From this section on we pass from the analysis of linear dependence and inde- pendence to the analysis of algebraic dependence and independence.
In this section in particular we define the system Algebraically Closed Field Atomic Dependence Logic (ACFADL) and then prove its soundness and com- pleteness. The syntax and deductive apparatus of this system are the same as those of ADL.
4.4.1
Semantics
In the context of algebraically closed fields we think of the atom =(x, y) as expressing that each element inyis bound to the elements inxvia the existence of a polynomial with coefficients from the subfield generated by the elements in
x. We do not define =(x, y) by saying that each variable iny is a polynomial or an algebraic expression of the elements in x, although we could and that would perhaps be worth studying. The reason for the adopted concept, which is also the concept of dependence used in algebra, is that there famously are polynomial equations of even as low degree as five, which cannot be solved in terms of radicals.
Definition 4.4.1. LetKbe an algebraically closed field ands: dom(s)→K
withx y ⊆dom(s)⊆Var. We say thatKsatisfies =(x, y) unders, in symbols
K|=s=(x, y), if for everyy∈y we have thats(y) is algebraic over the subfield
FofKgenerated by{s(x)|x∈x}, that is there exists a non-trivial polynomial
P(X) = Xn+ n−1
X
i=0
aiXi
with coefficients inFsuch thatP(s(y)) = 0.
Let Kbe an algebraically closed field,F a subfield ofKand a∈K. Ifa is not algebraic overF, then we say thatais transcendental overF.
Definition 4.4.2. Let Σ be a set of atoms and let s be such that the set of variables occurring in Σ is included in dom(s). We say thatKsatisfies Σ under
s, in symbolsK|=sΣ, ifKsatisfies every atom in Σ under s.
Definition 4.4.3. Let Σ be a set of atoms. We say that =(x, y) is a logical consequence of Σ, in symbols Σ|= =(x, y), if for every algebraically closed field
Kandssuch that the set of variables occurring in Σ∪ {=(x, y)}is included in
dom(s) we have that
if K|=sΣ then K|=s=(x, y).
4.4.2
Soundness and Completeness
Theorem 4.4.4. Let Σ be a set of atoms, then
Σ|= =(x, y) if and only if Σ`=(x, y).
Proof. (⇐) We prove only the soundness of rules (a1.) and (c1.) . LetKbe an
algebraically closed field andsan appropriate assignment.
(a1.) We want to show that K|=s =(x, x). If x=∅ this is trivially true, suppose then that x 6= ∅ and let x ∈ x and s(x) = a. Let P(X) = X−a, then clearlyP(a) = 0. Notice also thatP has coefficients in the subfieldFofK generated by{s(x)|x∈x} becausea∈F.
(c1.) First we state a useful characterization of algebraic dependence and a
lemma about field extensions, for the proofs of these results see [30, Proposi- tion 1.30 and Proposition 1.20].
Proposition 4.4.5. LetFbe a subfield of a fieldKandF(a) be the subfield of
Kgenerated overFby{a}. Thenais algebraic overFif and only if [F(a) :F] is
finite, where [F(a) :F] denotes the degree ofF(a) overF, that is the dimension
ofF(a) considered as a vector space overF.
Lemma 4.4.6. LetK, FandEbe fields with KextendingFand Fextending
E. ThenK/E is of finite degree if and only if K/Fand F/Eare both of finite
degree.
Suppose now that K|=s =(x, y) and K |=s =(y, z), we want to show that K|=s =(x, z). Let z∈ z, s(z) =c, s(x) =a ands(y) =b. Let then F be the
subfield ofKgenerated byaandF0 be the subfield ofKgenerated byb.
By assumption every b ∈b is algebraic overF, so by Proposition 4.4.5 for every b ∈ b we have that [F(b) : F] is finite and hence by Lemma 4.4.6 we have that [F(b) : F] is finite. Furthermore c is algebraic over F0 so clearly it
is algebraic overF0(a) and hence by Proposition 4.4.5 [F0(ac) : F0(a)] is finite.
Notice now thatF0(a) =F(b) because this field is nothing but the subfield ofK
generated bya b. Thus we conclude that both [F(b) : F] and [F(bc) : F(b)] are
finite and hence by Lemma 4.4.6 we have that [F(bc) :F] is finite.
Suppose now thatcis transcendental overF, then by Proposition 4.4.5 [F(c) : F] is infinite and hence by Lemma 4.4.6 [F(bc) : F] is also infinite, which is a
contradiction. Thusc is algebraic overFand henceK|=s=(x, y).
(⇒) Suppose Σ0=(x, y). LetV ={z∈Var|Σ`=(x, z)}andW = Var\V.
First notice that y 6=∅, indeed if not so, by the syntactic constraints that we put on the system, we have thatx, y = ∅ and so by the admitted degenerate case of rule (a1.) we have that Σ`=(x, y). Furthermore y∩W 6=∅, indeed if
y∩W =∅then for every y∈y we have that Σ`=(x, y) and so by rules (d1.),
(e1.) and, if necessary, (f1.)3 we have that Σ`=(x, y).
LetCbe the field of complex numbers and lets: Var→Cbe the following
assignment
s(v) =
(
0 ifv∈V π ifv∈W.
We claim that K6|=s =(x, y). In accordance to the semantic we then have
to show that there isy∈y such thats(y) is transcendental over the subfield of
C generated by{s(x)|x∈x}. Let y ∈ y∩W and x 6=∅, we then have that
s(y) =πand s(x) = 0 for everyx∈x, because forx∈xwe have Σ`=(x, x). Indeed by rule (a1.) `=(x, x) and so by rule (b1.) `=(x, x).
3Notice that (f
Notice that the subfield ofCgenerated by 0 is the fieldQof rational numbers and clearly πis transcendental over it. Finally if x=∅ then we are also done because the subfield ofCgenerated by the empty set is again the fieldQ.
Let now =(x0, y0)∈Σ, we want to show that K|=s=(x0, y0). Ify0 =∅ then alsox0=∅and so triviallyK|=s=(x0, y0). Noticed this, for the rest of the proof
we assumey06=∅.
Case 1. x0 =∅.
Suppose that K6|=s=(∅, y0), then there exists y0 ∈y0 such that s(y0) =π,
so Σ0=(x, y0). Notice though that Σ`=(∅, y0), so by rule (b1.) Σ`=(∅, y0)
and hence again by rule (b1.) Σ`=(x, y0).
Case 2. x0 6=∅ andx0 ⊆V. If this is the case, then
∀x0∈x0 Σ`=(x, x0) =⇒ Σ`=(x, x0) [by rules (d1.), (e1.) and (f1.)]
=⇒ Σ`=(x, y0) [by rule (c1.)]
=⇒ ∀y0∈y0 Σ`=(x0, y0) [by rule (b1.)]
=⇒ y0 ⊆V.
If x0 ⊆V then for every x0 ∈ x0 we have that s(x0) = 0 so again the subfield of C generated by {s(x0)|x0 ∈x0} is Q. Let now y0 ∈ y0, then we have that
s(y0) = 0 and clearly 0 is algebraic overQ.
Case 3. x0∩W 6=∅.
If this is the case, then there existsx0∈x0such that Σ0=(x, x0), so we have
x0∈x0such thats(x0) =π. Hence the subfield ofKgenerated by{s(x)|x∈x}
is Q(π) and then in both cases s(y0) = π and s(y0) = 0 we have algebraic
dependence.