2.5 Fundamental Theorem of General Algebra

2.5.1 Existence of a root in a suitable extension: Let a be any element of a field F and f(x) a polynomial of the ring F[x]. The division algorithm yields

f(x) = (x - a)f(x) + b,

where f(x) is a polynomial of the ring F[x] and b is an element of F. Setting x=a (cf. 2.3.5), one finds f(x) = b, whence

f(x) = (x - a)f(x) + f(a). (1)

Hence f(x) is divisible by (x - a) if and only if f(a)= 0. In particular, if f(x) is an irreducible polynomial of a degree > 1, it cannot be divisible by any factor of degree 1, whence there is no element a in F which satisfies the equation

f(a) = 0. (2)

In order to solve (2), on must consider a field F₁ of which F is a sub-field. Every polynomial f(x) of F[x] is also a polynomial of F₁[x], but if f(x) is an irreducible polynomial of F[x], nevertheless it may be a reducible polynomial of F₁[x]; in particular, f(x) may be divisible by a factor x - a, where a is an element of F₁. In this case, a satisfies Equation (2). If F is a sub-field of F₁, then F₁ is called an extension of F, and if a satisfies Equation (2), then a is said to be a root of the polynomial f(x). In order to find a root of an irreducible polynomial f(x) of F[x], one has to extend F to a field F₁, which contains such an element a that f(x) = (x - a)f₁(x), where f₁(x) is a polynomial of F₁[x]. For example, let F be the field of the rational numbers; then f(x) = (x² - 2) is irreducible, but f(x) is also a polynomial of F₁[x], when F₁ is the field of the real numbers. As a polynomial of F₁[x], f(x) = (x - Ö2)(x + Ö2) is reducible. Let F₂ be the field of the complex numbers, then x² + 1, which is an irreducible polynomial of F₁[x], is a reducible polynomial of F₂[x]. A fundamental question of General Algebra is whether, given a polynomial f(x), it is always possible to extend F to F₁ in such manner that f(x) has a root in F₁. This question is answered in the affirmative by the

Fundamental Theorem of General Algebra: If f(x) is a polynomial of F[x], then there exists an extension F_i of F which contains the root a of f(x).

Proof: Since f(x) is a product of (one or more) irreducible polynomials and the roots of these polynomials are also roots of f(x), there occurs no loss of generality if it is assumed that f(x) = a₀ + a₁x₁ + ··· + a_n xⁿ is irreducible. The classes of residues of F[x] modulo f(x) form a field F₁' and the classes characterized by elements a of F form a sub-field which is isomorphic to F. By this isomorphism, the class a₀ + k(x)f(x) corresponds to the element a₀ of F (cf. 2.4.7 Theorem 5). Hence one can extend the field F to a field F₁ which is isomorphic to F₁'. Let a be the element of F₁ which corresponds to the polynomial x, then a₀ + a₁x₁ + ··· + a_n xⁿ corresponds to the class containing the polynomial a₀ + a₁x₁ + ··· + a_n xⁿ = f(x), but this class is the class (0), whence f(a) = 0 and a is a root of f(x).

2.5.2 Extensions of F containing a root of f(x): Let f(x) be a polynomial of degree n which is irreducible in F[x]. Consider again the field F₁ which is isomorphic to the field of the classes of residues of f(x) in F[x]. Denote by f_n(x) polynomials of F[x] of degree < n; then every class of residues of f(x) contains one and only one element of the type f_n(x) and the elements of F₁ can therefore be represented by

Different polynomials f_n'(x) belong to different classes of residues and correspond therefore to different elements of F₁. Hence every element of F₁ can be represented in one and only one manner by (1) and it corresponds (1, 1) to the ordered set

of elements of F. Those elements for which b₁ = ··· = b_n-1 = 0 correspond to the elements of F. In order to add (subtract) two elements of F₁ , one must add (subtract) the corresponding coefficients (2). The addition and subtraction is therefore performed as if the elements were vectors, the corresponding results holds for multiplication with elements of F.

Let now f₁(a) and f₂(a) be any pair of elements of F₁. By 2.4.7 Theorem 1, one can find by the algorithm of division a polynomial f₃(x) satisfying

whence

can be obtained by a calculation consisting of a finite number of elementary operations in the field F. Let f₁(a) ¹ 0, then f₁(x) ¹ 0. Since f(x) is a primitive element of the ring F[x] and therefore relatively prime to f₁(x), the highest common factor (f(x),f₁(x) ) = 1 can be obtained by the algorithm of the h.c.f., whence a polynomial f₄(x) of degree < n satisfying

can be found by elementary operations in F and

Hence the considerations leading to the fundamental theorem allow to extend the field F to a field F₁ which contains a root of f(x) and in which the elementary operations (addition, subtraction, multiplication, division) can be performed by methods based only on the elementary operations in F. Thus, if F is given in the sense that one can perform the elementary operation in every case in a finite number of steps, the same is true for F₁. On the other hand, F₁ is not uniquely determined, but only in the sense of isomorphism. Obviously, there are extensions of F which contain a root of f(x) and are non-isomorphic to F₁; for example, every extension of F₁ has that property. It will be shown now that every extension of F which contains a root of f(x), say a, contains a sub-field which is an extension of F and is isomorphic to F₁.

Theorem: Let f(x) be an irreducible polynomial of F[x] and E any extension of F which contains any root, say a, of f(x); then the meet of all the sub-fields of E, which contain F and a, form a field which is isomorphic to the field F'₁ of the classes of residues of f(x) in F[x].

Proof: Every sub-field of E, which contains F and a, contains also the elements

where n is assumed to be the degree of f(x) and the coefficients run independently over all the elements of F and f(x) again denotes a polynomial of degree < n. Obviously, the elements (7) form a module M. Let f₁(ax), f₂(ax) be two elements of M; if f₃(x) is determined by (3), then f₃(a)=f₁(a)f₂(a). Hence M is a ring which is homomorphic to the field F'₁. A has been shown in 2.2.3, the module M is a field and is isomorphic to F'₁. On the other hand, every sub-field of E, which contains F, and a must contain M, whence it is the meet of all the sub-fields of E which contain M and the theorem has been proved.

2.5.3 Factorization of f(x) into linear factors. General remarks: It has been shown in 2.5.2 that the fundamental theorem of general algebra is far more than a mere theorem on the existence of roots. The fields, in which those roots exist, can actually be constructed, and the operations of addition, subtraction, multiplication and division can be performed practically in the extended field. By the last theorem, it has been shown that, if the polynomial f(x) under consideration is irreducible in F(x), there exists one extension F₁ of F which is uniquely determined in the sense of isomorphism such that every extension of F containing a root of f(x) is isomorphic to an extension of F₁. Obviously, there exist an infinite number of extensions of F which are isomorphic to F₁; one can even arrange them in such a way that they have no common elements other than elements of F. In each of these fields, there exist roots of f(x). It is therefore of no use to speak of the roots of f(x), but only of the roots of f(x) in a particular field. The number of roots of a polynomial of degree n in a field is the subject of the

Theorem: Let f(x) be a polynomial in F[x] of degree n.
1. If F ' is any extension of F, then f(x) has not more than n roots in F'.
2. There exists an extension, say F* of F, such that in F the function f(x)=a(x-a₁) ··· (x-a_n), where a belongs to F*, a₁, ··· , a_n belong to F, and this representation of f(x) in F* is unique up to an arbitrary permutation of a₁, ··· , a_n .

Proof 1: Let a₁, ··· , a_m be roots of f(x) in F'. Then f(x) is divisible by x - a_j, where j = 1, ··· , m; since the factorization in F'[x] is unique, f(x) is divisible by (x-a₁) ··· (x-a_m) which is a polynomial of degree m. Hence m £ n.

Proof 2: If n = 1, the proposition holds for F*= F. Assume that by mathematical induction the proposition holds for degrees < m, and let n = m. Then, by the fundamental theorem, F has an extension F₁ which contains a root, say a_m, then f(x) is divisible in F₁(x) by x - a_m=, say f(x) = f₁(x)(x - a_m). Now f₁(x) is of degree m - 1, and by the assumption, made for mathematical induction, it follows that f₁(x) = a(x-a₁) ··· (x-a_m), whence f(x) = a(x-a₁) ··· (x-a_m-1). The uniqueness of the representation follows from the uniqueness of the factorization of f(x) in F*[x]. Hence the proposition follows by mathematical induction. This theorem shows that the number of roots of a polynomial does not exceed the degree in any field and that it is equal to the degree (certain roots possibly being counted several times) in a suitably chosen extension. It is however often of great interest to know the exact number of roots in a particular field, e.g., in the field of real numbers or of complex numbers. These problems are not solved by the fundamental theorem of general algebra; they require investigations of a special kind. For the field of complex numbers, the question is completely resolved by the Fundamental Theorem of Classical Algebra. Methods of determining the number of the real solutions of any polynomial with real coefficients will be presented in 5-2. The fundamental theorem of general algebra and other theorems derived from it are of a more general nature than the classical investigations on real and complex roots. In the general theory, one does not assume that the coefficients of the polynomials are numbers; they may be elements of any field, e.g., they may be polynomials in an indeterminate y with complex coefficients

Thus, it follows from the fundamental theorem of General Algebra that a suitable extension of the quotient field of the polynomials f(y) contains an element a for which F(a,y) =0. This consideration is the very basis of the theory of algebraic functions. On the other hand, it is obvious that the field of real numbers and the field of complex numbers are of special interest as they have an important role in nearly every branch of mathematics. Thus, the classical question regarding the roots of a polynomial in these fields is of general importance far beyond its historical interest. When enquiring regarding the roots in the field of real (complex) numbers, one is, in general, not satisfied with knowing the number of the roots in intervals (domains) of the real axis (the complex plane). By choosing these intervals (domains) suitably small, one obtains approximate values of the roots. A determination of the magnitude of a root means nothing but a construction of a path leading to an approximation of the root with an error less than any predetermined number. Of course, an irrational number cannot be determined otherwise than by a method approximating it by rational numbers. Even the most familiar formulae expressing irrational numbers (e.g., Ö2) are only rules for an approximation in abbreviated form. The importance of an approximating calculation of roots is obvious, especially for problems of applied mathematics. On the other hand, if the elements of a field are not put into a definite order, there are no intervals or domains in that field. For this reason, there does not arise at all in General Algebra the question regarding the magnitude of a root .

Thus the problem of solving an algebraic equation demands some more detailed specification. If one wants to find a root in any suitable extension, then one has to apply only the methods of 2.5.2. Again, if the roots in any particular field are required (e.g., the field of real numbers, of complex numbers), then methods appropriate to the special nature of that field are necessary. For the field of complex numbers, the fundamental theorem of classical algebra (cf. 3.8) gives the most important information. Sometimes, the problem is put also in a different way. Is there a root of the given polynomial in a suitable field out of a particular class of fields? To this class of problems belongs the investigation of those roots which can be expressed by a finite number of radicals (square roots, cubic roots, etc.), Again an interesting special case concerns those roots which can be expressed by successive calculations of square roots; every problem of Planimetry which can be solved by the help of a ruler and compass leads to a root of this class.

2.6. Extension of a field

2.6.1 Vector spaces over a field: The representation of the extension F₁ of F as performed in 2.5.2 reminds us of the vector spaces considered in Chapter 1; this similarity has already been referred to in 2.5.2. The only essential difference is that the coordinates are elements of F which is supposed to be an arbitrary field, whereas in Chapter 1 the coodinates were supposed to be numbers. It has been mentioned at the end of 1.6 that—except for 1-7—the property of the coordinates to be numbers is readily disposed of. Thus, a more general definition of a vector space will now be given. Let M be a module and F a field; the elements of F will be denoted by characters a, b, c, ··· , the elements of M by Greek characters a, b, d, ···and the common null-element of M and of f by 0. Assume that the elements of M can be multiplied by the elements of F, the products being elements of M, and that this multiplication is governed by the laws:

Then M is said to be a module over F. Now,

0 a = (c - c)a = ca - ca = 0.

Similarly, it follows from (1) that c 0 = 0, whether the factor 0 in c0 is regarded as the zero-element of F or of M.

In particular, let M be a module over F, where there exists a base of n elements of M

such that every element a of M can be represented by

and that

then M is a vector space over F of Rank n and the elements of M are called vectors. If

then

whence it follows from (3) that a₁ - b₁ = ··· = a_n - b_n = 0. The representation of a vector of M by (2) is therefore unique. Thus, there exists a (1,1) mapping of the vectors of M onto the ordered sets of n elements of F

the addition of vectors and the multiplication of vectors and elements of F being determined by

Hence follows the

Theorem: Every vector space of Rank n over F is isomorphic to the system of all the ordered n-tuplets (4), where the operations of addition of vectors and of multiplication of a vector with an element of F are determined by (5).

In particular, a vector space W of rank n over the field of the real (or complex) numbers is isomorphic to the "vector space V of Rank n of Chapter 1. Every vector of W is represented by an n-vector (x₁, ..., x_n) of V. But W is not identical to V, nor are the vectors of W identical to the n-vectors which represent them, as, by definition, the latter are n-tuplets. Of course, if a different base of W is used, the vectors of W are mapped in a different way onto the n-vectors of V. As a consequence of the isomorphism between W and V, the formulae established for V can be applied to W and it is often convenient to identify V and W. On the other hand, whenever different representations of the same vector will be considered (e.g. in 6.2), it is necessary to distinguish between a vector of w and its representation by an n-vector of V. In a corresponding manner, the notion of n-vectors will be used in connection with vector spaces over any field K.

A linear transformation of a vector space W over K is a mapping of W onto itself leaving addition of vectors and their multiplication by elements of K invariant. This definition tallies with a characteristic property of linear transformations of spaces V of n-vectors (cf. 1.1.1 Theorem 2). Thus x ® x' implies Sc_ix_i ® Sc_ix'_i when c₁, ··· , c_m are elements of K. In the same manner as Theorem 2 of l.l.l has been proved, one shows easily that the n-vectors x₁, ··· , x_m representing the vectors of W are transformed by linear equations

with coefficients aⁱ_k out of K, when W is transformed by a linear transformation. On the other hand, every transformation (6) corresponds to a linear transformation of W when any base of W is selected.

2.6.2 Extension of the results of Chapter 1 to vector space over an arbitrary field: If F is the field of the real (or complex) numbers, a vector space over F is isomorphic to a vector space as considered in Chapter 1. However, it was only a matter of convenience that in Chapter 1 the coordinates of a vector were assumed to be numbers. With the only exception of 1.7, no other property of numbers has been used than that they form a field. For this reason it was stated in 1.1.6 that the notion of number could be understood to be a real or a complex number. The reader may check that 1.2.2 to 1.2.6 and 1.8 to 1.11 hold without any further change if the notion of number is automatically replaced by element of a field F".

In 1.7, it has been assumed that number should mean real number. This section forms a portion by itself; it is essential for the methods used there that 0 cannot be represented as a sum of squares. This supposition is not satisfied in fields of characteristic p and not even in every field of characteristic 0. The condition holds in the field of all real numbers and in every sub-field of it, but it is not satisfied in the field of all complex numbers. The main result of these considerations will be stated now in the

Theorem: Given any field F, then the investigations of 1.2 to 1.6 and 1-8 to 1.11 hold without any further change if the notion of vector space is replaced by vector space V over F", n-vector by vector of V, and if the coefficients of the linear equations, the coordinates of the matrices and the terms of the determinants are supposed to be elements of F.

For example, a vector a is considered to depend on the vectors b₁, ··· ,b_m, if a= b₁b₁ + ··· + b_mb_m holds, and the vectors b₁, ··· ,b_m are independent, if c₁b₁+ ··· + c_mb_m = 0 implies c₁ = ··· = c_m = 0. The Rank n of a vectors space V is equal to the maximum number of independent vectors in V and n is therefore independent of the choice of the base of V.

2.6.3 Finite extensions: If a module R over a field F is itself a ring, then R is said to be a ring over V. In Chapter VI, rings of matrices will be considered which are rings over the field of the coefficients of the matrices. A special case of great interest is when F is a sub-ring of the ring R over F. In this case, the unit element of F is also the unit element of R.

Let K be an arbitrary field, x be an indeterminate (the letter x not being used for denoting elements of K), and the polynomials in x with coefficients of K form an integral domain K[x] which is a ring over K and contains K as a sub-ring. This ring is not a vector space, since the powers of x form an infinite set of independent elements; thus there exists no maximum number of independent elements in K[x] and therefore K[x] has no base. The quotient field of K[x] will be denoted by

K(x). (1)

This field is an extension of K, but it is not a vector space over K, since it contains an infinite number of independent elements. Every extension of K which contains x must contain every element of K(x), and every ring over K which contains x must contain every element of K[x]. This notion can be generalized when x is an element which is not necessarily an indeterminate. Let L be an extension of K and a any element of L. If f(x) runs over all the polynomials of K[x], then the elements f(a) of L form a ring

K[a] (2)

and its quotient field will be denoted by

K(a). (3)

Hence K[a] is the meet of all the rings containing K and a ; similarly, K(a) is the meet of all the extensions of K which contain a. The correspondence by which f(x) is mapped onto f(a), when f(x) runs over K[x], is a homomorphism; thus K[a] is homomorphic to K[x]. If the field L is a vector space of Rank, say n, over K -, i.e., if there exists a maximum number n of independent elements in L - then L is said to be finite over K. The Rank n is denoted by

n = [L : K]. (4)

In this case, there exists in L a base a₁, ··· ,a_n of L over K such that every element a of L can be represented in one and only one manner by a₁a₁+···+a_na_n with the aid of the elements a₁, ··· , a_n of K. If an element a of L is a root of a polynomial of K[x], then a is said to be algebraic over K, otherwise a is transcendental over K. If every element of L is algebraic over K, the extension L of K is said to be algebraic over K. The link between these notions is shown in the

Theorem:

1. If L is finite over K, it is algebraic over K.
2. If a is algebraic over K, then K(a) is finite (and therefore algebraic) over K, and isomorphic to the field, formed by the classes of residues of the irreducible polynomial f(x) of K[x] of which a is a root; moreover, K(a) = K[a]. The Rank [K(a) : K] = n, which is equal to the degree of f(x), is said to be the degree of a over K; 1, a, ···, a^n-1 form a base of L over K - 3. If a is transcendental over K, K[a] is isomorphic to K[x] and therefore it is not a field.

Proof: 1. Let L be finite over K, say of Rank n, and b be any element of L. Then 1, b, ··· , b ⁿ cannot be independent; hence there exists a relation c₀+c₁b+···+c_nb ⁿ , where the coefficients c₀, c₁, ··· ,c_n belong to K and at least one of them is not 0. Hence, c₀ + c₁x + ··· + c_nx ⁿ = f(x) is not the polynomial 0 and f (b) = 0. Hence every element b of L is algebraic over K and therefore L is algebraic over K.
2. Let L be any extension of K and a be an element of L which is algebraic over K. Then there exists a polynomial f (x) in K[x] such that f (a) = 0. f (x) is a product of irreducible factors f (x) = f(x), f₁(x), ··· , f_m(x), whence f(a), f₁(a), ··· , f_m(a) = 0 and as the factors on the left hand side are elements of the field L, one of the factors is zero, say f(a) = 0, where f(x) is irreducible and of degree, say n. The ring K[a] is homomorphic to K[x]. The sub-ring R of K[x] which is mapped onto the zero-element of K[a] contains f(x) and all the polynomials divisible by it, but no element of K besides 0. Since K[a] is a Euclidean domain, R contains the h.c.f. of any two of its elements, and, since R does not contain 1, it cannot contain any element which is relatively prime to f(x); thus R contains those and only those elements of K[a] which are divisible by f(x). Hence two elements of K[x] are mapped onto the same element of K[a], if and only if their difference is divisible by f(x). Hence K[a] is isomorphic to the ring of the classes of residues of f(x) in K[x]. Since f(x) is irreducible in K[x], this ring is a field (cf. 2.4.7 Theorem 4). Hence K[a]¹K[a] and [K(a) : K] = n. There exists in every class of residues one and only one polynomial of degree < n, say b₀ + b₁x + ··· + b_n-1x ^n-1 (cf. 2.4.7). The elements of K[a] can therefore be represented in one and only one manner by b₀ + b₁a + ··· + b_n-1a ^n-1, whence 1, a, ··· , a ^n-1 is a base of K[a] over K.

3. Let a be transcendental over K. K[a] is homomorphic to K[x], The ring R of the elements of K[x] mapped onto 0 does not contain any element of K other than 0. Let R contain a polynomial f(x) of degree > 0, then f(a) = 0 and this implies that a is algebraic over K. Since a is supposed to be transcendental, R contains only the element 0, whence the homomorphism is an isomorphism. Hence follows the theorem.

2.6.4 Rank of a field over a field:

Theorem: Let L be a finite extension of K and M a finite extension of L, then M is a finite extension of K and

[M : K] = [M : L][L : K]. (1)

Proof: Let a₁, ···, a_n be a base of L over K, and b₁, ···, b_m be a base of M over L, then n = [L : K] and m = [M: L). Any element g of M can be represented by g = Sl_i b_i , where l₁ , ··· , l_m are elements of L and therefore l_i = Scⁱ_ja_j, where each cⁱ_j is an element of K. Hence g = Scⁱ_ja_j b_i . In order to prove that the n m elements of a_j b_i form a base of M over K, one must therefore only prove that they are independent. Suppose now that

0=Sdⁱ_ja_jb_i=Sb_iS dⁱ_ja_i.

Since in the last sum the coefficients of b _i are elements of L and b₁ , ··· , b_m are independent, these coefficients are equal to zero. Similarly, Scⁱ_ja_j = 0 implies dⁱ_j = 0, whence the n m elements are independent and form a base of M over K. Hence follows the theorem.

Corollaries:

1. If L is a sub-field of M and an extension of K, and M is finite over K, then M is finite over L, L is finite over K and (1) holds.

Proof: As M is finite over K, say of Rank q, and the elements of L belong to M, no set of more than q elements of L can he independent over K. Hence L is finite over K. Similarly, every set of more than q elements of M is connected by a linear homogeneous equation, the coefficients being elements of K - and therefore elements of L - of which at least one is other than 0. Hence M is finite over L, and (1) follows from the theorem above.

2. If [L : K] = q, then the degree over K of any element a of L is a factor of q.

Proof: [L : K(a)] [K(a) : K] = q.

3. If f(x) is a polynomial of K[x], and [K(a) : K] = q, then [K(f(a) : K] is a factor of q.

4. If [K(a) : K] = p is a prime number and 0 < degree f(x) < p, then K(f(a))=K(a).

Proof: The inequality yields that f(a) is not an element of K, whence [K(f(a))~ : K] > 1, whence it is equal to p. Hence [K(a) : K(f(a))] = 1 and the proposition is proved.

2.6.5 Highest common factor and extension of a field: If F' is an extension of F and f₁(x) is a polynomial of F[x], then it is also a polynomial of F'(x). If a polynomial f(x) of F[x] is a factor of f₁(x) in the ring F[x], it is also a factor of f₁(x) in F '[x]; on the other hand, if f(x) is a factor of f₁(x) in F'[x], one gets the quotient of the polynomials by the algorithm of division; hence these coefficients belong to F₁ and f₁(x) is also divisible by f(x) in F[x]. However, f₁(x) may have factors which are polynomials in F'[x] without belonging to F[x]. Let f₁(x) and f₂(x) be two polynomials of F[x]. The highest common factor ( f₁(x), f₂(x)) can be calculated by the algorithm of the h.c.f.; its coefficients are obtained by rational operations and belong therefore to F; a common factor of those coefficients, which is any element ¹ 0 of F, remains arbitrary. If one considers f₁(x) and f₂(x) to be elements of F '[x], then the algorithm furnishes the same polynomial ( f₁(x), f₂(x)), but a common factor of the coefficients remains arbitrary which is an element ¹ 0 of F', whence

Theorem: Let F' be an extension of F and f₁(x), f₂(x) and f(x) be elements of F[x]. A highest common factor of f₁(x), and f₂(x) in the ring of polynomials F[x] is also an h.c.f. of those polynomials in F '[x]. If f(x) is a factor of f₁(x) in F[x], it is also a factor of f₁(x) in F[x] and conversely.

2.6.6 Multiple roots: Let a be a root of a polynomial f(x) of K[x] and K(a)=L, then f(x) can be represented in L[x] by

f(x) = (x - a)f₁(x).

Hence f(x) is divisible by (x - a)², if and only if f₁(a) = 0. Denoting the derivatives in the usual manner:

f '(x) = f₁(x) + (x - a)f₁(x),

whence

f '(a) = f₁(a).

Thus f '(a) = 0 is the necessary and sufficient condition for f (x) to be divisible by (x - a)². If

Except for the case when f (x) is 0, the degree of f '(x) is less than the degree of f (x). Recall that the factor j means a sum of j terms, each equal to the unit element 1 of K (cf. 2.2.5 (3)); thus, if the characteristic of K is zero, j = 0 implies j =0, but if the characteristic is p, the element j is equal to zero if and only if j is divisible by p. Hence, in the case of a characteristic p ¹ 0, the degree of f '(x) may differ by more than 1 from the degree of f (x). Especially f'(x) = 0 if j a_j= 0 for j =0, 1, ··· , n. In the case when the characteristic is a prime number p, this condition means that only the coefficients of the terms x^pq can differ from zero. One can formulate this result in a manner which applies to both cases:

Theorem 1: f '(x) = 0 if and only if

f (x) = f (x^m), (1)

where m is the characteristic of K.

In fact, for m = 0, the condition means that f(x) is of degree < 1, whereas the theorem has been proved just above for the prime number characteristic.

Let f (x) be irreducible and of degree > 0 and consider the h.c.f. (f (x),f '(x)); two cases must be distinguished:

1. f '(x) ¹ 0, when (f (x), f '(x)) = 1,
2. f '(x) = 0, when (f (x), f '(x)) = f (x).

If again a is a root of f (x), then f '(a) ¹ 0 in the first case and f '(a) = 0 in the second case. Hence follows

Theorem 2: Let f(x) be an irreducible polynomial of K[x] and a be a root of it. Then f (x) is divisible by (x - a)² if and only if K has a prime number characteristic p and f (x) is a polynomial in x^p over K.

If f (x) is not divisible by a factor (x—a)² in any extension of K, then it is said to be separable, otherwise non-separable. Irreducible polynomials over a field of characteristic 0 are therefore separable, while irreducible polynomials over a field K of characteristic p are non-separable if and only if they belong to K[x^p].

2.6.7 Non-Separable Polynomials: Let K be a field of characteristic p and f(x) be an irreducible polynomial over K and of degree n. Then there exists a uniquely determined integral number e ³ 0 such that f (x) belongs to but not to Thus, f(x) is separable if and only if e = 0. In any case,

The polynomial y(y) of K[y] is irreducible as y(y) = y₁(y)y₂(y) implies Moreover y(y) cannot belong to K[y^p], since otherwise f (x) must belong to Hence y(y) is irreducible and separable. Let q be the degree of y(y), then

n = q p^e.

In a suitable extension L of K, the polynomial y(y) can be represented by

y(y) = a(y - b₁) ··· (y - b_q), (2)

where b₁, ··· , b_q belong to L and a belongs to K. Hence

Since y(y) is separable, b₁, ··· , b_q are q different elements.

In a suitable extension M of L, there exist elements g_j such that

Since the characteristic of M is p (cf. 2.2.6 (2) ),

whence

Since the q elements b_j are different, it follows from (4) that g₁, ··· , g_n are different. Hence follows

Theorem: An irreducible polynomial f (x) over a field K of characteristic p, which is of degree n, has exactly q = n: p^e different roots in a suitable extension of K and it can be represented by (5). The integral number e ³ 0 is uniquely determined by the condition that f (x) belongs to but not to

2.7 Repeated extension of a field

2.7.1 Extension of a field to a ring and field in a finite number of steps: Let K be a sub-field of L,

be elements of L and

Thus K₁ is the meet of all extensions of K which contain a₁ (cf. 2.6.3) and K₂ is the meet of all the extensions of K₁ which contain a₂. Hence K₂ is an extension of K containing a₁ and a₂. On the other hand, every extension K' of K which contains a₁ and a₂ is an extension of K₁, whence K₂ is a sub-field of K'; hence K₂ is the meet of all the extensions of K which contain a₁ and a₂. Similarly, K_m is the meet of all those extensions of K which contain all the elements (1). One therefore introduces the notation

where the elements in the brackets can be interchanged arbitrarily.

Let f(x₁, ··· , x_m) run over all the polynomials of K[x₁, ··· , x_m] (cf. 2.3.6), then the elements f(a₁, ··· , a_m) form a ring

which is homomorphic to K(x₁, ··· , x_m) and which is contained in K_m. The quotient-field of (4) is also contained in K_m. On the other hand, since the quotient-field is an extension of K containing the m elements (1), it must be identical with K_m. Hence every element of K_m can be represented in the form

f(a₁, ··· , a_m) : f(a₁, ··· , a_m), (5)

where the numerator runs over over all the elements of (4) and the denominator over those elements of (4) which are not zero. Although K[a₁, ··· , a_m] is homomorphic K[x₁, ··· , x_m], the quotient-field K(a₁, ··· , a_m) is, in general, not homomorphic to K(x₁, ··· , x_m), since a field cannot be homomorphic to a field unless they are isomorphic. Naturally, the correspondence

cannot be extended to the correspondence

if there are polynomials f which differ from the null-polynomial and for which neverthelessf(a₁, ··· , a_m) = 0. If there is no such polynomial, then K(x₁,···,x_m) and K[a₁, ··· , a_m] are isormorphic.

Theorem: If a₁, ··· , a_m are algebraic over K, then

Proof: (by mathematical induction). For m = 1, the theorem has been proved in 2.6.3. Assume that K[a₁, ··· , a_m-1] = K(a₁, ··· , a_m-1) = L. Then a_m is algebraic over L, and K[a₁, ··· , a_m] = L[a_m] = L(a_m) = L.

Exercises: Let R be the field of rational numbers.

1. Consider L = R(Ö2, Ö3). Construct a base of L and show that L=R(Ö2+Ö3).
2. Investigate

2.7.2. Primitive element of an extension: If

then a is said to be a prime element of the extension L of K. The examples given in the exercises above show that even an extension generated by more than one step may have a primitive element. As a matter of fact, a finite extension of a field of characteristic 0 can always be performed with the aid of a primitive element. The corresponding statement holds for a large class of fields of characteristic p; that this class includes all the finite fields, will be shown by a different method in 3.2.1. For an investigation of the case when K has an infinite number of elements, use the

Lemma: Let f(x) and g(x) be polynomials of K[x] and in a suitable extension of K

if c is an element of K such that for i = 1, ··· , n and k = 2 , ··· , m the n(m - 1) inequalities

are valid, then

Proof: Set K(g) = K' ; then f(x) =f(g - cx) is a polynomial of K'[x] which has a root b in common with g(x). If b_k were a common root of f(x) and g(x), then y-cb_k= a_i, in contradiction to (2). Hence the h.c.f. (f(x), g(x) = x - b₁, whence b₁ is an element of K'. Moreover, a₁ = y - cb_k belongs to K', and so does every element of K(a₁, b₁). Since, on the other hand, g belongs to K(a₁, b₁), the lemma has been proved.

Theorem: Let K be a field containing an infinite number of elements, a algebraic over K and b, ··· ,k be roots of separable polynomials of K[x], then there exists a primitive element l for the extension K(a,b,···,k) of K.

Proof: To start with, prove that K(a,b)= K(a'). Set a = a₁, b = b₁ and let f(x) and g(x), as given by (1), be the irreducible polynomials in K[x] with the roots a and b, respectively. Since g(x) is supposed to be separable, the roots b₁,b₂,···, b_m are all different. In order to determine a', one has to find an element b of K such that

These conditions are satisfied, if b is not a root of anyone of the linear equations

Since b₁ - b_k ¹ 0, each of these equations has exactly one solution in a suitable extension of K, and therefore at most one solution in K. Now K contains an infinity of elements, whence there exists an element b in K which does not satisfy anyone of the n(m —1) equations (4). Hence a' is a primitive element of the extension K(a, b) of K. Since a and b are algebraic over K, the extension K(a) is finite over K, and K(a') = K(a, b) is finite over K(a) and therefore finite over K. Hence a' is algebraic over K. Thus L = K(a ', ····· ,k ), where a ' is algebraic over K, and the other elements in the bracket are roots of separable polynomials. The procedure can therefore be repeated till the number of the elements in the brackets is reduced to one element

If, in particular, the characteristic of K is 0, then K has an infinite number of elements and every irreducible polynomial is separable, whence follows the

Corollary: If K is a field of characteristic 0 and a, b, ··· , k are algebraic over K, then there exists a (primitive) element l such that K(a, b, ··· , k) = K(l).

2.7.3 Extension by roots of irreducible polynomials: In 2.7.1 and 2.7.2, extensions of a field K have been considered which are generated by elements a, b, ··· of any field of which K is a sub-field. Already, in an earlier section, extensions of a different kind have been used. One can extend a field K to a field K(a), where a is not given, but has to be created in such a way that it is a root of a polynomial f(x) of K[x]. It has been proved in 2.5.1 that this extension is always possible, and in 2.5.2 it has been shown that if f(x) is irreducible, the extension is determined uniquely in the sense of isomorphism.

The first, statement can be generalized without difficulty to the case of more than one polynomial, Given polynomials f₁(x), f₂(x), ··· , f_k(x) in K(x), then one can construct by repeated extension a field K(a, b, ··· , k) such that 0=f₁(a)=f₂(b) = ··· , f_k(k) Now let the k polynomials be irreducible in K, then K(a) is determined uniquely in the sense of isomorphism, but f₂(x), although it is irreducible in K, may be reducible in K(a), Let f₂(x) = f₁(x)f₂(x) and, moreover, b be a root of f₁(x), and b ' a root of f₂(x), then it is not certain whether K(a, b) and K(a, b ') are isomorphic or not. That both the cases occur is shown by the following examples of extensions of the field R of rational numbers:

Hence R(a, b) and R(a, b ') are isomorphic. Note that these two isomorphic fields of numbers are different fields; the first is a field of real numbers only, whereas the second contains complex numbers.

K₂ consists of real numbers, whence 0 cannot be represented as a sum of squares of elements of K₂ which are different from 0 nor can 0 be represented in that manner in any field which is isomorphic to K₂. However, in K₂', ,

whence K₂' is non-isomorphic to K₂. Thus, it is possible that a field K can be extended to two non-isomorphic fields by roots of the same two polynomials, both of which are irreducible in K.

2.7.4 Normal extension of a field: A case of special interest is when a field K is to be extended by n different roots of a single polynomial of degree n. Then there applies a theorem of uniqueness (in the sense of isomorphism) and one is led to the normal extension of a field which has a very important role in algebra. These investigations require the three definitions:

Definition 1: Let I ' be an isomorphism mapping a field K onto a field L; then a sub-ring R of K is mapped onto a sub-ring L of L. This mapping of R onto L is an isomorphism of R to L, say the isomorphism I. Then the isomorphism I ' is called an extension of the isomorphism I.

The problem of finding an extension I' to a given isomorphism I occurs frequently

Definition 2: Let f(x) be a polynomial of K[x], then there exists an extension M of K, in which f(x) is a product of linear factors. Let a₁, ··· , a_n be the roots of f(x) in M, then f(x) can be represented as a product of linear factors only in those sub-fields of M which contain K(a₁, ··· , a_n]. This field is therefore said to be a smallest extension of K admitting the complete reduction of f(x).

Theorem 1: Let K and L be isomorphic, f(x) and f(x) corresponding polynomials of K[x] and L[x], respectively. Let K' be a smallest extension of K admitting the complete reduction of f(x) and L' a smallest extension of L admitting the complete reduction of f(x); then every isomorphism I of K and L can be extended to an isomorphism I' of K' and L'.

Proof: Obviously, the theorem applies if [K': K] = 1, since then K = K', and L=L'. In order to prove the theorem by mathematical induction, it will be assumed to hold for [K' : K] < m. Let [K' : K] = m and f₁(x) be an irreducible factor of f(x) of degree > 1. Let f₁(x) be the polynomial of L[x] isomorphic to f₁(x), a a root of f₁(x) in K' and b a root of f₁(x) in L'. We can extend the isomorphism of the classes of residues of f₁(x) in K[x] and of f₁(x) in L[x], and therefore to an isomorphism I₁ of K(a) and L(b). Every extension of K(a) admitting the complete reduction of f₁(x) is an extension of K admitting the complete reduction f(x), whence K' is a smallest extension of K(a) admitting this reduction. For the same reason, L' is a smallest extension of L(b) admitting the complete reduction of L(b). As [K' : K(a)] < m, the isomorphism I₁ can be extended to an isomorphism I' of K and L' and, since I' is an extension of I, the theorem applies.

This theorem can be applied also to cases when K and L are identical and I maps every element of K onto itself, whence one arrives at the important

Corollary: Any two smallest extensions of K admitting the complete reduction of a polynomial f(x) of K[x] are isomorphic. The isomorphism can be chosen in such a way that every element of K is represented by itself.

Definition 3: Let N be an extension of K with the property that every irreducible polynomial of K[x] which has a root in N can be represented in N[x] as a product of polynomials of degree 1, then N is called a normal extension of K.

Theorem S: If K' is a smallest extension of K. admitting the complete reduction of an arbitrary polynomial f(x) of K[x], then K' is a normal extension of K.

Proof: Let a₁, ··· , a_n be the roots of f(x) in K', g(x) irreducible in K, b and b' roots of g(x) and b belong to K'. It will be shown that b ' also belongs to K'. If this is not so, then b ' belongs to a suitable extension of K', whence K₁ = K(b ) and K₂ = K(b ') are isomorphic and there exists an isomorphism I of these fields, by which every element of K' corresponds to itself and corresponds to b '. By Theorem 1, one can extend I to an isomorphism I' of K₂(a₁, ··· , a_n) and K₁(a₁, ··· , a_n) = K(a₁, ··· , a_n) = K', By I', every root of a polynomial with coefficients from K will be represented by a root of the same polynomial, whence the elements a_i will only be interchanged. Let b = F(a₁, ··· , a_n), where the coefficients of F are elements of K, whence b = and therefore b ' belongs to K.'. Since b ' is supposed to be an arbitrary root of g(x), the theorem applies.

2.7.4.1*. Generalizations of the theorem on normal extensions: Consider now two different generalizations of this theorem.

*You may omit this section at a first reading,

Theorem 1: Let f₁(x), f₂(x), ··· be a sequence of polynomials of K[x], finite or infinite in number; the smallest extension of K admitting the complete reduction of all these polynomials is a normal extension of K.

Proof: Let F_m(x) = f₁(x) f₂(x) ··· f_m(x) and K_m be the normal extension obtained by extending K with the roots of F_m(x), m = 1, 2, ···. Now K₁ is a sub-field of K₂, again K₂ a sub-field of K₃, etc,; the smallest extension of K admitting the complete reduction of all the f_m(x) is the join of the fields K_m, i.e., the set of all those elements which belong to any field K, for m = 1, 2, ···. This set is a field K*. If therefore a polynomial g(x) has a root in K*, this root is an element of a suitable f₁(x), and since K is a normal extension of K, the polynomial g(x) is a product of linear polynomials in K_m and in every extension of K_m, in particular in K*, whence the theorem applies.

The second generalization concerns the factorization of g(x) in a field N, which is normal and algebraic over K when g(x) is irreducible in K[x]. Let y₁(x),···y_m(x) be irreducible factors of g( x) in N, b₁ a root of y₁(x) and b₂ a root of y₂(x) in a suitable extension of N. The coefficients of y₁(x) are roots of polynomials f₁(x), ··· , f_t(x) irreducible in K. As N is normal over K, all the roots of every f_i(x) belong to N. Let a₁, ··· , a_n be these roots; then K Í K(a₁, ··· , a_n) = L Í N. L is a finite extension of K and normal over K; the factorization of g(x) in L is the same as in N.

Since b₁ and b₂ are roots of g( x), irreducible in K, the fields K(b₁) and K(b₂) are isomorphic and there is an isomorphism by which b₁ is represented by b₂ and the elements of K do not change. Hence no f₁(x) is changed by it. We can extend this isomorphism to an isomorphism of K(b₁,a₁, ··· , a_n) = L(b₁) and K(b₂,a₁, ··· , a_n) = L(b₂). By this isomorphism, every f_i(x) remains invariant; hence its roots are only interchanged, whence an element of L is represented by an element of L. The polynomial y₁(x), which is irreducible in L and a factor of the polynomial g(x), irreducible in K, must therefore be represented by a factor of g(x), irreducible in L. Since the root b₁ of y₁(x) is represented by the root b₂ of y₂(x), the image of y₁(x) is y₂(x), and as these two polynomials are arbitrary, irreducible factors of g(x), one has

Theorem 2: If g(x) is irreducible in K, and N is normal and algebraic over K, every irreducible factor of g(x) in N can be transformed into every other one by a suitable automorphism of a certain field over K, whence these factors are all of the same degree.

If one of the irreducible factors is of degree 1, the others are also linear, whence 2.7.4 Theorem 2 is a special case of this theorem.

2.7.4.2 Automorphism of a normal extension: Let N == K(a₁) be a normal extension of K and [N : K] = n. Then 1, a₁, ··· ,a^n-1₁ form a base of N over K, whence a₁ is the root of a polynomial of degree n which is irreducible in K[x],

Since N is a normal extension of K and contains a root of f(x), this polynomial is factorized in N[x] by

Since K(a_j) is isomorphic to K(a₁), then [K(a_j):K] = n; moreover, a_j and therefore every element of K(a_j) is contained in K(a₁), whence [K(a₁):K(a_j)]=l. Hence

There exists an isomorphism mapping K(a₁) onto K(a_j) by which the elements of K remain unaltered. This isomorphism is an automorphism of N. On the other hand, if there is an automorphism of N for which the elements of K remain fixed, f(x) is transformed into itself and therefore every root of f(x) is mapped onto a root of f(x). The roots of f(x) are therefore undergoing a permutation by the automorphism, and the automorphism is uniquely determined by the condition that a₁ is to be mapped onto a₁; since this condition implies that the base 1, a₁, ··· ,a^n-1₁is mapped onto the base 1, a_j, ··· ,a^n-1_j, there exist exactly n automorphisms of N for which the elements of K remain invariant. For n > 2, not every permutation of the roots of f(x) corresponds to one of these automorphisms. There exists one and only one automorphism of N for which the elements of K remain invariant and a particular root of f(x),say a₁, is transformed into any particular root, say Uja_j. For n = 1, this automorphism is the identity. In particular, if K is of characteristic 0 and M is a finite and normal extension of K, then there exists a primitive element a [cf. 2.7.2] such that M = K(a) and the above statements on automorphism apply.

Moreover, let K be an arbitrary field, [L : K] = 2 and b₁ be an element of L which does not belong to K. Then L = K(b₁) [cf. 2.6.4 Corollary 4]. b₁ is a root of a polynomial of degree 2, say f(x), which is irreducible in K, but reducible in L

Hence b₂ belongs to L, An extension of Rank 2 is therefore always a normal extension. Besides this identity, there exists one automorphism A of L for which K remains invariant. A interchanges b₁ and b₂, and, since 1, b₁ is a base of L over K, one can express A by

where a and b run over K, The elements a and are said to be conjugate. Conjugacy is a symmetric relationship, because every element is the conjugate of its conjugate. The elements of K are the only ones which are self-conjugate. The remaining elements of A consist of pairs of conjugate elements.

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