In this section we develop some of the theory of independent sets in algebraically closed fields and then define a ternary independence relation between a tuple of elements, a subset and a subfield of an algebraically closed fieldK. In Section 4.7
we will use this relation to give an algebraically closed field interpretation of the independence atomx⊥y.
LetKbe an algebraically closed field andD,E,Fsubfields ofK.
Lemma 4.5.1. LetA={a0, ..., an−1} ⊆K. The following are equivalent.
i) For every a ∈ A we have that a is transcendental over the subfield of K
generated byA\ {a} overF.
ii) A is algebraically independent overF, that is for every non-trivial polyno-
mialP(X0, ..., Xn−1)∈F[X0, ..., Xn−1] we have that
P(a0, ..., an−1)6= 0.
Proof. i) ⇒ ii) LetP(X0, ..., Xn−1)∈F[X0, ..., Xn−1] be a non-trivial polyno-
mial such that P(a0, ..., an−1) = 0. Suppose that P(X0, ..., Xn−1) is a con-
P(X0, ..., Xn−1) = 0, in the first case6 P(X0, ..., Xn−1) is trivial and in the
second 06=P(X0, ..., Xn−1) =P(a0, ..., an−1) = 0. Let theni∗be such thatXi∗
occurs in P(X0, ..., Xn−1) and let A\ {ai∗} =ak
1, ..., akn−1 . Then we have
thatP(Xi∗, ak
1, ..., akn−1) :=Q(Xi∗)∈F(A\ {ai∗})[Xi∗] andQ(ai∗) = 0 soai∗
is algebraic overF(A\ {ai∗}).
ii) ⇒ i) Let a ∈ A and A\ {a} =
ak1, ..., akn−1 . Suppose that a is
not transcendental over F(A\ {a}), then there exists a non-trivial Q(X0) ∈
F(A\ {a})[X0] such thatQ(a) = 0. Let Q(X0) = X p 0 + p−1 X i=0 ciX0i. Now Q(X0) ∈ F(A\ {a})[X0] so each
ci∈F(A\ {a}), which means that
ci= mi−1 X j=0 qj1,i···jn−1,ia j1,i k1 · · ·a jn−1,i kn−1
whereqj1,i···jn−1,i ∈Ffor everyj∈ {0, ..., mi−1}, for this see [30, Lemma 1.24]. So Q(X0) = X0p+ p−1 X i=0 mi−1 X j=0 qj1,i···jn−1,ia j1,i k1 · · ·a jn−1,i kn−1 X i 0.
Consider now the following polynomial
P(X0, Xk1, ..., Xkn−1) = X p 0+ p−1 X i=0 mi−1 X j=0 qj1,i···jn−1,iX j1,i k1 · · ·X jn−1,i kn−1 X i 0, P(X0, Xk1, ..., Xkn−1)∈F[X0, ..., Xn−1],P(X0, Xk1, ..., Xkn−1) is non-trivial and
P(a, ak1, ..., akn−1) = 0. Hence A is not algebraically independent overF.
If A ⊆ K is infinite, then we say that A is independent over F if every finite subset of A is independent over F. Using the above characterization of algebraic independence and the fact that if a∈K is algebraic over F(A) then
a is algebraic overF(A0) with A0 ⊆fin A, it is possible to prove the following
proposition.
Proposition 4.5.2. LetA⊆K, thenAis independent overFif and only if for
every a∈A we have thata is transcendental over the subfield ofKgenerated
byA\ {a} overF.
We denote by Pthe subfield of Kgenerated by the empty set, we call this field the prime field of K. We say that a set A ⊆ K is independent if it is independent overP.
Lemma 4.5.3 (Exchange Principle). LetA⊆K andb ∈K. If ais algebraic overF(A∪ {b}) butais not algebraic overF(A), thenbis algebraic overF(A∪
{a}).
Let Fbe a subfield ofE. We say that Eis algebraic overF or thatEis an algebraic extension ofFif every element of Eis algebraic overF.
Lemma 4.5.4 (Transitivity of Algebraic Dependence). If Dis algebraic over E, andEis algebraic overF, then Dis algebraic over F.
Proof. See [30, Lemma 8.7].
Definition 4.5.5. We say that B⊆E is atranscendence basis forEover Fif
B is algebraically independent overFandEis algebraic overF(B).
Proposition 4.5.6. LetB ⊆E, the following are equivalent: i) B is a maximally independent overFsubset ofE;
ii) B is a basis forEoverF;
iii) B is a minimal subset ofE such thatEis algebraic overF(B).
Proof. i) ⇒ ii). Suppose that there existsa ∈ E such that a is not algebraic over F(B). We claim that B∪ {a} is independent over F. Suppose not, then there exists b ∈B∪ {a} such that b is algebraic over F((B∪ {a})\ {b}). By hypothesisa is not algebraic over F(B) sob6=a and henceb is algebraic over F((B\{b})∪{a}). Notice now thatBis independent overF, sobis not algebraic
over F(B\ {b}). Hence by the Exchange Principle we have that ais algebraic
overF(B), a contradiction.
ii) ⇒iii). Suppose there existsB0 (B such thatEis algebraic overF(B0).
Letb∈B\B0, thenbis algebraic overF(B0) and sobis algebraic overF(B\{b})
sinceB0⊆B\ {b}. Hence B is not independent overF.
iii)⇒i). Suppose there existsb∈Bsuch thatbis algebraic overF(B\ {b}),
thenF(B) is algebraic overF(B\ {b}). By hypothesisEis algebraic overF(B) so by Transitivity of Algebraic Dependence we have that E is algebraic over
F(B \ {b}) which contradicts the minimality of B. Thus B is algebraically
independent overF. Let nowa∈E\Band suppose thatB∪ {a}is independent overF, thenais not algebraic overF(B) and soEis not algebraic overF(B).
Proposition 4.5.7. Let A1⊆K and A0⊆A1 independent over F. Then A0
can be extended to a maximally independent overFsubset ofA1. In particular
for every subfieldFandEofK, there existsA⊆E such thatAis a basis forE overF.
Proof. See [30, Theorem 8.13]. Notice that the proof of this theorem requires Zorn’s Lemma.
It is possible to show that any two (possibly infinite) transcendence bases forEoverFhave the same cardinality. The cardinality of a transcendence basis for E over F is called the transcendence degree of E over F and denoted by trdg(E/F).
The following two lemmas are not of particular interest but they will be relevant in the proof of Theorem 4.7.4, this is the reason for which we state them here.
Lemma 4.5.8. IfAis independent overF, then trdg(F(A)/F) =|A|.
Proof. We show thatAis a minimal subset ofF(A) such thatF(A) is algebraic overF(A), by Proposition 4.5.6 this suffices. Suppose that there isB(Asuch that F(A) is algebraic overF(B). Letb ∈A\B, thenbis algebraic over F(B) sobis algebraic overF(A\ {b})), sinceB⊆A\ {b}. ThusAis not independent
overF.
Lemma 4.5.9. Let A ⊆ K be independent over F. Let D0, D1 ⊆ A and
D0∩D1=∅, then
i) D0is independent overF(D1);
ii) trdg(F(D0)/F(D1)) = trdg(F(D0)/F).
Proof. i) Suppose thatD0is not independent overF(D1), then there existsd∈
D0such thatdis algebraic overF(D1∪(D0\ {d})). By hypothesisD0∩D1=∅,
sodis algebraic overF((D1∪D0)\ {d}). ThusD0∪D1 is dependent overF, a
contradiction.
ii) By i)D0 is independent overF(D1) and thus independent overF, hence
by Lemma 4.5.8 we have that trdg(F(D0)/F(D1)) =|D0|= trdg(F(D0)/F).
Proposition 4.5.10. trdg(F(a)/F)≤ |a|.
Proof. Leta0 ⊆abe independent over F and such thatF(a) is algebraic over
F(a0). Thena0 is a basis forF(a) overF.
Proposition 4.5.11. IfFis a subfield ofE, then trdg(D/E)≤trdg(D/F).
Proof. LetB be a basis for D overF, then B is independent over F and Dis
algebraic over F(B). Let B0 ⊆ B be such that B0 is independent overE and F(B) is algebraic overE(B0). By choiceB0 is independent overE, furthermore Dis algebraic overE(B0) becauseDis algebraic overF(B) andF(B) is algebraic
overE(B0). HenceB0 is a basis forDoverE.
The notion of transcendence degree allow us to define a notion of indepen- dence with many desirable properties.
Definition 4.5.12. We say thatais independent fromB overFif
trdg(F(a)/F(B)) = trdg(F(a)/F).
We writea^|tr
F B.
Proposition 4.5.13. The following are equivalent: i) a^|tr
FB;
ii) every basis forF(a) overFis a basis forF(a) overF(B);
iii) every maximally independent over F subset of F(a) is independent over F(B);
iv) ifA⊆F(a) is independent overF, thenAis independent overF(B).
Proof. i)⇒ii) LetC be a basis forF(a) over F, thenC is independent overF andF(a) is algebraic overF(C). LetC0⊆Cbe such thatC0is independent over
F(B) and F(C) is algebraic overF(B∪C0). By choiceC0 is independent over F(B), furthermore F(a) is algebraic over F(B ∪C0) because F(a) is algebraic
over F(B) andF(B) is algebraic over F(B∪C0). Hence C0 is a basis for F(a)
overF(B). Now, by hypothesis we have that trdg(F(a)/F) = trdg(F(a)/F(B))
and by Proposition 4.5.10 we have thatC is finite. Hence C=C0. ii) ⇒iii) Immediate from Proposition 4.5.6.
iii)⇒iv) Suppose that there existsDindependent overFbut not overF(B).
By Proposition 4.5.7,Dcan be extended to aD0 maximally independent overF
so there exists a maximally independent overFsubset ofF(a) that is dependent
overF(B).
iv)⇒i) LetD be a basis forF(a) overF, thenDis independent overFand so by the hypothesis it is independent overF(B). FurthermoreF(a) is algebraic overF(B∪D) becauseF(a) is algebraic overF(D). ThusD is a basis forF(a) overF(B) and hence trdg(F(a)/F(B)) = trdg(F(a)/F).
Lemma 4.5.14 (Monotonicity). Ifa^|tr FB andC⊆B, thena^| tr F C. Proof. By Proposition 4.5.11, trdg(F(a)/F(B))≤trdg(F(a)/F(C))≤trdg(F(a)/F). So if trdg(F(a)/F(B)) = trdg(F(a)/F), then trdg(F(a)/F(C)) = trdg(F(a)/F). Lemma 4.5.15 (Transitivity). a^|tr F b cif and only ifa^| tr Fbanda^| tr F(b)c. Proof. By Proposition 4.5.11, trdg(F(a)/F(b c))≤trdg(F(a)/F(b))≤trdg(F(a)/F). Thus trdg(F(a)/F(b c)) = trdg(F(a)/F) m trdg(F(a)/F(b)) = trdg(F(a)/F) and trdg(F(a)/F(b c)) = trdg(F(a)/F(b)).
Lemma 4.5.16 (Symmetry). Ifc^|trFb, then b^|trF c.
Proof. LetD be a basis forF(b) overF and C ={c0, ..., cm−1} be a basis for
F(c) over F. Notice that if D = ∅ or C = ∅, then b ^|trA c. In the first case we have that trdg(F(b)/F) = 0 = trdg(F(b)/F(c)). In the second we have that trdg(F(c)/F) = 0, which implies that F(c) is algebraic over F and hence that trdg(F(b)/F) = trdg(F(b)/F(c)) because if an element is transcendental over a field it is also transcendental over an algebraic extension of it.
Suppose that b^6 |trF c, then there existsd∈D such thatdis algebraic over
F(c∪(D\ {d})). By hypothesisF(c) is algebraic overF(C) so we can conclude
Now D is independent over F so there exists p∈ {0, ..., m−1} such that
d is not algebraic over F({c0, ..., cp−1} ∪(D\ {d})) but d is algebraic over
F({c0, ..., cp} ∪(D\ {d})).
But then by the Exchange Principle we have thatcpis algebraic overF(D∪
{c0, ..., cp−1}) and hence algebraic over F(b ∪ {c0, ..., cp−1}). Thus C is not
independent overF(b) and hencec^6 |trFb.
Corollary 4.5.17. a b^|tr Fc if and only ifa^| tr F candb^| tr F(a)c. Proof. a b^|tr Fc ⇔ c^| tr Fa b [by symmetry] ⇔ c^|tr Faandc^| tr F(a)b [by transitivity] ⇔ a^|tr Fcandb^| tr F(a)c [by symmetry]. Corollary 4.5.18. Ifa^|tr Fb anda b^| tr Fc, thena^| tr F b c. Proof. a^|tr Fb anda b^| tr Fc ⇓ a^|tr Fb andb a^| tr Fc ⇓ a^|tr Fb anda^| tr F(b)c [by Corollary 4.5.17] ⇓ a^|tr Fb c [by Transitivity].