Infinite Field Extensions
Chapter 7 was almost entirely devoted to field extensions of finite degree and con-
centrated on Galois theory. However, even an introductory account should make
some mention of infinite field extensions, and we shall discuss them in the present
chapter, including transcendental extensions (Section 11.3) and infinite Galois
theory (Section 11.8). The notion of algebraic dependence has similarities to linear
dependence, which are described in abstract form in Section 11.1 and applied in
Section 11.2. In addition there are concise accounts of topics that are useful in com-
mutative ring theory and algebraic geometry, besides being of independent interest:
separability (Sections 11.4 and ll.S), the interactions of two or more subfields
(Sections 11.6 and 11.7), applications of Galois theory (Section 11.9) and abelian
extensions of finite exponent (Section 11.1 0).
11.1 Abstract Dependence Relations
The notion of algebraic dependence is similar to that oflinear dependence in a vector
space, but there is not the same intuitive picture. For this reason it is advisable to
begin by presenting an account of abstract dependence relations; this concept also
occurs elsewhere, e.g. in some geometrical constructions, and in this section we
shall derive their general properties.
Let S be a set; by a dependence relation on S we understand a rule which associates
with each finite subset X of S certain elements of S, said to be dependent on X, and
subject to the following conditions:
D.O If X = {Xl, ... ,xnL then each Xi is dependent on X.
D.1 (Transitivity) If z is dependent on Y and each member ofY is dependent on X, then
z is dependent on X.
D.2 (Exchange property) If y is dependent on {Xi> ... , xnL but not on {X2, ... , xnL
then Xl is dependent on {y, X2, ... , xn}.
We note that by D.O and D.1, if Y is dependent on X' and X'S; X, then y is
dependent on X.
The notion of dependence can be extended to infinite sets by saying that y is
dependent on an infinite subset X of S if there is a finite subset X' of X such that
y is dependent on X'. Then it is easily seen that if the exchange property D.2
397
,398 Infinite Field Extensions
holds for finite subsets, it also holds for infinite subsets. Similarly D.O and D.I hold
for arbitrary subsets.
As an example let us consider linear dependence in a vector space over a field or,
more generally, in a module M over a ring R. Given a left R-module M and a subset X
of M, an element y of M is said to be linearly dependent on X if
y = L CiXi for some Ci E R, where Xi EX. (11.1.1)
The element y is said to satisfy a non-trivial relation with the elements of X if
by + L aiXi = 0 for some ai, b E R, b i= 0, where Xi EX. (11.1.2)
It is clear that (11.1.1) implies (11.1.2) (with b = 1, ai = -Ci). Conversely, if (11.1.2)
holds with an invertible element b, then (11.1.1) follows by taking Ci = -b - I ai. This
shows in particular that (11.1.1) and (11.1.2) are equivalent for vector spaces over a
field.
Now D.O-D.2 are easily verified for linear dependence as defined by (11.1.1) or
(11.1.2) over a field. We prove D.2 as an example. If y is dependent on x[, ... , Xn
then we have a relation (11.1.2), and if y is not dependent on X2, ••• , Xn then al
cannot be zero, so (11.1.2) shows XI to be dependent on y, X2, ••• , X n •
For modules over a ring, (11.1.1) and (11.1.2) are no longer equivalent. The rela-
tion (11.1.1) always satisfies D.O and D.I but not always D.2. For example, in Z2 (as
Z-module) (2,2) depends on (1,0) and (0,1) but not on (0,1) alone, yet (1,0) does
not depend on (2,2) and (0,1), i.e. there is no relation (1,0) = m(2, 2) + nCO, 1),
m, n E Z. Similarly the relation (11.1.2) satisfies D.O and D.2 but not necessarily
D.I (see Exercise 2).
Given a dependence relation on a set 5, we define for each X S; S the span of X as
the set (X) of all elements of 5 dependent on X; we also say that (X) is spanned by X.
From D.O we see that(X) 2 X and D.I shows that every element dependent on (X)
itself belongs to (X).
A family X = {xiii E I} of elements of 5 is said to be independent if no Xi is depen-
dent on {xjlj i= i}; otherwise X is called dependent. Clearly an independent family
consists of distinct elements. Moreover, being independent is a property of finite
character, for a family is dependent whenever this is true of some finite subfamily.
An independent family which spans 5 is called a basis of S. This of course generalizes
the usual definition of a basis for a vector space; as in that case, we can recognize
bases in the following way:
Proposition 11.1.1. Let S be a set with a dependence relation. Then for any subset X of
5 the following conditions are equivalent:
(a) X is a maximal independent subset of 5,
(b) X is a minimal spanning set,
(c) X is a basis of S.
Proof. (a) {} (c). Let X be a maximal independent subset of 5; we have to show that
every element of 5 is dependent on X. Given YES, if Y E X, then y is dependent on X
,11.1 Abstract Dependence Relations 399
by 0.0, and if Y ¢ X, then X U {y} is dependent, by maximality, so some element is
dependent on the rest, say x E X is dependent on X' U {y}, where X' = X\{x}. Since
X is independent, x is not dependent on X', hence by 0.2, Y is dependent on
X' U {x} = X, and this shows that X spans 5. Conversely, if X is a basis, it is indepen-
dent, but every element not in X is dependent on X, so X is maximal independent.
(b) <=} (c). Let X be a minimal spanning set; we have to show that X is indepen-
dent. If X were dependent, say x E X is dependent on the rest of X, then we could
omit x from X and still have a spanning set, by 0.1. This contradicts the minimality,
hence X is a basis. Conversely, if X is a basis, it is independent, so no element is
dependent on the rest, and hence X is a minimal spanning set. •
It is instructive to go through this proof in a particular case, and the reader is
advised to do this, say for vector spaces over a skew field, taking first the case
where the basis is finite and then the general case.
Our main aim will be to show that every set with a dependence relation has a basis,
and that any two bases of a given set have the same number of elements. This will
again show that a vector space, even infinite-dimensional, has a basis, and a uniquely
defined dimension. We first obtain an important consequence of 0.2.
Lemma 11.1.2 (Exchange lemma). Let 5 be a set with a dependence relation. If X is an
independent subset and Y is a spanning set of 5, then we can complete X to a basis of 5
by elements from Y, i.e. there is a subset Y' ofY such that X nY' = 0 and XU Y' is a
basis of 5. Moreover, if Y is finite, then X is finite and
IX U Y'I ::: IYI· (11.1.3)
Proof. Consider the collection f of all independent subsets Z of 5 such that
X~ZUY. (11.1.4)
Since independence is a property of finite character, f is inductive, and so by Zorn's
lemma f has a maximal member B, i.e. a maximal independent set B such that
X ~ B ~ XU Y. This shows that B consists of X together with some elements of
Y, thus B = X U Y'; by omitting from Y' elements that are also in X we ensure
that X nY' = 0. Now every element of Y depends on B, by maximality, hence
(B) 2 (Y) = 5, therefore B is a basis, as claimed.
Assume now that Y is finite. We shall show that starting from any finite subset X'
of X we can obtain a basis by adjoining at most IYI - IX'I elements of Y. This will
show that IX'I ::: IYI and for X' = X we reach the desired conclusion (11.1.3). We
shall use induction on IX'I; when IX'I = 0, we can by the first part choose a
subset of Y which is a basis. Now suppose that IX'I = r > 0 and that we have
a basis Br- l = {Xl, ... ,Xr-l, Yr, ... ,Yn}, where Xi EX', Yj E Y. With another ele-
ment Xr of X' we form the spanning set Br- l U {xr} which contains the independent
subset {Xl, ... , x r}. By the first part it contains a basis Br including Xl, ... ,X" but
Br cannot include all of y" ... ,Yn because the family {Xl, ... , Xr, Yr, ... ,Yn} is
dependent eXr is dependent on the rest); hence Br has the required form and it
contains at most n elements. By induction we obtain a basis of at most IYI elements,
, 400 Infinite Field Extensions
consisting of X' together with some elements of Y. In particular, IX'I ::: IYI and since
X' was any finite subset of X, it follows that X is finite and if the basis is X U Y', then
IX U Y'I ::: In •
By the last part, the number of elements in an independent family is bounded by
the number of elements in a finite spanning set. In particular, given two bases B, e, if
B is finite, then lei::: IBI and so by symmetry, IBI = Ic!. Thus we have
Corollary 11.1.3. Let S be a set with a dependence relation. If S has a finite spanning
set e, then any independent family has at most IC! members and any two bases have
the same number of elements. •
Our next task is to show that any two bases, finite or not, have the same cardinal.
For the case of infinite bases this is true in a rather wider context, which we shall now
explain, as it is sometimes useful.
Suppose that instead of a dependence relation we have a relation satisfying only
D.O and D.I (not necessarily D.2); let us call this a spanning relation. We can as
before define the span of a subset X as the set (X) of all elements dependent on X.
If we examine the proof of Proposition 11.1.1 we find that for a spanning relation
(c) {} (b) ~ (a), but a maximal independent set need no longer be a basis. In fact
the term 'basis' in this context tends to mislead and we shall only speak of minimal
spanning sets. The example of the rational numbers Q (as Z-module) shows that
minimal spanning sets need not exist. But when they do, we have the following
relation, generalizing the situation for modules (see Proposition 4.6.4).
Proposition 11.1.4. Let S be a set with a spanning relation and suppose that S has
a minimal spanning set X. If X is infinite, of cardinal ex, then any spanning set of S
has at least ex elements. In particular, S has no finite spanning set and any two minimal
spanning sets of S have the same cardinal.
Proof. This is similar to the module case and again uses the relation ~o{3 = {3 for infi-
nite cardinals (Proposition 1.2.7). Let Y be any spanning set of S, of cardinal {3. Each
yE Y is dependent on X and hence is dependent on a finite subset Xy of X. We claim
that
(11.1.5)
For the span of the right-hand side contains each y E Y and so must be S. Thus UXy
is a subset of X spanning S, and (11.1.5) follows by the minimality of X. If Y were
finite, then (11.1.5) would express X as a finite union of finite sets, but this contra-
dicts the fact that X is infinite. We thus have
Hence ex ::: {3; if Y is also minimal, we can interchange the roles of X and Y and so
conclude that ex = {3. •
Chapter 7 was almost entirely devoted to field extensions of finite degree and con-
centrated on Galois theory. However, even an introductory account should make
some mention of infinite field extensions, and we shall discuss them in the present
chapter, including transcendental extensions (Section 11.3) and infinite Galois
theory (Section 11.8). The notion of algebraic dependence has similarities to linear
dependence, which are described in abstract form in Section 11.1 and applied in
Section 11.2. In addition there are concise accounts of topics that are useful in com-
mutative ring theory and algebraic geometry, besides being of independent interest:
separability (Sections 11.4 and ll.S), the interactions of two or more subfields
(Sections 11.6 and 11.7), applications of Galois theory (Section 11.9) and abelian
extensions of finite exponent (Section 11.1 0).
11.1 Abstract Dependence Relations
The notion of algebraic dependence is similar to that oflinear dependence in a vector
space, but there is not the same intuitive picture. For this reason it is advisable to
begin by presenting an account of abstract dependence relations; this concept also
occurs elsewhere, e.g. in some geometrical constructions, and in this section we
shall derive their general properties.
Let S be a set; by a dependence relation on S we understand a rule which associates
with each finite subset X of S certain elements of S, said to be dependent on X, and
subject to the following conditions:
D.O If X = {Xl, ... ,xnL then each Xi is dependent on X.
D.1 (Transitivity) If z is dependent on Y and each member ofY is dependent on X, then
z is dependent on X.
D.2 (Exchange property) If y is dependent on {Xi> ... , xnL but not on {X2, ... , xnL
then Xl is dependent on {y, X2, ... , xn}.
We note that by D.O and D.1, if Y is dependent on X' and X'S; X, then y is
dependent on X.
The notion of dependence can be extended to infinite sets by saying that y is
dependent on an infinite subset X of S if there is a finite subset X' of X such that
y is dependent on X'. Then it is easily seen that if the exchange property D.2
397
,398 Infinite Field Extensions
holds for finite subsets, it also holds for infinite subsets. Similarly D.O and D.I hold
for arbitrary subsets.
As an example let us consider linear dependence in a vector space over a field or,
more generally, in a module M over a ring R. Given a left R-module M and a subset X
of M, an element y of M is said to be linearly dependent on X if
y = L CiXi for some Ci E R, where Xi EX. (11.1.1)
The element y is said to satisfy a non-trivial relation with the elements of X if
by + L aiXi = 0 for some ai, b E R, b i= 0, where Xi EX. (11.1.2)
It is clear that (11.1.1) implies (11.1.2) (with b = 1, ai = -Ci). Conversely, if (11.1.2)
holds with an invertible element b, then (11.1.1) follows by taking Ci = -b - I ai. This
shows in particular that (11.1.1) and (11.1.2) are equivalent for vector spaces over a
field.
Now D.O-D.2 are easily verified for linear dependence as defined by (11.1.1) or
(11.1.2) over a field. We prove D.2 as an example. If y is dependent on x[, ... , Xn
then we have a relation (11.1.2), and if y is not dependent on X2, ••• , Xn then al
cannot be zero, so (11.1.2) shows XI to be dependent on y, X2, ••• , X n •
For modules over a ring, (11.1.1) and (11.1.2) are no longer equivalent. The rela-
tion (11.1.1) always satisfies D.O and D.I but not always D.2. For example, in Z2 (as
Z-module) (2,2) depends on (1,0) and (0,1) but not on (0,1) alone, yet (1,0) does
not depend on (2,2) and (0,1), i.e. there is no relation (1,0) = m(2, 2) + nCO, 1),
m, n E Z. Similarly the relation (11.1.2) satisfies D.O and D.2 but not necessarily
D.I (see Exercise 2).
Given a dependence relation on a set 5, we define for each X S; S the span of X as
the set (X) of all elements of 5 dependent on X; we also say that (X) is spanned by X.
From D.O we see that(X) 2 X and D.I shows that every element dependent on (X)
itself belongs to (X).
A family X = {xiii E I} of elements of 5 is said to be independent if no Xi is depen-
dent on {xjlj i= i}; otherwise X is called dependent. Clearly an independent family
consists of distinct elements. Moreover, being independent is a property of finite
character, for a family is dependent whenever this is true of some finite subfamily.
An independent family which spans 5 is called a basis of S. This of course generalizes
the usual definition of a basis for a vector space; as in that case, we can recognize
bases in the following way:
Proposition 11.1.1. Let S be a set with a dependence relation. Then for any subset X of
5 the following conditions are equivalent:
(a) X is a maximal independent subset of 5,
(b) X is a minimal spanning set,
(c) X is a basis of S.
Proof. (a) {} (c). Let X be a maximal independent subset of 5; we have to show that
every element of 5 is dependent on X. Given YES, if Y E X, then y is dependent on X
,11.1 Abstract Dependence Relations 399
by 0.0, and if Y ¢ X, then X U {y} is dependent, by maximality, so some element is
dependent on the rest, say x E X is dependent on X' U {y}, where X' = X\{x}. Since
X is independent, x is not dependent on X', hence by 0.2, Y is dependent on
X' U {x} = X, and this shows that X spans 5. Conversely, if X is a basis, it is indepen-
dent, but every element not in X is dependent on X, so X is maximal independent.
(b) <=} (c). Let X be a minimal spanning set; we have to show that X is indepen-
dent. If X were dependent, say x E X is dependent on the rest of X, then we could
omit x from X and still have a spanning set, by 0.1. This contradicts the minimality,
hence X is a basis. Conversely, if X is a basis, it is independent, so no element is
dependent on the rest, and hence X is a minimal spanning set. •
It is instructive to go through this proof in a particular case, and the reader is
advised to do this, say for vector spaces over a skew field, taking first the case
where the basis is finite and then the general case.
Our main aim will be to show that every set with a dependence relation has a basis,
and that any two bases of a given set have the same number of elements. This will
again show that a vector space, even infinite-dimensional, has a basis, and a uniquely
defined dimension. We first obtain an important consequence of 0.2.
Lemma 11.1.2 (Exchange lemma). Let 5 be a set with a dependence relation. If X is an
independent subset and Y is a spanning set of 5, then we can complete X to a basis of 5
by elements from Y, i.e. there is a subset Y' ofY such that X nY' = 0 and XU Y' is a
basis of 5. Moreover, if Y is finite, then X is finite and
IX U Y'I ::: IYI· (11.1.3)
Proof. Consider the collection f of all independent subsets Z of 5 such that
X~ZUY. (11.1.4)
Since independence is a property of finite character, f is inductive, and so by Zorn's
lemma f has a maximal member B, i.e. a maximal independent set B such that
X ~ B ~ XU Y. This shows that B consists of X together with some elements of
Y, thus B = X U Y'; by omitting from Y' elements that are also in X we ensure
that X nY' = 0. Now every element of Y depends on B, by maximality, hence
(B) 2 (Y) = 5, therefore B is a basis, as claimed.
Assume now that Y is finite. We shall show that starting from any finite subset X'
of X we can obtain a basis by adjoining at most IYI - IX'I elements of Y. This will
show that IX'I ::: IYI and for X' = X we reach the desired conclusion (11.1.3). We
shall use induction on IX'I; when IX'I = 0, we can by the first part choose a
subset of Y which is a basis. Now suppose that IX'I = r > 0 and that we have
a basis Br- l = {Xl, ... ,Xr-l, Yr, ... ,Yn}, where Xi EX', Yj E Y. With another ele-
ment Xr of X' we form the spanning set Br- l U {xr} which contains the independent
subset {Xl, ... , x r}. By the first part it contains a basis Br including Xl, ... ,X" but
Br cannot include all of y" ... ,Yn because the family {Xl, ... , Xr, Yr, ... ,Yn} is
dependent eXr is dependent on the rest); hence Br has the required form and it
contains at most n elements. By induction we obtain a basis of at most IYI elements,
, 400 Infinite Field Extensions
consisting of X' together with some elements of Y. In particular, IX'I ::: IYI and since
X' was any finite subset of X, it follows that X is finite and if the basis is X U Y', then
IX U Y'I ::: In •
By the last part, the number of elements in an independent family is bounded by
the number of elements in a finite spanning set. In particular, given two bases B, e, if
B is finite, then lei::: IBI and so by symmetry, IBI = Ic!. Thus we have
Corollary 11.1.3. Let S be a set with a dependence relation. If S has a finite spanning
set e, then any independent family has at most IC! members and any two bases have
the same number of elements. •
Our next task is to show that any two bases, finite or not, have the same cardinal.
For the case of infinite bases this is true in a rather wider context, which we shall now
explain, as it is sometimes useful.
Suppose that instead of a dependence relation we have a relation satisfying only
D.O and D.I (not necessarily D.2); let us call this a spanning relation. We can as
before define the span of a subset X as the set (X) of all elements dependent on X.
If we examine the proof of Proposition 11.1.1 we find that for a spanning relation
(c) {} (b) ~ (a), but a maximal independent set need no longer be a basis. In fact
the term 'basis' in this context tends to mislead and we shall only speak of minimal
spanning sets. The example of the rational numbers Q (as Z-module) shows that
minimal spanning sets need not exist. But when they do, we have the following
relation, generalizing the situation for modules (see Proposition 4.6.4).
Proposition 11.1.4. Let S be a set with a spanning relation and suppose that S has
a minimal spanning set X. If X is infinite, of cardinal ex, then any spanning set of S
has at least ex elements. In particular, S has no finite spanning set and any two minimal
spanning sets of S have the same cardinal.
Proof. This is similar to the module case and again uses the relation ~o{3 = {3 for infi-
nite cardinals (Proposition 1.2.7). Let Y be any spanning set of S, of cardinal {3. Each
yE Y is dependent on X and hence is dependent on a finite subset Xy of X. We claim
that
(11.1.5)
For the span of the right-hand side contains each y E Y and so must be S. Thus UXy
is a subset of X spanning S, and (11.1.5) follows by the minimality of X. If Y were
finite, then (11.1.5) would express X as a finite union of finite sets, but this contra-
dicts the fact that X is infinite. We thus have
Hence ex ::: {3; if Y is also minimal, we can interchange the roles of X and Y and so
conclude that ex = {3. •
