Canonical-p-bases^{∗}

Martin Ziegler

14.3.2003

The purpose of this note is to give a proof of a remark^{1} in [1]:

Theorem 1. Every ω–saturated strict D–field has a canonical p–basis.

I will use the definitions and notation of [1]. As there, all fields have characteristic p. We start the proof with a couple of Lemmas.

In our application the following lemma, except of its last sentence, can be replaced by Lemma 3.

Lemma 2. Let K be a field, d_{1}, …, d_{e} be a sequence of commuting derivations of K, and C = C_{1} ∩ ⋯ ∩ C_{e}, where C_{i} is the field of constants of d_{i}. Assume that

a) d_{i}^{p} = 0 for i = 1, …, e

b) (K : C) = p^{e}

Then there are elements b_{1}, …, b_{e} such that d_{i}(b_{j}) = δ_{i,j}. Each such sequence generates K over C.

Proof. The proof of [1, Lemma 2.1] shows that, for every i, C is a proper subfield of F_{i} = ⋂_{j≠i} C_{j}, which is closed under d_{i}. Choose b_{i} ∈ F_{i} with d_{i}(b_{i}) = 1. Consider the sequence

K = B_{0} ⊃ B_{1} ⊃ ⋯ ⊃ B_{e} = C,

where B_{i} = C_{1} ∩ ⋯ ∩ C_{i}. b_{i} generates B_{i−1} over B_{i}, so C(b_{1}, …, b_{e}) = K.

Note that K^{p} ⊂ C. If C = K^{p}, the b_{i} form a p–basis of K.

Lemma 3. Let K and d_{1}, …, d_{e} as in Lemma 2. For any sequence x_{1}, …, x_{e} of elements of K the following are equivalent:

1. There is a y ∈ K such that d_{i}(y) = x_{i} for i = 1, …, e.

∗Revision : 1.3

1After Lemma 4.1

2. a) d_{i}^{p−1}(x_{i}) = 0 for all i.

b) d_{i}(x_{j}) = d_{j}(x_{i}) for all i, j.

Proof. That 1 implies 2 is clear. We prove the converse by induction on e.

Case e = 1: d = d_{1} is a C-linear map, its kernel has dimension 1. This implies that the dimension of d(K) is p − 1 and the dimension of ker d^{p−1} at most p − 1. Since d(K) ⊂ ker d^{p−1}, we have d(K) = ker d^{p−1}.

Case e > 1: Since (K : C_{e}) = p, we can apply the first case to obtain an element z ∈ K with d_{e}(z) = x_{e}. Set x′_{i} = x_{i} − d_{i}(z). The x′_{i} again satisfy our assumption. They belong to C_{e}, since d_{e}(x′_{i}) = d_{i}(x′_{e}) = d_{i}(0) = 0. We apply the induction hypothesis to C_{e}, with derivations d_{1}, …, d_{e−1}, and x′_{1}, …, x′_{e−1}. This gives us a y′ ∈ C_{e} such that d_{i}(y′) = x′_{i} for i = 1, …, e−1. Finally we set y = y′ +z.

Lemma 4. Let K be a strict D–field and n > 0. Assume that we have an element a such that for all m < n

D_{i,p^{n}}D_{j,p^{m}}(a) = 0  (1)

for all i, j. Then there is an a′ in K such that for all j D_{j,p^{n}}(a′) = 0 and

D_{j,p^{m}}(a′) = D_{j,p^{m}}(a)

for all m < n.

Proof. Set x_{i} = D_{i,p^{n}}(a). If we can find a y in

F = {z ∈ K | D_{j,p^{m}}(z) = 0, for all j and all m < n} = K^{p^{n}}

such that D_{i,p^{n}}(y) = x_{i} for all i, a′ = a − y will do the job.

We observe first, that the x_{i} belong to F, because for all j and m < n

D_{j,p^{m}}x_{i} = D_{j,p^{m}}D_{i,p^{n}}(a) = D_{i,p^{n}}D_{j,p^{m}}(a) = 0.

The field F together with the derivations D_{i,p^{n}} satisfies the conditions of Lemma 3. So it remains only to check the conditions on the x_{i}:

D_{i,p^{n}}^{p−1}(x_{i}) = D_{i,p^{n}}^{p}(a) = 0
D_{i,p^{n}}(x_{j}) = D_{i,p^{n}}D_{j,p^{n}}(a) = D_{j,p^{n}}D_{i,p^{n}}(a) = D_{j,p^{n}}(x_{i})

Proof of Theorem 1: Let K be a strict D–field and n a natural number. Choose a p–basis b_{1}, …, b_{e} by Lemma 2 such that D_{i,1}(b_{j}) = δ_{i,j}. Now for every i, if we start with a = b_{i} and apply Lemma 4 n-times, we get an element b′_{i} such that for all 0 < m ≤ n D_{j,p^{m}}(b′_{i}) = 0 and D_{j,1}(b′_{i}) = D_{j,1}(b_{i}) for all j. (Note that (1) holds trivially, since all D_{j,p^{m}}(a) are 0 or 1.)

The b′_{i} form a canonical p–basis “of depth p^{n+1}”, i.e. we have for all 0 < m < p^{n+1}

D_{i,m}(b′_{j}) = \begin{cases}1 & if m = 1 and i = j \\ 0 & otherwise \end{cases} .

References

[1] Martin Ziegler. Separably closed fields with Hasse derivations. J. Symbolic Logic, 68:311–318, December 2003.