative reduced dimension of YK,y over y
is ≤ e2 . We may choose U 0 so that its projection is contained in U .
Let q be a power of p sufficiently large so that the conclusion holds for RU 0 → U 0 and for SU → U ,
with constants b1 and b2 . Let L be a perfect field of characteristic p > 0, and y ∈ Mq (Y )(L),
y ∈ U (L). Then (Specσ SU )q,y has at most b2 q e2 points, hence at most that many points which are
in U 0 (L). For each z ∈ U 0 (L) ∩ Zq,y (L), Specσ RU 0 )q,z (L) has at most b1 q e1 points; hence there are
at most b1 b2 q e1 +e2 points x ∈ Xq,y (L) which satisfy x ∩ S ∈ U 0 (L).
By Noetherian induction, the result holds for the closed subscheme of R defined by x∩S ∈ Z \U 0 .
Note that it contains the closed subscheme of R defined by x ∩ D ∈ Y \ U .
The claim allows us to reduce the proof to the case of R = D[a]σ , a a single element.
Case 1. a is transformally transcendental over K.
(This case does not occur in (2)). Then for every y ∈ Mq (Y )(L), the set Xq,y is simply A1L , of
dimension 1, since the ring Mq (R) is generated over Mq (D) by one element.
Case 2. a is transformally algebraic over K.
Let P ∈ Specσ R be minimal, and work in the difference domain R1 = R/P , let X1 = Specσ R1 .
We know that tr.deg (FracR1 /FracD) = r for some r, and r ≤ e in case (2). Let G(T ) ∈ D[T ]σ
be irreducible, non-zero, and of minimal order r such
that G(a) ∈ P . (So, in (2), we have r ≤ e
Pm
because tr.deg (FracR1 /FracD) = r). Write G(T ) = i=1 ci T νi (σ) , with ci ∈ D \ {0}, νi (σ) ∈ N[σ],
and assume that the νi (σ) form an increasing sequence in N[σ]. We choose N sufficiently large so
that if q ≥ N , then the νi (q), 1 ≤ i ≤ m form a strictly increasing sequence, and νm (q) ≤ bq r ,
where b = (cm + 1). We now consider the open subset U of Specσ D defined by cm 6= 0. Then if
q > N , L an algebraically closed field, and all y ∈ Mq (U )(L), the algebraic set Xy,q has at most
νm (q) points, and therefore ≤ bq r points. Indeed, the kernel of the map D → L corresponding to
y, is a transformally prime ideal of D, which does not contain cm , and therefore the image of G(T )
by yMq will be a non-zero polynomial with leading term yMq (cm ) 6= 0.
COMMENTS. The first part does not seem to be used anywhere.
2
2
Easy algebraic lemmas
Lemma 2.1. Let K be a field, R a finitely generated K-algebra, of Krull dimension 0. Then R is
a finite dimensional K-vector space.
Proof. Every prime ideal of R is maximal, and the intersection of the finitely many maximal ideals of
R is the ideal I of nilpotent elements of R. As R is Noetherian, this ideal I is finitely generated, and
nilpotent. Clearly R/I is a finite-dimensional K- vector space, say with basis A. Let {b1 , . . . , bm }
be a set of generators of the ideal I, and C = {αbi11 · · · bimm | α ∈ A}; as each bi is nilpotent, C
is finite. Moreover every element of I can be written as a K-linear combination of elements of C,
which gives the result.
Definition 2.2. Let X = Spec R be an algebraic variety of dimension 0, P a closed point in X,
RP the localized ring, M = P RP and k = RP /M . Then MultP X = supt dimk (RP /M t ).
Lemma 2.3. Let k be a field of characteristic p > 0, with [k : k p ] = pe (e = 0 if p = 0), and let S
be a k-algebra generated by m elements, M a maximal ideal of S with dimk (S/M ) < ∞. Then M
is generated by at most m + e + 1 elements.
{H-9.10}
Proof. Let h : S → K be an epimorphism with kernel M , and let s1 , . . . , sm be generators of the
k-algebra S.
Let us first assume e = 0. Then K is a separable extension of k of finite degree. Let s0 ∈ S be
such that h(s0 ) = a generates K over k. Let F (x) ∈ k[x] be the monic minimal polynomial of a
over k, and G1 (x), . . . , Gm (x) ∈ k[x] such that h(si ) = Gi (a). Then M is generated by
F (s0 ), si − Gi (s0 ), i = 1, . . . , m.
In the general case, K is generated over k by ≤ e + 1 elements. Indeed, let k1 = K ∩ k s , then
k1 = k(a) for some a; as K/k is finite, we know that [K : K p ] = [k : k p ]; if a1 , . . . , ae is a pbasis of K, then K = k1 (a1 , . . . , ae ) = k(a, a1 , . . . , ae ). Let F (x) ∈ k[x] be the minimal monic
polynomial of a over k, and for i = 1, . . . , e, let Hi [x, x1 , . . . , xi ] ∈ k[x, x1 , . . . , xi ] be such that
Hi (a, a1 , . . . , ai−1 , xi ) is the minimal monic polynomial of ai over k[a, a1 , . . . , ai−1 ]. Then
K ' k[x, x1 , . . . , xe ]/(F (x), H1 (x, x1 ), . . . , He (x, x1 , . . . , xe )).
We reason in the same manner as in the first case: if a = h(s0 ), ai = h(ti ) and h(sj ) =
Gj (a, a1 , . . . , ae ), then
M = (F (s0 ), H1 (s0 , t1 ), . . . , He (s0 , t1 , . . . , te ), sj − Gj (s0 , t1 , . . . , te ), i = 1, . . . , m).
COMMENT: the case e 6= 0 seems to be used nowhere.
{H-9.11}
Lemma 2.4. let k be a field, R a finite dimensional local k-algebra with maximal ideal M . Let N
be a finitely generated R-module. Then
dimk N ≤ (dimk R)(dimk (N/M N )).
Proof. Let A be a k-subspace of N of dimension dimk (N/M N ) and such that A + M N = N .
Let T be the subspace of N generated by RA. Then T is an R-submodule of N , and dimk (T ) ≤
dimk (R) dimk (N/M N ). As T + M N = N , Nakayama’s lemma gives T = N .
3
{H-9.12}
Lemma 2.5. Let f : X → Y be a morphism of finite schemes over Spec k, let b ∈ Y , Xb the fiber
of X above b, and a ∈ Xb . Then
Multa X ≤ (Multa Xb )(Multb Y ).
Proof. Let R = OY,P and S = OX,Q . Then OXP ,Q = S/M S, where M = P R. For any n > 0,
S/Qn S is an (R/P n R)-module of finite type. By 2.4, we have MultQ (XP ) = dimk (S/Qn S) ≤
dimk (S/M S) dimk (R/P n R), which gives the result.
{H-9.13}
Lemma 2.6. Let S be a local ring with maximal ideal M , which is finitely generated. Assume that
S/SM r has length n < r. Then M n = (0).
Proof. For some i ≤ n, we have SM i = SM i+1 ; by Nakayama, SM i = 0.
Notation. Recall that φq denotes the q-Frobenius map. If I is an ideal of the ring R of characteristic
p, then φq (I) will therefore denote {aq | a ∈ I} (not necessarily an ideal), while I q will denote the
usual ideal generated by products of q members of I.
{H-9.14}
Remark 2.7. Let S be a ring, with ideal M generated by s1 , . . . , sb . Assume p = 0 in S, and q is
a power of p. Then M bq ⊂ Sφq (M ) ⊂ M q .
P
mb
1 m2
Proof. The generators of M bq are of the form sm
with
mi = bq. One of the mi ’s
1 s2 · · · sb
i
must therefore be ≥ q, and sm
∈
φ
(M
).
q
i
Let S be a local ring of characteristic p > 0, with maximal ideal M . Assume that M is finitely
generated. Then φq (S) has a unique maximal ideal, φq (M ), with residue field φq (S)/φq (M ) =: k.
As S is a φq (S)-module, the ring S/Sφq (M ) is a k-vector space.
Lemma 2.8. Let S, M be as above. If dimk (S/Sφq (M )) = n < q, or if dimk (S/M q ) = n < q, then
the S-module S has length ≤ n.
{H-9.15}
Proof. Note that Sφq (M ) ⊆ M q . By 2.6, M q = 0, and S = S/M q .
{H-9.16}
Lemma 2.9. Let S be a local ring of characteristic p and with maximal ideal M generated by b
elements. Let r and s be powers of p. We assume that the length of the φr (S)-module φr (S/Sφs (M ))
(= the subring of S/φs (M ) obtained by applying φr ) has length n, and that rbn < s.
Then M rbn = 0, φr (S)φr (M )n = 0, and the φr (S)-module φr (S) has length ≤ sup{n, bn − 1}.
Proof. We have the following isomorphism of φr (S)-modules:
φr (S/Sφs (M )) = (φr (S) + Sφs (M ))/Sφs (M ) ' φr (S)/(Sφs (M ) ∩ φr (S)).
Hence the φr (S)-module φr (S)/(Sφs (M ) ∩ φr (S)) has length n. For each i, consider the φr (S)module
Ni = (M s ∩ φr (S)) + φr (S)φr (M )i .
The Ni form a decreasing sequence of φr (S)-modules between φr (S) and Sφs (M ) ∩ φr (S) (use
Sφs (M ) ⊂ M s ). As the length of the φr (S)-module φr (S)/(Sφs (M ) ∩ φr (S)) is n, there is some
i ≤ n such that Ni = Ni+1 . By Nakayama (in the ring φr (S)/(M s ∩ φr (S)), we have Ni =
M s ∩ φr (S), i.e., φr (S)φr (M )i ⊂ M s ∩ φr (S). By 2.7, we have M br ⊂ Sφr (M ) ⊂ M r , which gives
4
SM bri ⊂ Sφr (M )i ⊂ M ri ; from bri ≤ brn < s we obtain M s = M brn = 0, and Sφr (M )n = 0 =
φr (S)φr (M )n .
As M is generated by b elements, so is φr (M ), and therefore each φr (S)φr (M )` is generated by
`
≤ b elements. So, φr (S)φr (M )` /φr (S)φr (M )`+1 is a homomorphic image of the cartesian product
of b` copies of φr (S)/φr (M ), and has therefore length ≤ b` . So, the length of φr (S) is bounded by
lgth(φr (S)/φr (M )) +
n−1
X
lgth(φr (S)φr (M )` /φr (S)φr (M )`+1 ),
`=1
i.e., by 1 + b + · · · + bn−1 ≤ sup{n, bn − 1}.
{H-4.10}
{H-4.12}
Lemma 2.10. Let k be a difference field, R a difference k-algebra, and K a difference field extending
k. Assume t.dim (R) = `.
(1) Then t.dim (R ⊗k K) ≤ `.
(2) There is a difference field k̄ extending k, with underlying field k a , and such that t.dim (R ⊗k
k̄) = `.
(3) t.dim (R ⊗k k inv ) = `.
Proof. (1) If P is an algebraically prime difference ideal of R ⊗k K, then P ∩ R is an algebraically
prime difference ideal of R, and we have an inclusion R/(P ∩ R) → (R ⊗k K)/P .
(2) Let P be an algebraically prime difference ideal of R such that tr.deg k (R/P ) = `; it then suffices
to find k̄ such that t.dim ((R/P ) ⊗k k̄) = `, and we may therefore assume that R is algebraically
prime. Let L = FracR, and La an algebraic closure of L, k a ⊂ La the algebraic closure of k.
The ring R0 = R[k a ] is integral over R, and the ring homomorphism σ : R → L extends to a
ring homomorphism σ̃ : R0 → La . We have σ̃(R) ⊂ R and σ̃(k a ) ⊂ k a , so that σ̃(R0 ) ⊂ R0 . So σ̃
defines an endomorphism on R0 , which agrees with σ on R, and restricts to an endomorphism σ̄ of
k a extending σ. We let k̄ be the difference field (k a , σ̄). Then R0 is a quotient of the difference ring
R ⊗k k̄; as tr.deg k (FracR0 ) = tr.deg k (L) = `, we obtain the result.
(3) As in (2), we may assume that R is algebraically integral. By (1), it is enough to show that
t.dim (R ⊗k k̄ inv ) ≥ ` (where the difference field structure on k̄ is given by (2)).
By (2), R ⊗k k̄ has a quotient R0 which is an algebraically integral difference ring, with field of
fractions of transcendence degree ` over k̄. As k̄ is algebraically closed, R0 ⊗k̄ k̄ inv is algebraically
integral, and its field of fractions has transcendence degree ` over k̄ inv . Hence R ⊗k k̄ inv ' (R ⊗k
k̄) ⊗k̄ k̄ inv has an algebraically integral quotient with field of fractions of transcendence degree `
over k̄ inv .
Lemma 2.11. Let X = Specσ OX be a well-mixed difference scheme of finite type over an inversive
difference field K. Assume that the relative total dimension of X → Bm (X) is n < m. Then for
any difference field L, point P ∈ Bm (X)(L) and Q ∈ X(L) above P , the total dimension (over
σ n (L)) of σ n (OX,Q ⊗P L) is 0.
Proof. We may assume X = Specσ R, with R a well-mixed difference ring. By 2.10, we may also
assume that L = Linv . Let S be the quotient of RQ ⊗P L by its well-mixed radical. We need to
show that t.dim (σ n (S)/σ n (L)) = 0.
5
{H-5.11}
3
Proof of the second bounding lemma: number of points
counting multiplicities
Definition 3.1. (1) Let f : X → Y be a morphism of difference schemes. The reduction multiplicity of X over Y is ≤ e if for some B ∈ N, for any sufficiently large prime power q, for any
perfect field L, y ∈ Mq (Y )(L) and z ∈ Xq,y = Mq (X) ×y L, Multz Xq,y ≤ Bq e .
(2) Let X be a difference scheme of finite type over a finitely generated difference field K. We say
that X is generically of e-bounded reduction multiplicity over K if there is finitely generated
difference ring D ⊂ K and a difference scheme X0 of finite type over D, such that
X ' X0 ×Specσ D Specσ K,
and X0 7→ Specσ D has reduction multiplicity ≤ e.
The proof of the following lemma is fairly long. It uses two kinds of Noetherian inductions:
one on the Noetherian rank of a topological Noetherian space:we assume that the result holds for
all proper closed subsets; and the other on the Noetherian rank of a Noetherian ring (see the last
section of [Box 3]): we assume that the result holds for all proper quotients of our ring.
{H-9.19}
Lemma 3.2. Let X be a difference scheme of finite type over a Noetherian difference scheme
Y . Assume that X is transformally radicial over Y , and t.dim (X/Y ) ≤ e. Then the reduction
multiplicity of X over Y is ≤ e.
Proof. We may assume that Y is irreducible, Y = Specσ R where R is a difference domain. We
will use Noetherian induction on Y , and assume that the lemma holds for XZ → Z, Z any proper
closed subscheme of Y . In particular we will be able to localize R as much as we want, and we will
do this all the time.
For X we cannot use Noetherian induction, since multiplicity on a subvariety of X can be smaller
than on X. However, writing X as a union of open subsets Xi , it suffices to show the lemma on
each Xi . This allows us first to assume that X = Specσ S. It will also appear in the proof later.
We may also assume that S is well-mixed: this is because Mq (Xy ) will always be well-mixed (see
???); however it is not always reduced and so we cannot assume that S is algebraically reduced.
As S is transformally radicial over R, and finitely generated over R as a difference ring, it is a
finitely generated R-algebra. Hence there are a1 , . . . , an ∈ S such that
S = R[a1 , . . . , an ], σ(a1 ) ∈ R, and σ(ai ) ∈ R[a1 , . . . , ai−1 ] for i > 1
(∗).
We let L = FracR and SL = S ⊗R L. We know that SL is generated over L by (the images of)
a1 , . . . , an , and is a Noetherian ring, hence is ranked by Nrk.
We will therefore consider all tuples (R, S, L, SL ) as above, and do an induction on the set of
tuples
(t.dim (S/R), dimN(R), n(S/R), Nrk(SL ))
ordered lexicographically, where n(S/R) denotes the minimal number n such that there are a1 , . . . , an
satisfying (∗), and Nrk the Noetherian rank, see exp3. Before starting the proof, we will show a
claim which will be used several times:
Claim. Let R0 be a subalgebra of S, which is finitely generated as an R-module. If the Lemma
holds for X/Specσ R0 , then it holds for X.
6
Proof. Let d be the (minimal) number of generators of R0 as an R-module, Z = Specσ R0 . If y ∈
Mq (Y )(L) and z ∈ Zq,y (L), then Multz (Zq,y ) ≤ d. By 2.5, we have Multz (Xq,y ) ≤ dMultx (Xq,z );
the hypothesis on R0 then gives the result.
Step 1. We may assume S → SL is injective.
Let I ⊂ L[X1 , . . . , Xn ] be the ideal of polynomials vanishing at (a1 , . . . , an ), so that SL ' L[X1 , . . . , Xn ]/I.
Since L[X1 , . . . , Xn ] is Noetherian, I is finitely generated, and there is some 0 6= b ∈ R such that I
is generated by I ∩ R[b−1 ]σ [X1 , . . . , Xn ]. So, the map S ⊗R R[b−1 ]σ → SL is injective.
Note. The fact that S → SL is injective implies that if c 6= 0 is in SL , then for some r ∈ R, we
have rc ∈ S, rc 6= 0. In particular, any ideal of SL is generated by its intersection with S.
Step 2. We may assume that that there are no a ∈ SL such that σ(a) = a2 = 0.
Let IL be the ideal of SL generated by A = {a ∈ SL | σ(a) = a2 = 0}. Since SL is Noetherian, this
ideal is finitely generated, say by b1 , . . . , bm , which may be chosen in S by step 1.
Let R0 = R[b1 , . . . , bm ]; it is an R-difference subalgebra of S (because σ(b) = 0), and it is of
P`
finite dimension as an R-module, since every bi is nilpotent: if b = j=1 aj , with aj ∈ A, then
b`+1 = 0 and σ(b) = 0.
The ideal J = IL ∩ R0 is nilpotent, and R0 /J ' R; this implies that Specσ R = Specσ R0 ,
and so R and R0 have the same Noetherian dimension, and t.dim (S/R) = t.dim (S/R0 ), since
any prime ideal of R0 will contain J. Furthermore n(S/R0 ) = n(S/R). Let L0 = Frac(R0 /J). The
inclusion R → R0 induces an isomorphism L → L0 , and the natural map SL → S ⊗R0 L0 has kernel IL . Hence Nrk(S ⊗R0 L0 ) < Nrk(SL ), and we are done by induction hypothesis, and by the claim.
Note that σ −1 (0) contains no nilpotent elements.
Step 3. We may assume that if c ∈ SL is a zero-divisor, then σ(c) ∈
/ L.
Let c ∈ SL a zero-divisor such that σ(c) ∈ L. Multiplying c by some element of R (see step 1) we
may assume that σ(c) ∈ R. If σ(c) 6= 0, then replacing R by R[σ(c)−1 ]σ , we may assume that σ(c)
is invertible in R. But then cd = 0 implies σ(d) = 0, and therefore there is some d ∈ SL , such that
σ(d) = 0 and d 6= 0. By step 1 (and the note immediately after), there is such a d in S. We replace
c by d, and so we have σ(c) = 0, c ∈ S a zero-divisor.
Let J = Ann(c) (in SL ). Then both J and cSL are difference ideals (because be assumed S
well-mixed, whence SL is well-mixed, and because σ(c) = 0). We also have J ∩ S 6= (0), by the
remark after Step 1. By induction hypothesis, the lemma holds for the R-algebras S1 = S/(S ∩ J)
and S2 = (S/cS).
Moreover cS ∩ J = (0): if d ∈ J ∩ cSL , then d2 = 0 = σ(d), and d = 0 by Step 2. Hence, the
natural map S → S1 × S2 is an injection, and X is covered by the open sets Specσ S1 and Specσ S2 .
Hence S satisfies the conclusion of the lemma.
Case 1. Assume that there is a non-zero polynomial F (x) ∈ R[x] which vanishes at a = a1 .
Replacing R by some R[b−1 ]σ , we may assume that this polynomial F is monic. Then R[a] is a finite
dimensional R-module, and Specσ R[a] and Y have the same Noetherian dimension. By induction
hypothesis (on n), the lemma holds for S/R[a1 ]; by the claim it holds for S/R.
7
Case 2. Not case 1.
Then a = a1 is transcendental over L (by Step 1). Moreover, if f (x) ∈ R[x], then f (a) is not a
zero-divisor, since σ(f (a)) ∈ R and by step 3. As t.dim (S/R) = e, replacing R by a localization,
we may assume that the total dimension of each fiber of Specσ S → Specσ R[a] is ≤ e − 1: since this
is true generically, there is b ∈ R[a] such that this is true on Specσ R[a, b−1 ]σ (see ???). Replace R
by R[σ(b)−1 ]σ : no algebraically prime difference ideal of S ⊗R R[σ(b)−1 ]σ can contain b.
Let now Z = Specσ R[a]. For any q, and y ∈ Mq (Y )(L), z ∈ Zq,y (L), we have Multz (Zq,y ) ≤ q
(because σ(a) ∈ R). By induction hypothesis on e (wouldn’t n be enough?), there is some B such
that for any q 0, z ∈ Z(L), x ∈ Xq,z , Multx (Zq,z ) ≤ Bq e−1 . By 2.5 we obtain Multx (Xq,y ) ≤ Bq e ,
which finishes the proof.
{H-9.20}
Lemma 3.3. Let X be a difference scheme of finite type over a finitely generated difference domain
D. Assume that X is of generic transformal multiplicity ≤ n. Then there exists a non-empty open
set U ⊂ Specσ D and a bound b0 such that for all inversive difference fields L, and y ∈ U (L),
x ∈ Bn+1 (Xy,L )(L), for every local ring R of (Xy )x , σ n (R) is a finite dimensional vector space
over L, of dimension ≤ b0 .
Proof. Without loss of generality, X = Specσ S, where S = D[a]σ is a D-difference algebra.
Let K = FracD, and y ∈ Specσ D the corresponding prime. Let SK = S ⊗D K. Then Xy =
Specσ SK . Let x ∈ Bn+1 (Xy )(K). By 2.11, if P ∈ Spec SK is a difference ideal above x, and R
the corresponding local ideal (for the difference scheme Xy ), then t.dim (σ n (R)/L) = 0, and by
2.3 in exp 3, σ n (R) is a finite-dimensional σ n (K)-vector space. By the above, there is some m and
some finite b ⊂ σ n (a) ∪ · · · ∪ σ m+n (a) such that every element of σ n (a) ∪ σ(b) is an σ n (K)-linear
combination of elements of b; let c ∈ D be such that every element of σ n (a) ∪ σ(b) is an σ n (D)[b−1 ]linear combination of the elements of b. ???? NOT QUITE CORRECT - NEEDS TO BE FIXED.
H-5.11v needs to be changed as well.
{H-9.21}
Lemma 3.4. Let f : X → Y be a morphism of difference schemes of finite type, of transformal
relative multiplicity ≤ n. Then there is B ∈ N such that for any q > B, for any perfect field K and
y ∈ Mq (Y )(K), any closed point x ∈ Xq,y has multiplicity ≤ Bq n on Xq,y .
Proof. We may assume X = Specσ R, Y = Specσ D, with R a D-difference algebra which is finitely
generated. Using Noetherian induction on Y and 3.3, we may assume that D is a difference domain,
and that for any inversive difference field L and y ∈ Y (L), z ∈ Bn+1 (Xy )(L), if A is the local ring
of Xy in z, then σ n (A) is an L-vector space of dimension ≤ b0 . Let m be the number of generators
of R as a difference D-algebra.
Let q be a power of the prime p satisfying q > b0 (m+1), let K be a perfect field of characteristic p,
y ∈ Mq (Y )(K) and x ∈ Xq,y (K). Let C be the local ring of Mq (R)⊗Mq (D) L at x (ring corresponding
to Xq,y ×y L), M its maximal ideal, and let z = σ n+1 (x) ∈ Bn+1 (Xq,y ). Then x ∈ (Xq,y )z , and its
local ring is
A ⊗σn+1 (A) (σ n+1 (A)/????
{H-9.24}
{H-9.23}
3.5.
Theorem 3.6. Let X0 be a difference scheme of finite type over a difference field K. Assume that
t.dim (X) = d, and that no algebraic component of X0 of total dimension d has positive transformal
multiplicity. Then there is a finitely generated D ⊂ K, and a difference scheme X of finite type
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over Y = Specσ D, such that X0 = X ⊗Y Specσ K, and some B such that: for any q 0,
for any perfect field L of characteristic p, and y ∈ Mq (Y )(L), the difference scheme Xq,y (L) has
≤ Bq d points. Moreover, the set of points of Xq,y (L) having multiplicity > B has size counted with
multiplicities ≤ Bq d−1 , and if X 0 is a subscheme of X with t.dim (X 0 /Y ) < d, then the number of
0
points in Xq,y
(L) counted with multiplicities, is O(q d−1 ).
References
[Box 3] Box
3:
What
is
needed
in
difference
http://www.logique.jussieu.fr/∼zoe/Frob/exp3.pdf
algebra.
Available
at:
[Co]
R.M. Cohn, Difference algebra, Tracts in Mathematics 17, Interscience Pub. 1965.
[H]
E. Hrushovski, The elementary theory of the Frobenius automorphism, preprint (24 July
2012).
9