3.4 lrreducibility of polynomials: In nearly every application of the methods of General Algebra to a particular problem, one is faced with the task of establishing the irreducibility of a polynomial. As it depends on the coefficients of the polynomial as well as on the field for which it should be proved, the problem needs special investigation for single cases. Criteria have been developed especially for the field of rational numbers. Some of the principles used there can be generalized for a larger class of fields.
3.4.1* A general method: Consider at first a method which, although it is applicable to every field, it is nevertheless of little practical use. A polynomial f(x) of degree n will be shown to be irreducible in K[x] if and only if a homogeneous equation of the nth degree, F(x1, ··· , xn) = 0, has no solution a1,···,an = 1 in K. The coefficients of F belong to the ring generated by the coefficients of f(x), and they can be calculated by elementary methods.
* This section may be omitted at a first reading.
However, it is, in general, not easier to show that F(x1, ··· , xn) has no solution in K than to establish the irreducibility of f(x). On the contrary, the irreducibility of f(x), when proved, helps to show that F has no solution in K. The investigation follows:
The classes of residues of a polynomial f(x) in K[x] form a ring which is a field if and only if f(x) is irreducible in K[x] (cf. 2.4.7). Let f(x) be of degree n, and a be a root of f(x); an, an+1, ··· can be linearly expressed by 1, a, ··· , an-1, the coefficients depending on those of f(x) only. Hence
where
ajik are elements of K.
The elements 1, a, ··· , an-1 form a base of a field K(a) with Equation (1) determining the multiplication if and only if there exists for every pair of systems (x1, ···, xn) and (z1, ···, zn) of elements of K there a a uniquely determined system (y1, ···, yn). Hence the equation
must have a uniquely determined solution. The necessary and sufficient condition is det (djk) ¹ 0. But the left hand side of this inequality is a homogeneous polynomial F of degree n in x1, ···, xn, whence
is for every system (x1, ···, xn) of elements of K the necessary and sufficient condition for f(x) to be irreducible in K(x).
Exercise:. There corresponds to every polynomial x³ + px + q, which is irreducible in R[x], a curve of the third class Saijkuiujuk = 0 such that none of its tangents (u1, u2,u3) passes through more than one point with rational coordinates. Compute aijk.
3.4.2 Reduction of the problem to investigation of irreducibility in D[x]: In many cases, it is possible to replace an investigation of irreducibility in K[x] by one on reducibility in D[x], where D )is an integral domain. This reduction of the problem to an easier one is based on the
Lemma: Let D be an integral domain with unique factorization, K = Q(D), the quotient field of D, f(x) and f(x) polynomials of D[x], and f(x) a primitive polynomial ; if f(x) is a factor of f (x) in K[x], then it is also a factor of f(x) in D[x]; if f(x) is irreducible in D[x], it is also irreducible K[x].
Proof: Let f(x) = f(x)y(x), where y(x) is a polynomial of K[x] and f(x) is primitive in D[x] (i.e., the common factors of its coefficients are unities of D only); then there exists an element a of D such that ay(x) belongs to D[x]. Hence af (x) is divisible by f(x) in D[x]. By the Corollary in 2.4.8, f (x) is divisible by f(x) in D[x]. Now, let f (x) be divisible in K[x] by any polynomial f1(x) of degree m, where 0 <m< degree f (x). Then f1 (x) = (a : b)f (x), where a and b are elements of D, whereas f(x) is a primitive polynomial of D[x] and of degree m. Hence f (x) is divisible by f (x) in D[x]. Irreducibility in D[x] therefore implies irreducibility in K[x]. Hence follows the lemma.
3.4.2.1 Irreducibility in R[x] When the reducibility of a polynomial f(x) of R[x] has to be investigated (as usual, R denotes the field of rational numbers), one may assume without loss of generality that the coefficients of f (x) are integral numbers. By the preceding lemma, it is sufficient to ask whether f (x) is reducible in I [x], where I is the ring of the integral numbers. In principle, the method below is applicable to every case as it leads to a decision after a finite number of steps. However, its practical application demands very skilful handling, since otherwise that finite number may become unreasonably large.
Let f (x) = y (x)y1(x), where the coefficients are
integral numbers and f (x) is of degree 2n or 2n+1. Without loss
of generality, let the degree of y(x) be £ n. Let a0,
a1, ··· , an
be arbitrary different integers, then f(a1)=y(a1)y1(a1),
whence y(a1) is a factor of f(a1).
Let y(x)=y0+y1x+···+ynxn
and 
be the different factors of f(ai).
The integers y must therefore satisfy one of the following
systems of n + 1 linear, non-homogeneous equations:
There exists for every system n0,
··· , nn a system of equations and for
every system of equations one solution (y0,
··· , yn), since the
determinant Dn of the homogeneous
systems is ¹ 0. In order to prove the last statement,
mathematical induction may be employed. For n = 1, D1
= a1 - a0 ¹ 0. In the
general case, apply the method of sweep-out to the
first row by multiplying the first column consecutively by 
and
subtracting it from the second, third, ··· , (n + l)-th
columns, respectively. Hence 
where
as one finds readily by successive column addition. Thus, by mathematical induction, Dn ¹ 0 (a different proof is given in 3.5.3). One has to consider among the solutions y0, ··· , yn only those which consist of integral numbers; finally, check by the algorithm of division whether the polynomials y0+ y1 x + ··· +yn xn are factors of f(x), otherwise f(x) is irreducible. This method applies to every integral domain, where the factorization is unique and there exists a method of factorizing any element in a finite number of steps.
3.4.3 Method of homomorphism: The method of 3.4.2 becomes more powerful if used together with considerations concerning homomorphism. Let
be the three polynomials belonging to D[x]. D is mapped by a suitable homomorphism onto an integral domain D1; Equation (1) is transformed by the same homomorphism into
Hence, if f1(x) is irreducible, one of its factors, say y1(x), must be associated to a unity of D, whence f(x) and f(x) are mapped by the homomorphism into associated polynomials. In this manner, it can often be shown that irreducibility of f1(x) in D1[x] implies the irreducibility of f(x) in D[x]. The criteria of irreducibility obtained by this method furnish conditions which are sufficient, but not necessary for irreducibility, since the reducibility of f1(x) does not imply that of f(x). It is essential for the application of this method that D is not a field, as any ring which is homomorphic to a field is isomorphic to it.
3.4.31 Eisenstein's Theorem: The method of 3.4.3 will now be applied to the proof of a theorem from which many more special criteria of irreducibility have been derived. It will be presented here in its original form, although it can easily be generalized to any integral domain with unique factorization.
Esenstein's Theorem. Let a0,
··· , an be. integers, a0,
··· , an-1 be divisible by an
arbitrary prime number p, an
not divisible by p and a0 not by pn,
then 
is
irreducible in R[x].
Proof: Represent the domain I of the integral numbers and I[x] by the classes (mod p). I is mapped homomorphically onto a field GFp and I[x] onto GFp[x] which is a ring with unique factorization. Let f(x) = f1(x)f2(x), where the degree of f1(x) = s > 0 and the degree of f(x) = t > 0. By the homomorphism,
(where the brackets are used to denote the classes of residues as in 2.1.3). Since (x) is a prime element in D[x] and the factorization in this domain is unique, it follows that
c and d not divisible by p, whence
Hence the term a0 in the product f1(x)f2(x) is divisible by p², in contradiction to the assumption, whence f(x) is irreducible in R[x
3.4.32 A special case: The same homomorphism will be used now to prove the irreducibility of
where p is a prime number of the type 4m + 1 and d is a quadratic non-residue of p. A. polynomial of degree < 4 cannot be congruent to y(x) mod p. Hence it is sufficient to prove that x4 + d is irreducible in GFp[x]. Since -1 is a quadratic residue, there exists an integer e such that (e)2 = - 1,(e)4 = 1; moreover, -d is quadratic non-residue, whence x4 + d º 0 (mod p) has no solution, and therefore (x4 + d) has no factor of degree 1 in GFp. In order to prove that it has no factor of degree 2, extend GFp by a root d of x4 + d. In GFp(d),
Each factor of degree 2 has therefore a coefficient d ²en+m. Since d does not belong to GFp, x4 + d is irreducible in GFp, whence y(x) is irreducible in I and therefore also irreducible in R.
3.4.33 Irreducibility of the cyclotomic polynomials in R[x]: The same method has to be applied in a somewhat more subtle manner to prove the irreducibility of the cyclotomic polynomials in R[x].
Theorem: Cyclotomic polynomials are irreducible in the field R of rational numbers.
Proof: It is enough to prove the irreducibility in I[x], where I is the ring of the integral numbers. Consider xn-1 - 1 as a polynomial in I[x] and let a be a primitive root of it in a suitable extension R(a). Then one obtains all the primitive roots in the form am, where (m, n) = 1. Hence one has to prove that if a is a root of a primitive polynomial f(x) which is irreducible in I[x], then am is also a root. It is sufficient to prove this for the prime numbers p which are relatively prime to n; the general case follows from it by a trivial mathematical induction to be taken over the number of the prime factors of m. As f(x) is a primitive polynomial, it follows from the lemma that it is a factor of xr - 1 in I[x]. The coefficient of the highest term is therefore a unity, say
The elements 1, a, ··· , ar-1 form a base of R(a). The representation of the elements of R(a) by
with rational coefficients bi is therefore unique. R(a) has a sub-module M consisting of those elements for which the coefficients bi are integral numbers. The module M contains
Hence M is a ring. As the coefficients of f(x) are integral numbers, f(x) can also be considered as a polynomial in GFp[x], where p is any prime number which is relatively prime to n. Let b be a root of f(x) in GFp[x]; so it is a primitive root of xn — 1. Map the ring M onto GFp[b] by the correspondence
then there corresponds to every element of M exactly one element of GFp[b]. Addition, subtraction and multiplication are invariant for the representation (3), whence it is a homomorphism. Different elements of M may correspond to the same element of GFp[b], but not conversely. There correspond to a j, for j = 1, ··· , n, the elements b j, and as all of these n elements are different, no two different elements a j can correspond to the same power of b , whence (3) places a,···,an and b,···,b n in a (1,1)-correspondence. The roots of f(x) in R[a] are powers of a, those in GFp[b] powers of b ; as f(x) is unaltered by (3), the exponents must be the same. By It follows from the corollary in 3.2.2 that b p must be a root of f(x) in GFp[b]. Hence, a p is a root of f(x), where p was supposed to be any prime number which is relatively prime to n. Hence follows the theorem.
It is remarkable that homomorphisms mapping R[x] onto different rings GFp[x] help to prove the irreducibility of cyclotomic polynomials in R[x], although they may be reducible in each of these rings.
Exercise: Show that the cyclotomic polynomial for n = 8 is reducible in every GFp. Discuss the factorization for the different classes of prime numbers p. In particular, find the distribution of the roots for p = 3, 5 and 7, and show that there follows the irreducibility of the polynomial in R[x] from the difference of the distribution of the roots in the three cases
3.4.4 Irreducibility of determinants: As an example of a proof of the irreducibility of a polynomial in more than one indeterminate, consider the following interesting
Theorem: Let K be an arbitrary field in which there occurs no term in x11,···,xnn and let n > 0, then the determinant X = det (xki) is irreducible in K[x11, ··· , xnn].
Proof. X is a linear function of each of the n² indeterminates, and it is a linear and homogeneous polynomial in the n indeterminates of the first row
Suppose X is reducible, then it must be the product of a linear polynomial in x11, ··· , x1n and a polynomial which is of degree 0 in these indeterminates, the latter being a common factor of A1, ··· , An. If n > 1, Ai is the determinant of the (n — l)² indeterminates x8t, where s ¹ 1, t ¹ 1; if n = 1, then A i = 1. Hence A i ¹ 0, and therefore it is not divisible by any polynomial in indeterminates other than xn. As there is no indeterminate which occurs in A1, A2 ,··· , An simultaneously, every common factor of these polynomials must be an element of K, whence X is irreducible.
3.5.1 Elementary symmetric polynomials: Let K. be an arbitrary field. A polynomial f(x1, ··· , xn) of K[x1, ··· , xn] is said to be a symmetric polynomial if f(x1, ··· , xn) is not altered by any permutation of x1, ··· , xn. The integral rational functions corresponding to symmetric polynomials are said to be integral symmetric functions.
As the polynomial
does not change with any permutation of the indeterminates xi, the coefficients
are symmetric polynomials. These are called elementary symmetric polynomials and represented by
The summation has to be taken over all the systems of different subscripts k1,···, ki. Let f(x) = xn + b1xn-1 + ··· + bn be a polynomial of K[x]. In a suitable extension of K
whence
The coefficients of f(x) are therefore syrnmetric functions of the roots.
Theorem: Let f(y1, ··· , yn) be a polynomial of K[y1, ··· , yn], ai the elementary symmetric polynomials, defined by (3), and F(x1, ··· , xn) = f(a1, ··· , an). Then F(x1, ··· , xn) is the polynomial 0 only if f(y1, ··· , yn) is the polynomial 0.
Proof: Obviously f(y1, ··· , yn) = 0 implies F(x1, ··· , xn)= 0. Now assume that F(x1, ··· , xn)= 0; it will be shown that f(y1, ··· , yn) = 0 follows from this assumption. In a suitable extension of K(y1, ··· , yn), the polynomial f(z)=zn-y1zn-1 + ··· + (-1)nyn has roots, say a1, ··· , an . The elements y1,···,yn are therefore the elementary symmetric functions of those roots
Replace x1, ··· , xn by a1, ··· , an, then
As y1, ··· , yn are indeterminates over K, the right hand side differs from 0, unless f is the polynomial 0. However, as F(x1, ··· , xn) = 0, so F(a1,···,an)=0, whence follows the theorem.
Corollary: Let the coefficients of f(y1, ··· , yn) be integral numbers and p be a prime number. Then f(a1, ··· , an) = F(x1, ··· , xn) º 0 (mod p) only if the coefficients of f are congruent to 0 (mod p).
Proof: The ring GFp[x1, ··· , xn] is homomorphic to the ring of the polynomials in x1, ··· , xn with integral coefficients. If F(x1, ··· , xn) º 0 (mod p), the corresponding polynomial of GFp[x1, ··· , xn] is the polynomial 0, whence f(x1,···,xn) corresponds to the polynomial 0 of GFp[y1, ··· , yn]. Therefore the coefficients of f are divisible by p.
3.5.2 The Main Theorem:
Theorem 1. A symmetric polynomial can be represented in one and only one manner as the sum of homogeneous symmetric polynomials of different degrees.
Proof: Every polynomial of K[x1, ··· , xn] can be represented as the sum of homogeneous polynomials of the same ring and one can choose the summands so that no two of them have the same degree. The difference of two such representations of the same polynomial is a representation of 0 as a sum of homogeneous polynomials, no two of them having the same degree, which is impossible. Therefore the representation is unique. A homogeneous polynomial is transformed by a permutation of the indeterminates into a homogeneous polynomial of the same degree, whence each of the homogeneous portions of a symmetric polynomial are transformed each themselves by every permutation and they are therefore symmetric homogeneous polynomials.
Let P be a permutation of 1, ··· , n and P' its inverse. If two terms
with the same system of exponents are transformed by P into equal terms, they will also be transformed by PP' into equal terms and therefore they are equal. Hence any permutation carries different terms into different terms, whence the polynomial
which denotes the sum of all different terms which are obtained by all the permutations of the lower subscripts of (1) is a symmetric polynomial. When taking the exponents in a non-decreasing order, (2) becomes
where every t1³ 0; the summation on the right side of (3) has to be performed as in (2). The degree of [t1, ··· ,tn ] is
Theorem 2: If f(x1,···,xn) is a homogeneous, symmetric polynomial of degree m, it can be represented by
where ci are coefficients of f and the sum is taken over the N different systems of non-negative integers t1, ··· , tn satisfying (4).
Proof: Since each term on the right hand side of (3) can be transformed by a suitable permutation of the indeterminates into every other one, and different terms are always transformed into different ones, either each of these terms occurs as a term of f or none of them does. If therefore one of the terms of (3) has in f the coefficient ct, then
is a symmetric polynomial in x1, ··· , xn, where none of the terms of [t1, ··· , tn], occurs. By repeating this procedure with all non-negative integral solutions of (4), , the difference of the two sides of (5) is proved after N steps to be equal to 0.
Main theorem of symmetric polynomials: Any symmetric polynomial f(x1,···,xn) of K[x1, ··· , xn] can be represented in one and only one manner by
where ai are the elementary symmetric polynomials and the coefficients of F belong to K.
Proof: The symmetric, homogeneous polynomials of degree m form a module M over K (cf. 2.6.1), generated by the polynomials [t1, ··· , tn]. The Rank of M is therefore not greater than N. The polynomials
belong to M, when the exponents satisfy (4), and it follows from 3.5.1 that the N polynomials (7) are independent, whence they form a base of M. Thus, every homogeneous symmetric polynomial can be represented by (6), and it follows from 3.5.2 that the same holds for any symmetric polynomial. If there are two such representations by different F1(a1, ··· , an) and F2(a1, ··· , an), then the difference must be 0 in contradiction to 3.5.1, whence follows the theorem.
Lemma: The polynomials (3) can be generated by addition and subtraction of polynomials (7).
Proof: Let R be the field of the rational numbers. Since every polynomial (3) can be considered as a polynomial of R[x1, ··· , xn], it can be represented as a linear homogeneous function of the N polynomials (7) of the same degree. Hence
where Sr1 = ··· = Ss1 = m and c1, ··· , cN and c > 0 are integers without a common divisor ¹ ± 1. We will show that c = 1. If it is not, let p be a prime
factor of c, then not every ci is divisible by p, and
in contradiction to the corollary in 3.5.1, whence c = 1 and the lemma holds.
Since the elements c in (5) are coefficients of f(x1, ··· , xn), a direct consequence of the lemma is
Theorem 3: The coefficients of F are elements of the ring generated by the coefficients of f.
3.5.3 Alternating polynomials: A polynomial f(x1, ··· , xn) of K[x1, ··· , xn] is said to be an alternating polynomial if every odd permutation of x1, ··· , xn causes it to take the factor - 1. Since an even permutation consists of two odd permutations, an alternating polynomial is not altered by any even permutation of the indeterminates. The product of two alternating polynomials is a symmetric polynomial, the product of an alternating and a symmetric polynomial is alternating. If an alternating polynomial is divisible by an alternating (a symmetric) polynomial, the quotient is a symmetric (an alternating) polynomial. Since every odd permutation is composed of an odd number of transpositions (cf. 0.3), a polynomial is alternating if it takes the factor -1 whenever a single transposition of its indeterminates is performed. When the characteristic of K is equal to 2, every alternating polynomial is symmetric and conversely; when the characteristic differs from 2, the polynomial 0 is the only polynomial which can be considered to be symmetric as well as alternating.
Theorem 1: If a polynomial f(x1, ··· , xn) of K [x1, ··· , xn] has the property that it is transformed by every permutation of the indeterminates into a polynomial which is divisible by f(x1, ··· , xn), then it is either symmetric or alternating.
Proof: Let cpq be the factor which is taken by f(x1, ··· , xn) when xp and xq are interchanged, then the polynomial takes cpq², when this transposition is repeated; but the twice performed transposition does not alter the polynomial, whence cpq² = ± 1, whence cpq = ± 1. In order to prove that in this equation the same sign + or - holds for every pair p, q of subscripts, consider the following identity between transpositions
It follows from this equation that
whence either every transposition leaves the polynomial invariant, so that it is symmetric, or it takes the factor -1 at every transposition and is therefore alternating.
Let f(x1, ··· , xn) be an alternating polynomial of K(x1, ··· , xn], where the characteristic of K differs from 2. Replace xk by xi and get a polynomial f(x1, ··· , xi, ··· , xi, ·· , xn) which is alternating, but nevertheless invariant for an interchange of the i-th and the k-th indeterminate. Since the characteristic is to differ from 2, the polynomial is 0. Now
f(x1, ··· , xn) = f(x1, ··· , xn) - f(x1, ··· , xi, ··· , xi, ·· , xn)
is divisible by xi - xk, as is seen by forming the differences of corresponding
terms. The n(n—l) : 2 polynomials xi - xk where i < k, are non-associated prime elements of K[x1, ··· , xn], whence f(x1, ··· , xn) is divisible by 
This product is itself an alternating
polynomial, whence the quotient is a symmetric polynomial and one
has
Theorem 2: Let f(x1, ··· , xn) be an alternating polynomial of K[x1, ··· , xn], where the characteristic of K differs from 2 and
then
where S is a symmetrical polynomial.
Corollary:
Proof: The determinant is obviously an alternating polynomial and therefore divisible by D. The second factor S is of degree 0; its value is found to be the unit element by comparing the coefficient of the diagonal term of the determinant with the coefficient of the corresponding term in D.
3.5.4 Symmetric rational functions:
Theorem 1: Let f(x1, ··· , xn) and g(x1, ··· , xn) be polynomials of K[x1,···,xn], (f, g) = 1 and f:g be a symmetric rational function, then and g are symmetric polynomials.
Proof: Let f be transformed into f1 and g into g1 by an .arbitrary permutation of the indeterminates; then fg1 = gf1. By 2.4.7 and the assumption of this theorem, f is divisible by f1, g is divisible by g1 and conversely. As this holds for every permutation of the indeterminates, it follows from the lemma of 3.5.3 that f and g are symmetric or alternating. If one of them is alternating, the other must also be alternating, because the quotient is symmetric; however, in this case the two polynomials must be divisible by D in contradiction to the supposition (f, g) = 1, whence f and g are symmetric.
Theorem 2: Let F(a1, ··· ,an) and G(a1, ··· ,an) be polynomials of K[a1,···,an], ai being the elementary symmetric polynomials of x1, ··· , xn and (F, G) = H(a1, ··· ,an), F = f(x1, ··· , xn), =g(x1, ··· , xn), H = h(x1, ··· , xn); then h = ( f, g).
Proof: Let h = (f, g). By the preceding theorem, f: h' and g: h' are symmetric, whence h' is symmetric and can be represented by H'(a1, ··· ,an). Since H' is a common factor of F and G, it is divisible by H and therefore h is a factor of h'; however, h' is also a factor of h, for h' = (f, g), whence h and h' are associated, and the theorem applies.
3.5.5 Power sums: The symmetric polynomials
are called power sums. It follows from (1) for m < n that
whence
Similarly for m > n,
Theorem: Let K be a field of characteristic 0; then am can be expressed by s1, ··· ,sm with coefficients from K, whence the symmetric polynomials of K[x1, ··· ,xn] are polynomials of K[s1, ··· ,sm].
Proof: a1 = s1; by mathematical induction, assume that a1, ··· , am-1 can be expressed in terms of s1, ··· , sm-1. By (2), am can be expressed in terms of s1, ··· , sm.
Note that this theorem does not hold if K has a characteristic p £ n, and that the elementary symmetric polynomials cannot be expressed in terms of power sums with the aid of integral coefficients only. The power sums offer a possibility to express D² in a very simple form, where D is the alternating function defined in 3.5.3. Squaring column by column the determinant in 3.5.3 (3) yields
Exercise: A special problem of the theory of numbers asks for the sets of integral numbers a1, ··· , an and b1, ··· , bn, which satisfy the m conditions
Prove that there exists for m = n no solution except the trivial solutions, where b1, ··· , bn is a permutation of a1, ··· , an.
3-6. Solution of special equations by radicals: Equations of the type
are called binomial equations. If a is a solution of (1) and e is a primitive root of xn - 1, then, for j = 1, ··· , n, the n roots of (1) are given by a e j whence
Hence K(a, e) is the smallest extension of K admitting the complete reduction of xn - a. A solution of (1) is said to be an n-th root or n-th radical of a, or a square (cubic root, biquadratic) root when n = 2 (n = 3, n = 4). lf a is a positive number, there exists exactly one positive number among the solutions of (1) and, in general, this solution is referred to if one speaks of the radical.*
* * The notion of radicals is also used in the theory of ideals and of hypercomplex systems, but in a completely different sense.
If a1 is a radical of an element of K, moreover a2 is a radical of an element of K(a1), etc. and an is a radical of an element of K(a1, ··· , an-1), then every element of K(a1, ··· , an) can be generated from the elements of K by addition, subtraction, multiplication, division and extraction of roots. Questions regarding algebraic extensions of this type are answered by the Galois theory of algebraic equations. This theory is to be discussed in the second volume. Here, only some special classes of equations, which can be solved by means of radicals, will be investigated.
There is little loss of generality in assuming that in the equation
the coefficient b1 equals zero. Since -b1 is equal to the sum of the roots, b1=-(a1+···+an). The transformation
converts (3) into an equation where the coefficient b1 is replaced by 0. The transformation (4) can be performed unless n is divisible by the characteristic of the field. For this reason, it is assumed for the rest of this section that the characteristic of K is other than 2 and 3.
3.6.1 Cubic equations: For n = 2, under the assumption just made, the general equation can be reduced to x² - a = 0. This equation can be solved by extraction of one square root.
Let n = 3. The reduced form of the equation is
Let K be a field containing p and q and in a suitable extension of K the roots of (1) be a1, a2, a3. The elementary symmetric functions of the roots are therefore
By the formulae of 3.5.4, one computes easily the power sums
Let
D = (a1 - a2) (a1 - a 3) (a2 - a3). (4)
This alternating function of the roots is called the discriminant. Its square is a symmetric function of the roots and therefore it belongs to K. Using the formula for D² in 3.5.3, one finds
and, substituting the values (3) for the power sums,
The derivative of f(x) is
Hence
Moreover,
Hence
are elements of K(D, a1). On the other hand, since D is contained in K(a1,a2,a3),
Assume that f(x) is irreducible in K[x]. Then a1 is of degree 3 over K, while it follows from (5) that D is of degree 2 or 1 over K, whence
[K(D,a1) : K(D)] = 3.
In order to to find a1 by extraction of only square and cubic roots, one requires the roots of the cyclotomic polynomial x² + x + 1. These are denoted by
Moreover, introduce the Lagrange's symbols
It follows from
that
By elementary manipulation, one finds
(7) shows that Lagrange's symbols can be obtained by extraction of cubic roots only after extraction of the square roots of - 4p² - 27q² and of - 3. By extracting a cubic root, there remains an arbitrary factor. The two arbitrary factors, which occur when the cubic roots of the right hand sides in (7) are extracted, are interconnected as is shown by (8). If (w,a) takes a factor we, then (w²,a) takes w2e. These factors generate an even permutation of the roots a1, a2, a3. Extraction of the square roots of - 4p² - 27q² and of - 3 leaves a factor ± 1 arbitrary. These factors may interchange Lagrange's symbols and generate a transposition of a2 and a3. Thus, Equations (5), (6), (7) and (8) determine the three roots uniquely up to an arbitrary permutation of them and the result is obtained by mere extraction of square and cubic roots.
3.6.2 Bi-quadratic equations: For n = 4, the reduced form of the equation is
Let a1, a2, a3 and a4 be the roots in a suitable extension of K.
Let
Since a transposition of any two of the roots interchanges only the elements b1, b2, b3, an arbitrary permutation of a1, a2, a3 and a4 interchanges also the elements b1, b2, b3, whence any symmetric function of b1, b2, b3 is not altered, whence it is a symmetric function of a1, a2, a3 and a4. Thus, there exists an equation x3 + b1 x2 + b2x + b3 = 0 with elements from K which has the roots b1, b2, b3. Of course,
Thus, it is possible to find b1, b2, b3 by extraction of roots only. In order to find a1, a2, a3 and a4, one has to take into consideration that
whence a1 + a2 and a3 + a4 are the roots of x2— b1 = 0.
For every root, a factor ± 1 remains arbitrary, but since
applies, only two of the factors are arbitrary. A choice of these factors corresponds to a permutation of a1, a2, a3 and a4, as is seen from the final formulae:
If you exclude the cases where K is of characteristic 2 or 3, the equations of degree £ 4 can therefore be solved by extraction of square and cubic roots only.
3.7 Resultants:. Consider the polynomials
where the coefficients belong to the same field. In a suitably chosen extension, f(x) has the roots a1, ··· , an and g(x) has the root b1, ··· , bm. The necessary and sufficient condition for f(x) and g(x) to have a common root is that the resultant
equals zero. It follows immediately from (2) that
Hence R(f, g) is a polynomial in a1, ··· , an with the coefficients belonging to the ring which is generated by b1, ··· , bm; this polynomial is symmetric. It follows from 3.5.2 that R(f, g) can be expressed by the elementary symmetric polynomials of a1, ··· , an with coefficients belonging to the ring generated by b1, ··· , bm, and since those elementary symmetric polynomials differ from the coefficients of f(x) only by factors ± 1 , the resultant is an element of the ring generated by a1, ··· , an, b1, ··· , bm, whence
The reason why the constant 1 is used in this notation will become obvious in 3.7.2. Let the coefficients of f(x) and g(x) be polynomials of K[y], then R(f, g) is a polynomial of that ring, say, R(f, g) = R(y). If R(y) is of degree >0, there exist in a suitable extension of K elements h1, ··· , hs such that R(hj)=0. The polynomials f(x) and g(x) can be considered to be elements of the ring K[x,y], say
Then f(x,hj) and g(x,hj) have common roots; thus
In this way, one can find the common solutions of two algebraic equations of two variables
with the aid of a resultant.
3.7.1 The case when the coefficients of the highest term are equal to 1: In order to find R(f, g) in terms of the coefficients aj and bk, consider a1, ··· , an, b1, ··· , bm as indeterminates; the coefficients aj are symmetric polynomials in the indeterminates an , and similarly the coefficients bk are symmetric in the indeterminates bm . Every term
is a homogeneous polynomial in the n + m indeterminates a1, ··· , bm of degree
The integral number w is said to be the weight of A. Let A1, ··· , An be different terms of the type (1); then, by 3.5.5,
cannot equal 0 unless all the N coefficients cj
are zero. If the terms Ai are not of the same
type and the coefficients ai differ
from zero, then B is the sum of homogeneous polynomials in (a1,
··· , bm) of different degrees with
non-vanishing coefficients, whence it is not a homogeneous
polynomial. Since R(f, g) is a
homogeneous polynomial of degree n m in the indeterminates, it is
the sum of terms (1) of weight n m each. It is seen from 
that
the term bmn has the coefficient 1;
moreover, R(f, g) has the property
that it is zero when f(x) and g(x)
have a common root. It will be proved now that these three
properties are characteristic for a resultant.
Theorem 1: Let S be the sum of terms (1) of weight n m each with the property that S = 0 when f(x) and g(x) have a common root; moreover, let the coefficient of bmn be equal to 1, then S = R(f, g).
Proof: Express S as a polynomial in a1, ··· , bm; then S can be considered to be a polynomial y(ai), its coefficients being polynomials in the remaining n+m-1 indeterminates. y(ai) - y(bk) is divisible by ai - bk. Since S is zero when ai and bk are equal, y(bk) = 0. Hence S = y(ai) is divisible by ai-bk. In the ring generated by the indeterminates a1, ··· , bm, the factorization is unique, and the n m irreducible polynomials ai, ··· , bk are non-associate. Now S is divisible by each of them, whence it is divisible by their product which is equal to R(f, g). Since the terms of S are all of weight n m, the quotient must be of weight zero. By assumption, the term bmn has the coefficient 1 in S, that is, the same coefficient as in R(f, g); the quotient is therefore equal to 1, whence follows the theorem.
A polynomial in a1, ··· , bm with the properties assumed in the last theorem can be found by the following considerations. If a is a common root of f(x) and g(x), then the following n + m equations hold, and they can be considered as a set of linear homogeneous equations in an+m, an+m-1 , ··· ,a :
when f(x) and g(x) have a common root. The diagonal element is bmn and this term does not occur any more in the determinant, whence the term bmn in the determinant has the coefficient 1. The diagonal element's weight equals n m. By interchanging any two columns, the weight of the diagonal element either does not change or the element is equal to zero; in this manner, one can prove easily that all the terms of the determinant have the same weight, whence the assumption of Theorem 1 is satisfied and one has
Theorem 2: The determinant (4) is equal to R(f, g).
3.7.2 The general case: Let
be polynomials, where the coefficients a0, ....an, b0, ...,bn are elements of the same field.
By definition, the determinant
is called the resultant of F and G.
For a0 = b0 = 1, this definition agrees with that given for R(f, g) in 3.7.1; moreover, R(F, G) = (- 1)mnR(G, F). For any further investigation, one has to distinguish 4 cases:
1. Let a0 ¹ 0, b0 ¹ 0, a1, ....an be the roots of F(x) and b1, ....bm the roots of G(x).
Set
Then
Hence R(f, g) = 0 is the necessary and sufficient condition for F(x) and G(x) to have a common root.
2. Let a0 = b0 = 0. Then R(F, G) = 0, independently of the existence of a common root.
3. b0 = b1 = ··· = bm = 0. Then R(F, G) = 0, but as every element satisfies the equation g(x) = 0, every root of f(x) is a root of g(x). Similarly, if f(x) is the polynomial 0.
4. a0 ¹ 0, b0 = ... = br-1 = 0, br ¹ 0. Set Gr(x)x) = br xm-r + ... + bm, then
in this case, the resultant equals zero if and only if there exists a common root. Similarly, if b0 ¹ 0, a0 = ... = as-1 = 0, as ¹ 0, whence one has the
Theorem: If F(x) and G(x) are a common root, then R(F,G) = 0. If on the other hand R(F,G) = 0, then F(x) and G(x) have a common root, unless a0=b0= 0.
It appears that the resultant R(F,G) is not completely determined by the two polynomials F(x) and G(x) themselves, but by the manner how they are represented. In the definition of polynomial. it was stated (see 2.3.2) that terms with coefficients zero should not matter, i.e., that polynomials differing by such terms are considered to be equal. By adding terms with coefficients zero to F(x) and G(x), one can always arrange that a0=b0= 0, and therefore R(F,G) = 0. This difficulty can be overcome by a rule that the polynomials F(x) and G(x) should be written in a standard form, where the highest coefficients are different from zero unless the polynomial is the zero-polynomial. However, a rule of this kind would imply another incongruity for the case that the coefficients are functions of a variable, say y; the resultant would then cease to be a resultant for such values of y, for which a0(y) = 0, b0(y) = 0. For this reason, preference has been given to the present notation. Thus, the resultant depends on the mode of representation of the polynomials, and it ceases to be a sufficient condition for a common root as soon as a0 = b0 = 0 applies.
Example: To find the solutions of
Consider the left hand side as polynomials in x with coefficients from R[y]. The resultant is
For y = 0, a0=b0= 0 and the second equation has no solution. For the four other roots of y(y), a0 ¹ 0, b0 ¹ 0. There corresponds to each of these roots a solution, whence these solutions are
3.7.3 Linear representation of a resultant:
Theorem: Let D be the integral domain generated by the coefficients of F(x) and G(x), then there exist in D[x] polynomials u(x) of degree < m and v(x) of degree < n such that
Proof: Consider the n + m equations
These form a system of linear equations in x0, ··· , xn+m-1; the determinant of the matrix is equal to R(F,G). Multiply each equation by the cofactor of the last element and add the results. Then one gets Equation (1), where
and u1, ··· , vn are the cofactors of the elements of the last column, whence follows the theorem.
3.8 Closed fields: A field C is said to be closed if it is impossible to extend it algebraically, i.e., if every polynomial of C[x] of degree > 1 is reducible. The main result now is that the field of all complex numbers is closed.
Theorem 1: Let K be & field of characteristic 0, [L ; K] = 2, every polynomial of odd degree of K[x] and also every quadratic polynomial of L[x] has a root in L; then L is & closed field.
The proof will be given in two steps. At first, it will be shown that any polynomial y(x) of K[x] has a root in L, and then the same will be proved for the polynomials in L[x].
Proof: 1. Let y(x) be a polynomial of K[x]. Without loss of generality, one can assume that y(x) is irreducible; as K is supposed to have characteristic 0, the polynomial y(x) is separable (cf. 2.6.5) and all its roots in any suitable extension are therefore different. Let n be the degree of y(x) and n = 2m, where u º 1 (mod 2), then the proposition holds for m = 0; let it be true for m < k and prove it for m = k > 0. In a suitable extension, y(x) has the roots
For an arbitrary element c of K, the number of the elements
is n(n - 1) : 2.The ordered pairs of element ai aj, ai + aj are all different and the non-ordered pairs ai , aj are uniquely determined by ai aj, ai + aj. As K contains an infinity of elements, one can choose c in such a way that all the elements (2) differ and
if
By an arbitrary permutation of the elements (1), the elements (2) will only be interchanged. Hence every symmetric function of the elements (2) is also a symmetric function of (1), whence it is an element of K. Thus, the elements (2) are the roots of a polynomial f (x) of K[x] of degree n(n—l):2==2m-1u(n—1). As n is even, u(n - 1) is odd, whence f (x) has a root in L. Let a1a2 + c(a1 + a2) be such a root. a1a2 is a root of an irreducible polynomial f(x) of K[x] and a1 + a2 a root of an irreducible polynomial g(x) of K[x]. All the roots b ' = a1a2, b", ··· , bn of f(x) are different as the characteristic of K is 0, and so are the roots g ' = a1 + a2, g ", ··· , g n of g(x). Hence, by (3), all the elements b i + c g k differ. We therefore can apply the Lemma of 2.7.2 to the field K(b ', g '). Thus b '+ cg ' is a primitive element of K(a1a2, a1 + a2) = K(a1a2 + c(a1 + a2)), whence a1a2 and a1 + a2 belong to L. a1 and a2 are the roots of a polynomial of L[x] of degree 2. Hence a1 and a2 are elements of L, and therefore y(x) has roots in L.
2. Let y(x) be a
polynomial of L[x]. In order to prove that y(x)
has a root in L, consider the automorphisms of L as an extension of K. As [L : K] = 2, any
primitive element, say a of L, is a root of a quadratic polynomial which is
irreducible in K[x], but reduced in L[x] to Hence L is a normal
extension of K (cf. 3.7.3) and there exists an automorphism A of L which interchanges
a and 
,
whereas the elements of K remain invariant (cf. 2.7.4.2). The polynomial y(x)
is transformed by A into 
and b into a conjugate element. Then y(x) = P(x)
is a polynomial of K[x] and has therefore a root, say b in L. Since b is a root
of P(x), it is a root of y(x) or a
root of ; if b is a root of , then its conjugate is a root of y(x).
At any rate, y(x) has a root in L, and as it is supposed to
be irreducible, it is of degree 1; therefore every root of y(x)
belongs to L, whence follows the theorem.
It is easy to show that the assumptions of
Theorem 1 are satisfied if K is taken as the field of all real
numbers and L as the field of all complex numbers. Of course, the
characteristic of both the fields equals 0, moreover L = K(i),
whence [L : K] == 2. Let f(x) be a polynomial of an odd
degree in K[x], say
; set
then there is f(c) > 0 and f(-c) < 0, whence there exists a real number b in the interval (- c, + c) for which f(b) = 0. Thus, every polynomial of an odd degree with real coefficients has a root in K. Finally, every quadratic polynomial of L[x] has roots in L, since one can find them by extraction of square roots. Hence the assumptions of Theorem 1 are satisfied and therefore one has the
Fundamental Theorem of Classical Algebra: The field of the complex numbers is closed.
Corollary: Every algebraic extension of the field K of real numbers is isomorphic to K(i).
Proof: Let L be an algebraic extension of K and a be an element of L not belonging to K. Since a is a root of a polynomial of K[x], K(a) is isomorphic to a sub-field of K(i) othere than K, and as [K(i) : K] = 2, K(a) is isomorphic to K(i), whence K(a) is closed; every element of L is algebraic to K and must therefore be an element of K(a). Hence L = K(a).
The Fundamental Theorem of Classical Algebra can also be expressed in the form: Every polynomial with complex coefficients (which, for example, may be real and, in particular, rational) has complex roots, and can therefore be represented as a product of linear polynomials with complex coefficients.
therefore