On the image of derivations Kazuo Kishimoto

On the image of derivations

Kazuo Kishimoto

Department of Mathematics, Shinshu University 390 Matsumoto, Japan, e-mail: a88603@sinet.ad.jp

Andrzej Nowicki

Faculty of Mathematics and Informatics, N. Copernicus University 87-100 Toru´n, Poland, e-mail: anow@mat.uni.torun.pl

Introduction. Let L be a field. An additive mapping d : L −→ L is said to be a derivation of L if d(ab) = d(a)b + ad(b) for any a, b ∈ L. If K ⊂ L is a field extension, then a derivation d of L is said to be a K-derivation if d(αa) = αd(a) for any α ∈ K and a ∈ L. If d is a derivation of L such that d(L) = L, then we say that d is epimorphic.

(a) Let K ⊂ L be a finite dimensional field extension and let d be a K- derivation of L. Then L has a direct sum decomposition L = Ker d ⊕ L⁰, as a K-module, where L⁰ is K-isomorphic to d(L). Thus dim_KL > dim_Kd(L) yields d(L) 6= L for any K-derivation d.

(b) Assume now that L is a purely transcendental extension with infinite transcendental basis {x_λ; λ ∈ Λ} over Q, the field of rational numbers. Since

|L| = |Λ| (the cardinalities of L and Λ), we may put L = {aλ; λ ∈ Λ}. Then the Q-derivation d of L defined by d(x_λ) = a_λ is epimorphic.

Thus it is natural to ask whether a field extension K ⊂ L has (or has not) an epimorphic K-derivation. The purpose of this paper is to study on this problem.

For an extension of a derivation of a field, the following are known ([1], The- orem 4.3.5). We use these facts in the proof of our results.

Let d be a derivation of K and let L = K(a) be a simple extension of K.

(1) If a is transcendental over K, then for each ν ∈ L, there is an extension D of d to a derivation of L such that D(a) = ν.

(2) If a is separably algebraic over K, then an extension D exists and is unique.

(3) Assume that K is of characteristic p > 0 and a^p ∈ K, a 6∈ K. Then there is an extension D of d to a derivation of L if and only if d(a^p) = 0. If this condition is satisfied, then for every ν ∈ L, there is such a derivation D with D(a) = ν.

1. The case of characteristic 0. In this section, we assume that all fields considered are of characteristic 0. The purpose of this section is to prove the following theorem.

Theorem 1.1. Let K ⊂ L be fields and let d : L −→ L be a K-derivation. If transcendence degree of L over K is finite, then d(L) 6= L.

First we note the following evident lemma.

Lemma 1.2. Let L be a field, d : L −→ L a derivation, a ∈ L r {0} and δ = ad.

Then d(L) = L if and only if δ(L) = L.

The next fact is a special case of our theorem.

Proposition 1.3. Let M = K(x₁, . . . , x_n) be the field of rational functions over a field K and let L be a finite field extension of M. If d : L −→ L is a K-derivation, then d(L) 6= L.

Proof. Let s + 1 = dim_ML and let us denote by R the polynomial ring K[x₁, . . . , x_n]. Since the extension M ⊂ L is finite separable, there exists an element γ ∈ L such that L = M [γ] is a simple algebraic field extension.

Let f = t^s+1+ a_st^s+ · · · + a₁t + a₀ be the minimal polynomial of γ over M and let q be a common multiple of all denominators of the fractions a₀, . . . , a_s. Multiplying by q^s+1 the equality γ^s+1+ a_sγ^s+ · · · + a₁γ + a₀ = 0 we obtain

(qγ)^s+1+ b_s(qγ)^s+ · · · + b₁(qγ) + b₀ = 0,

for some b₀, . . . , b_s ∈ R. Moreover, M [qγ] = M [γ]. Therefore we may assume that all coefficients a₀, . . . , a_s of the polynomial f belong to R. By this assumption we see that the set

S = {b₀+ b₁γ + · · · + b_sγ^s ; b₀, . . . b_s ∈ R}

is a subring of L.

Now let d : L −→ L be a K-derivation and suppose that d(L) = L.

If d(x₁) = · · · = d(x_n) = 0, then d is an M -derivation of L, and so d = 0 because the extension M ⊂ L is separable algebraic (see [1]). In this case we have a contradiction L = d(L) = 0.

Therefore there exists i ∈ {1, . . . , n} such that d(x_i) 6= 0. By a permutation of variables we may assume that d(x₁) 6= 0.

Let us consider the K-derivation d₁ = d(x₁)⁻¹d. Then d₁(x₁) = 1 and, by Lemma 1.2, d₁(L) = L. Moreover (since L = M [γ]):

d₁(x_i) =

s

X

j=0

A_ijγ^j, for i = 2, . . . , n,

d₁(γ) =

s

X

j=0

B_jγ^j, where all elements A_ij and B_j belong to M .

Denote by A a common multiple of denominators of all elements A_ij and B_j. Put δ = Ad1. Then δ is a K-derivation of L such that

δ(x₁) = A, δ(xi) =

s

X

j=0

Pijγ^j, for i = 2, . . . , n,

δ(γ) =

s

X

j=0

Q_jγ^j,

where A 6= 0, P_ij, Q_j are elements of R. Moreover, by Lemma 1.2, δ(L) = L and observe that δ(S) ⊂ S.

Since K is infinite, there exists an element a ∈ K such that (x₁ − a) - A (where ”|” means the divisibility in R). Denote y = x₁− a.

Now we shall show that 1/y 6∈ δ(L). Suppose that there exists u ∈ L such that 1/y = δ(u). Put u = w/b, w = w₀+ w₁γ + · · · + w_sγ^s, where b, w₀, . . . , w_s ∈ R and gcd(b, w₀, . . . , w_s) = 1. Then

1/y = δ(w/b) = (δ(w)b − δ(b)w)/b², so

b² = y(δ(w)b − δ(b)w),

hence y | b² because the element (δ(w)b − δ(b)w) belongs to S. Let b = y^mb1, where b₁ ∈ R, y - b1 and m > 1. Then

y^2mb²₁ = b²

= y(δ(w)b − δ(b)w)

= y(δ(w)y^mb₁− y^mδ(b₁)w − my^m−1Ab₁w)

= y^m+1(δ(w)b₁− δ(b₁)w) − my^mAb₁w,

and we see that yv = mAb₁w, for some v ∈ S. Let v = v₀+ v₁γ + · · · + v_sγ^s, where v₀, . . . , v_s∈ R. Then

(yv0) + (yv1)γ + · · · + (yvs)γ^s = (mAb1w0) + (mAb1w1)γ + · · · + (mAb1ws)γ^s, that is, y | Ab₁w₀, y | Ab₁w₁, . . . , y | Ab₁w_s. Since y - A, y - b1 and y = (x₁ − a) is an irreducible polynomial in R we have: y | w₀, . . . y | w_s. Moreover y | b. But this contradicts the assumption gcd(b, w₀, . . . , w_s) = 1. Therefore we showed that 1/y 6∈ δ(L) which completes the proof of our proposition.

The following lemma is given in [2], but we include here a proof for the sake of completeness.

Lemma 1.4. Let F ⊂ L be an algebraic field extension and let d : L −→ L be a derivation such that d(F ) ⊂ F . If δ(F ) 6= F , then d(L) 6= L, where δ = d | F .

Proof. Let u ∈ F r δ(F ). We shall show that u 6∈ d(L). Suppose that u = d(γ) for some γ ∈ L. Since γ is algebraic over F ,

γ^s+ a_s−1γ^s−1+ · · · + a₁γ + a₀ = 0, for some a0, . . . as−1 ∈ F . Assume that s is minimal. Then

0 = d(0) = (sγ^s−1+(s−1)a_s−1γ^s−2+· · ·+a₁)u+δ(a_s−1)γ^s−1+· · ·+δ(a₁)γ +δ(a₀), so, by minimality of s, su + δ(a_s−1) = 0. Hence u = δ(−a_s−1/s) ∈ δ(F ), a contradiction.

Now we are in a position to prove our theorem.

Proof of Theorem 1.1. Let {x₁, . . . , x_n} be a transcendental basis of L over K and denote M = K(x₁, . . . , x_n), F = M (d(x₁), . . . , d(x_n)). Then K ⊆ M ⊆ F ⊆ L and the extension M ⊆ F is finite, and the extension F ⊆ L is algebraic.

Observe that d(M ) ⊆ F . In fact, if w ∈ M then d(w) = ∂w

∂x₁d(x₁) + · · · + ∂w

∂x_nd(x_n) ∈ F.

Since M ⊂ F is finite, there exists γ ∈ F such that F = M [γ] is a simple algebraic field extension. Let

f = t^s+ a_s−1t^s−1+ · · · + a₁t + a₀ ∈ M [t]

be the minimal polynomial of γ over M. Then we have 0 = d(f (γ)) = f^d(γ) + f⁰(γ)d(γ), where

f^d= d(a_s−1)t^s−1+ · · · + d(a₁)t + d(a₀),

and f⁰ is the derivative of f . It is clear that f⁰(γ) 6= 0, f⁰(γ) ∈ F and f^d(γ) ∈ F (since d(M ) ⊆ F ). Therefore d(γ) = −f^d(γ)/f⁰(γ) is an element of F and hence d(F ) ⊆ F .

Denote δ = d | F . Then, by Proposition 1.3, δ(F ) 6= F and, by Lemma 1.4, d(L) 6= L.

As a consequence of Theorem 1.1, we have a slight generalization of (b) of the Introduction.

Corollary 1.5. Let k be an algebraic number field and let L be an extension field of K. Then there is an epimorphic K-derivation of L if and only if the transcendence degree of L over K is infinite.

Proof. If there is an epimorphic K-derivation, then tr.deg_KL (the transcen- dence degree of L over K) must be infinite by Theorem 1.1. Conversely, assume tr.deg_KL is infinite and S is a transcendental basis of L over K. Noting that

|K| = |Q|, we have |K(S)| = |S|, and |L| = |K(S)| since the extension K(S) ⊆ L is algebraic. Then we can construct an epimorphic K-derivation by the same way as in (b) of the Introduction.

2. The case of characteristic p > 0. In this section, we assume that K ⊆ L is an extension field of characteristic p > 0 and M = K(L^p), where L^p = {α^p; α ∈ L}.

Proposition 2.1. If dim_ML = λ is infinite, then there exists an epimorphic K-derivation d of L.

Proof. (i) case λ = ℵ0. We may assume that L = M [y1, . . . , yn, . . . ], where {y₁, . . . , y_n, . . . } is a p-basis of L over M . Then we have a sequence

M = M₀ ⊂ M₁ ⊂ · · · ⊂ M_n ⊂ . . .

of intermediate fields, where M_n = M [y₁, . . . , y_n] for n = 1, 2, . . . . Then we can define an M -derivation d of L by

d(y_i) = y_i−1, for i = 1, 2, . . . ,

where y0 = 1. It is easy to see that for any Mk, there exists a positive integer s = s(k) such that d^s(M_k) = 0. We shall show that d is epimorphic.

For α = y_i^k₁¹· · · y_i^k_nⁿ ∈ L with i1 < i2 < · · · < in, it suffices to prove that α ∈ d(L). First we note that F = M (y_i1, . . . , y_in) is contained in M_in, and hence d^m(F ) = 0 for some m.

Put α = yx where y = y_i^k1

1y_i^k2

2 · · · y_i^k_nⁿ⁻¹ and x = y_in. For z_j = y_in+j, d(yz1) = d(y)z1+ yd(z1) = d(y)z1+ α,

d(d(y)z2) = d²(y)z2+ d(y)z1. Thus

d(β2) = α − d²(y)z2,

where β₂ = yz₁− d(y)z₂. Repeating same procedures, we have d(β_m) = α ± d^m(y)z_m,

for some β_m ∈ L. Therefore α = d(β_m) ∈ d(L) since d^m(y) = 0.

(ii) case λ > ℵ₀. For a p-basis P of L over M , we may put P = P₁∪ P₂ where

|P₁| = ℵ₀, |P₂| = λ and P₁∩ P₂ = ∅. Let S be the subfield of L generated by P₂ over M . Then dim_SL = ℵ₀, and hence there exists an S-derivation (and so a K-derivation) d of L such that d(L) = L by (i).

Theorem 2.2. If K ⊂ L are fields of characteristic p > 0 and M = K(L^p), then d(L) 6= L for any K-derivation d of L, if and only if dimML is finite.

Proof. If d(L) 6= L for any K-derivation d of L, then dim_M L must be finite by Proposition 2.1. Conversely assume that dim_M L is finite. Since any K-derivation d of L is an M -derivation, we can see that d(L) 6= L as in the Introduction (remark (a)).

The following example shows that Theorem 1.1 is not valid in positive chara- cteristics.

Example 2.3. Let K = GF(p)(x₁, . . . , x_n, . . . ) be a field of rational functions over GF(p) with infinitely many variables and let L = K(y₁, . . . , y_n, . . . ) with y^p_i = xi for i = 1, 2, . . . . Then the extension K ⊂ L is algebraic (and hence tr.deg_KL = 0) and the equalities M = K(L^p) = K(y₁^p, . . . , y_n^p, . . . ) = K(x₁, . . . , x_n, . . . ) = K show that dim_ML is infinite. Thus there exists an epimorphic K-derivation of L (by Theorem 2.2) though tr.deg_KL is finite.

Acknowledgement. This note was written when the second author was a guest at Department of Mathematics of Shinshu University (Matsumoto, Japan).

He acknowledges for hospitality of the Department and excellent working condi- tions.

References

[1] M. Nagata, Field Theory, Marcel Dekker, Inc., New York and Basel, 1977.

[2] S. Suzuki Some types of derivations and their applications to field theory, J. Math. Kyoto Univ., 21(1981), 375 – 382.