Field Theory
Fields form one of the basic algebraic concepts, for which there is an extensive
theory, dealing mainly with the form taken by field extensions. In Section 7.1 field
extensions are described, and the special case of splitting fields is introduced in
Section 7.2, leading to the notion of algebraic closure (Section 7.3). In Section 7.4
we examine the problems arising in finite characteristic. One of the main tools in
this study is Galois theory and this forms the subject of Sections 7.5 and 7.6,
while Sections 7.10 and 7.11 bring its application to the solution of equations.
The special case of finite fields is studied in Section 7.8, using information on the
roots of unity (Section 7.7); Section 7.9 is devoted to generators and some invariants
of extensions.
7.1 Fields and their Extensions
We recall from Section 4.1 that a field is a commutative ring in which 1 =j:. 0 and
every non-zero element has an inverse. It follows that a field F has exactly two
ideals, 0 and F, and hence every homomorphism from a field to a non-trivial ring
is injective (because the kernel is a proper ideal). Obvious examples of fields are
the rational numbers Q and, for every prime number p, the field Fp = Zip of p
elements.
Let us also recall the notion of characteristic, basic in all that follows. In any field F
consider the multiples of 1 : 1, 1 + 1, 1 + 1 + 1, .... We abbreviate these expressions
as 1,2.1,3.1, ... and remark that two cases can arise:
(i) n. 1 =j:. 0 for all n > O. Then F is said to have characteristic O. In this case the
homomorphism n 1---+ n. 1 provides an embedding of Z in F; since F is a field,
we can invert the elements n. 1 and so form the sub field P generated by 1.
This is the prime subfield of F; clearly it is isomorphic to Q. It is contained in
every subfield of F and so is the least subfield of F.
(ii) For some n > 0 we have n. 1 = O. Since F has no zerodivisors, it follows easily
that the least such n is a prime p, say, which is called the characteristic of F in
this case. Now 0, 1,2.1, ... , (p - 1).1 already form a subfield P, isomorphic
to Fp ' which thus is the prime subfield.
189
,190 Field Theory
We see that the characteristic of F, often written char F, is either 0 or a prime
number p, and our results can be summed up as follows:
Theorem 7.1.1. Every field F has a least subfield P, the prime subfield of F, which is
contained in every subfield of F. Either (i) char F = 0 and P ~ Q, or (ii) char
F = p, a prime number, and then P ~ Fp. •
Let F be a field and k be a subfield; we shall often describe k as the ground field, F
as an extension of k and write F/k for the field F considered as an extension of k
(the risk of confusion with quotient rings RIa is small since fields have no non-
trivial quotients). Thus F is a k-algebra, in particular it is a vector space over k.
Its dimension is called the degree of F over k and is written [F: kJ; this notation
will sometimes be used even if F is only a vector space. The dimension is a positive
integer or 00; for example, [C : RJ = 2, [R : QJ = 00, and of course for every field k,
[k: kJ = 1. By a finite extension one understands an extension of finite degree. We
note that for every extension F/k, F and k have the same prime subfield, and hence
the same characteristic.
Let F/k be a finite extension, of degree n say. Then there is a basis UI, ... , Un for F
over k and relative to this basis each element a E F can be uniquely expressed in the
form
a =L CiiUi, where Cii E k. (7.1.1)
In order to know F completely we need only know k and the expressions (7.1.1) for
the n2 products UiUj, but it is not particularly easy to work with these n2 expressions
and we shall soon find simpler ways of describing the extension F/k. A k-homo-
morphism (of extensions of k) is a homomorphism which is k-linear, or equivalently,
one which reduces to the identity on k.
The following product formula is a basic tool in the study of field extensions.
Proposition 7.1.2. Let F/k be a field extension. Given any vector space V over F, we can
regard V also as a vector space over k; its dimension [V : kJ is finite if and only if both
[V : FJ and [F : kJ are finite, and when this is so, then
[V: kJ = [V: F][F: kJ. (7.1.2)
Proof. Suppose first that F/k is finite, with basis UI, ... , Urn say, and that V is finite-
dimensional over F, with basis VI, ... , v n • For clarity we shall use latin subscripts for
the range 1, ... , m of the first basis, and greek subscripts for the range 1, ... , n of the
F-basis of V. Thus the Ui form a basis for F/k and the VA form an F-basis of V; we shall
prove that the set of products UiVA is a k-basis for V. This will prove that
[V: kJ = mn, and (7.1.2) follows in this case.
Let x E V; since V is spanned by the VA over F, we have
(7.1.3)
,7.1 Fields and their Extensions 191
We express each a). in terms of the Ui : a). = L:a~)'iU where a).i E k and insert these
values in (7.l.3):
This shows that the mn products UiV). span V over k. To prove their linear indepen-
dence, let us assume a relation
(7.1.4)
On writing a). = L a).iUi, we see that this is of the form La). v). = 0, where a). E F,
and hence, by the linear independence of the v). we have a). = 0 (A = 1, ... , n). This
means that L a).iUi = 0 for all A, and since the Ui are linearly independent over k,
we conclude that a).i = 0 for all A, i. Hence the relation (7.104) is trivial and it follows
that the UiV), are linearly independent over k; this then shows that they form a k-basis
for V.
We observe that in order to prove that the UiV). are linearly independent we needed
only the linear independence of the sets {ud, {v).}, not the fact that they were bases.
This means that from any m linearly independent elements of F over k and any n
linearly independent elements of V over F we can form mn linearly independent
elements of V over k. Suppose now that V is of finite dimension N over k; then N
is a bound for the number of elements in a linearly independent set of vectors in
V over k. Hence mn :::: N and it follows that both [V: F] and [F: k] are finite.
Thus [V: k] is finite iff both [V: F] and [F: k] are finite, and when this is so,
(7.1.2) holds. •
We note particularly the case where V is an extension field E of F. Thus whenever
we have a tower of fields,
k S; F S; E, (7.1.5)
then we have the product formula:
[E : k] = [E : F][F : k], (7.1.6)
whenever either side is finite.
As in the case of Lagrange's theorem for groups (Theorem 2.1.3), this has the
useful consequence that for any tower of fields as in (7.1.5), the degree [F: k] is a
divisor of [E : k]. For example, an extension Elk of prime degree has no proper sub-
extensions. However, an extension without proper subextensions need not be of
prime degree; examples are easily constructed with the help of Galois theory, as
we shall see in Section 7.6.
Consider any field extension Flk. Given elements al, ... , an of F, we write
k[al,"" an] for the subring and k(al,"" an) for the subfield generated by
ai, ... , an over k. For an extension of finite degree we actually have equality:
(7.1.7)
, 192 Field Theory
for the right-hand side is a finite-dimensional k-algebra without zerodivisors, and so
is a field.
Let us take a closer look at a simple extension, n = 1. Such an extension has
the form k(a), where a E F. Consider first the subring k[a]; we have a unique
k-homomorphism A from the polynomial ring k[x] to k[a] which maps x to a. This
is clearly surjective and it gives rise to the exact sequence
0-+ ker A -+ k[x] -+ k[a] -+ O. (7.1.8)
There are two cases to consider:
(i) A is injective; it is then an isomorphism: k[ a] ~ k[x]. In this case we say that a is
transcendental over k. The field of fractions k(a) of k[a] is called a purely trans-
cendental extension of k.
(ii) A is not injective; in this case a is called algebraic over k. For complex numbers
these terms are used with reference to the rational numbers Q as ground field;
e.g. --/2 and --/ - 1 are algebraic numbers, while e and 7i are transcendental.
We observe that in (7.1.8) the kernel of A represents the set of all polynomials in x
which vanish for x = a. Thus a is algebraic precisely when it satisfies a non-trivial
polynomial equation over k. By (7.1.8) we have k[a] ~ k[x]/ker A; we claim that
k[a] is finite-dimensional over k. For ker A, being a non-zero ideal, contains a poly-
nomial f = coxn + CIX n- I + ... + Cn which is non-zero, say Co i= O. By definition we
have
coa n + ... + Cn = 0, (7.1.9)
and since we can divide by co, this equation shows an to be linearly dependent on
1, a, ... ,an-lOVer k. On multiplying (7.1.9) by a, we see that a n+ I is linearly
dependent on a, a 2 , ••• , an and hence on 1, a, ... , an-I. An easy induction shows
that all powers of a are linearly dependent on 1, a, ... ,a n- I and since k[a] is
spanned by the positive powers of a over k, it follows that k[ a] is finite-dimensional
over k. By the previous remark we see that k[a] is already a field. When (7.1.9) holds,
a is called a root of the equation f = 0; it is a zero (or also a root) of f
To study the kernel of A more closely, we recall that the polynomial ring k[x] is a
principal ideal domain. Explicitly, let a i= 0 be an ideal in k[x] and take a non-zero
polynomial p of least degree in a; then any f E a can be written as f = pq + r for
polynomials q, r, where deg r < deg p. Hence r = f - pq E a and by the minimality
of deg p, r must vanish, so that f = pq and a = (p). Hence every ideal of k[x] is
principal and so ker A is generated by the monic polynomial p (i.e. with highest
coefficient 1) of least degree with a as zero. If p has degree n, then k[a] is spanned
by 1, a, ... , an - I and these elements are linearly independent over k, for any depen-
dence would give an equation oflower degree satisfied by a. Thus 1, a, ... ,a n- I is a
basis of k[a] and we see that [k[a] : k] = n = degp. We note that p may always be
taken monic and it is then uniquely determined, for if a satisfied two different monic
equations of degree n, it would also satisfy their difference, an equation of lower
degree, contradicting the minimality of n. The monic polynomial p of least degree
satisfied by a is called the minimal polynomial for a over k, and deg p is called the
degree of a over k.
Fields form one of the basic algebraic concepts, for which there is an extensive
theory, dealing mainly with the form taken by field extensions. In Section 7.1 field
extensions are described, and the special case of splitting fields is introduced in
Section 7.2, leading to the notion of algebraic closure (Section 7.3). In Section 7.4
we examine the problems arising in finite characteristic. One of the main tools in
this study is Galois theory and this forms the subject of Sections 7.5 and 7.6,
while Sections 7.10 and 7.11 bring its application to the solution of equations.
The special case of finite fields is studied in Section 7.8, using information on the
roots of unity (Section 7.7); Section 7.9 is devoted to generators and some invariants
of extensions.
7.1 Fields and their Extensions
We recall from Section 4.1 that a field is a commutative ring in which 1 =j:. 0 and
every non-zero element has an inverse. It follows that a field F has exactly two
ideals, 0 and F, and hence every homomorphism from a field to a non-trivial ring
is injective (because the kernel is a proper ideal). Obvious examples of fields are
the rational numbers Q and, for every prime number p, the field Fp = Zip of p
elements.
Let us also recall the notion of characteristic, basic in all that follows. In any field F
consider the multiples of 1 : 1, 1 + 1, 1 + 1 + 1, .... We abbreviate these expressions
as 1,2.1,3.1, ... and remark that two cases can arise:
(i) n. 1 =j:. 0 for all n > O. Then F is said to have characteristic O. In this case the
homomorphism n 1---+ n. 1 provides an embedding of Z in F; since F is a field,
we can invert the elements n. 1 and so form the sub field P generated by 1.
This is the prime subfield of F; clearly it is isomorphic to Q. It is contained in
every subfield of F and so is the least subfield of F.
(ii) For some n > 0 we have n. 1 = O. Since F has no zerodivisors, it follows easily
that the least such n is a prime p, say, which is called the characteristic of F in
this case. Now 0, 1,2.1, ... , (p - 1).1 already form a subfield P, isomorphic
to Fp ' which thus is the prime subfield.
189
,190 Field Theory
We see that the characteristic of F, often written char F, is either 0 or a prime
number p, and our results can be summed up as follows:
Theorem 7.1.1. Every field F has a least subfield P, the prime subfield of F, which is
contained in every subfield of F. Either (i) char F = 0 and P ~ Q, or (ii) char
F = p, a prime number, and then P ~ Fp. •
Let F be a field and k be a subfield; we shall often describe k as the ground field, F
as an extension of k and write F/k for the field F considered as an extension of k
(the risk of confusion with quotient rings RIa is small since fields have no non-
trivial quotients). Thus F is a k-algebra, in particular it is a vector space over k.
Its dimension is called the degree of F over k and is written [F: kJ; this notation
will sometimes be used even if F is only a vector space. The dimension is a positive
integer or 00; for example, [C : RJ = 2, [R : QJ = 00, and of course for every field k,
[k: kJ = 1. By a finite extension one understands an extension of finite degree. We
note that for every extension F/k, F and k have the same prime subfield, and hence
the same characteristic.
Let F/k be a finite extension, of degree n say. Then there is a basis UI, ... , Un for F
over k and relative to this basis each element a E F can be uniquely expressed in the
form
a =L CiiUi, where Cii E k. (7.1.1)
In order to know F completely we need only know k and the expressions (7.1.1) for
the n2 products UiUj, but it is not particularly easy to work with these n2 expressions
and we shall soon find simpler ways of describing the extension F/k. A k-homo-
morphism (of extensions of k) is a homomorphism which is k-linear, or equivalently,
one which reduces to the identity on k.
The following product formula is a basic tool in the study of field extensions.
Proposition 7.1.2. Let F/k be a field extension. Given any vector space V over F, we can
regard V also as a vector space over k; its dimension [V : kJ is finite if and only if both
[V : FJ and [F : kJ are finite, and when this is so, then
[V: kJ = [V: F][F: kJ. (7.1.2)
Proof. Suppose first that F/k is finite, with basis UI, ... , Urn say, and that V is finite-
dimensional over F, with basis VI, ... , v n • For clarity we shall use latin subscripts for
the range 1, ... , m of the first basis, and greek subscripts for the range 1, ... , n of the
F-basis of V. Thus the Ui form a basis for F/k and the VA form an F-basis of V; we shall
prove that the set of products UiVA is a k-basis for V. This will prove that
[V: kJ = mn, and (7.1.2) follows in this case.
Let x E V; since V is spanned by the VA over F, we have
(7.1.3)
,7.1 Fields and their Extensions 191
We express each a). in terms of the Ui : a). = L:a~)'iU where a).i E k and insert these
values in (7.l.3):
This shows that the mn products UiV). span V over k. To prove their linear indepen-
dence, let us assume a relation
(7.1.4)
On writing a). = L a).iUi, we see that this is of the form La). v). = 0, where a). E F,
and hence, by the linear independence of the v). we have a). = 0 (A = 1, ... , n). This
means that L a).iUi = 0 for all A, and since the Ui are linearly independent over k,
we conclude that a).i = 0 for all A, i. Hence the relation (7.104) is trivial and it follows
that the UiV), are linearly independent over k; this then shows that they form a k-basis
for V.
We observe that in order to prove that the UiV). are linearly independent we needed
only the linear independence of the sets {ud, {v).}, not the fact that they were bases.
This means that from any m linearly independent elements of F over k and any n
linearly independent elements of V over F we can form mn linearly independent
elements of V over k. Suppose now that V is of finite dimension N over k; then N
is a bound for the number of elements in a linearly independent set of vectors in
V over k. Hence mn :::: N and it follows that both [V: F] and [F: k] are finite.
Thus [V: k] is finite iff both [V: F] and [F: k] are finite, and when this is so,
(7.1.2) holds. •
We note particularly the case where V is an extension field E of F. Thus whenever
we have a tower of fields,
k S; F S; E, (7.1.5)
then we have the product formula:
[E : k] = [E : F][F : k], (7.1.6)
whenever either side is finite.
As in the case of Lagrange's theorem for groups (Theorem 2.1.3), this has the
useful consequence that for any tower of fields as in (7.1.5), the degree [F: k] is a
divisor of [E : k]. For example, an extension Elk of prime degree has no proper sub-
extensions. However, an extension without proper subextensions need not be of
prime degree; examples are easily constructed with the help of Galois theory, as
we shall see in Section 7.6.
Consider any field extension Flk. Given elements al, ... , an of F, we write
k[al,"" an] for the subring and k(al,"" an) for the subfield generated by
ai, ... , an over k. For an extension of finite degree we actually have equality:
(7.1.7)
, 192 Field Theory
for the right-hand side is a finite-dimensional k-algebra without zerodivisors, and so
is a field.
Let us take a closer look at a simple extension, n = 1. Such an extension has
the form k(a), where a E F. Consider first the subring k[a]; we have a unique
k-homomorphism A from the polynomial ring k[x] to k[a] which maps x to a. This
is clearly surjective and it gives rise to the exact sequence
0-+ ker A -+ k[x] -+ k[a] -+ O. (7.1.8)
There are two cases to consider:
(i) A is injective; it is then an isomorphism: k[ a] ~ k[x]. In this case we say that a is
transcendental over k. The field of fractions k(a) of k[a] is called a purely trans-
cendental extension of k.
(ii) A is not injective; in this case a is called algebraic over k. For complex numbers
these terms are used with reference to the rational numbers Q as ground field;
e.g. --/2 and --/ - 1 are algebraic numbers, while e and 7i are transcendental.
We observe that in (7.1.8) the kernel of A represents the set of all polynomials in x
which vanish for x = a. Thus a is algebraic precisely when it satisfies a non-trivial
polynomial equation over k. By (7.1.8) we have k[a] ~ k[x]/ker A; we claim that
k[a] is finite-dimensional over k. For ker A, being a non-zero ideal, contains a poly-
nomial f = coxn + CIX n- I + ... + Cn which is non-zero, say Co i= O. By definition we
have
coa n + ... + Cn = 0, (7.1.9)
and since we can divide by co, this equation shows an to be linearly dependent on
1, a, ... ,an-lOVer k. On multiplying (7.1.9) by a, we see that a n+ I is linearly
dependent on a, a 2 , ••• , an and hence on 1, a, ... , an-I. An easy induction shows
that all powers of a are linearly dependent on 1, a, ... ,a n- I and since k[a] is
spanned by the positive powers of a over k, it follows that k[ a] is finite-dimensional
over k. By the previous remark we see that k[a] is already a field. When (7.1.9) holds,
a is called a root of the equation f = 0; it is a zero (or also a root) of f
To study the kernel of A more closely, we recall that the polynomial ring k[x] is a
principal ideal domain. Explicitly, let a i= 0 be an ideal in k[x] and take a non-zero
polynomial p of least degree in a; then any f E a can be written as f = pq + r for
polynomials q, r, where deg r < deg p. Hence r = f - pq E a and by the minimality
of deg p, r must vanish, so that f = pq and a = (p). Hence every ideal of k[x] is
principal and so ker A is generated by the monic polynomial p (i.e. with highest
coefficient 1) of least degree with a as zero. If p has degree n, then k[a] is spanned
by 1, a, ... , an - I and these elements are linearly independent over k, for any depen-
dence would give an equation oflower degree satisfied by a. Thus 1, a, ... ,a n- I is a
basis of k[a] and we see that [k[a] : k] = n = degp. We note that p may always be
taken monic and it is then uniquely determined, for if a satisfied two different monic
equations of degree n, it would also satisfy their difference, an equation of lower
degree, contradicting the minimality of n. The monic polynomial p of least degree
satisfied by a is called the minimal polynomial for a over k, and deg p is called the
degree of a over k.
