Galois Groups An algebraic field is, by definition, a set of elements (numbers) that is closed under the ordinary arithmetical operations of addition, subtraction, multiplication, and division (except for division by zero). For example, the set of rational numbers is a field, whereas the integers are not a field, because they are not closed under the operation of division (i.e., the result of dividing one integer by another is not necessarily an integer). The real numbers also constitute a field, as do the complex numbers. It is possible to construct other examples of fields by means of extensions. For example, given the field Q of rational numbers, we can augment the set Q with a particular number e that is not in Q, and this automatically implies that every rational function of e is also in the extended field, e.g., The set of numbers that can be formed from the rationals and e by means of ordinary arithmetical operations constitute a new field E, which is an extension of the original field Q. An important special case of a field extension is when the number e is algebraic, i.e., the root of a polynomial with coefficients in the base field. In that case, division of one polynomial in e by any other can always be expressed as a simple polynomial with coefficients in the base field. (Hereafter "polynomial" refers to one with coefficients in the base field.) To prove this, let e be a root of f(e) = 0 for some polynomial f of degree d, and suppose we wish to evaluate the ratio g(e)/h(e) for any two polynomials g and h. Our claim is that this ratio equals a polynomial q(e) of degree no greater than d. This is true if and only if g(e) = h(e)q(e) where q is of degree no greater than d. By applying the identity f(e) = 0 we can reduce both sides of this putative equation to degree no greater than d, and then equate the coefficients of like powers to determine d rational conditions on the d coefficients of q(e), so we are assured of a solution. A field extension of the rationals Q based on an algebraic number is therefore denoted as Q[e], signifying that the elements are all polynomials in e with rational coefficients. (It's easy to see that these representations are unique, provided e is not in Q.) An important and interesting attribute of any given field is the set of automorphisms of the field. An automorphism is a one-to-one mapping from the set to itself such that the operations of addition and multiplication are preserved. In other words, if we let M(x) denote the image of x under the mapping M, this mapping is an automorphism if and only if it is one-to-one and satisfies the relations M(x+y) = M(x) + M(y) and M(xy) = M(x)M(y). If there are more than one automorphisms of a field, it's clear the composition of two or more automorphisms is also an automorphism, and these form a group. For the field Q (i.e., the rational numbers) it's easy to see that the only automorphism is the identity mapping I(x) = x. However, for some fields there exist non-trivial automorphisms. As an example, consider the field , which consists of all numbers of the form where p and q are rational numbers. In addition to the trivial identity mapping, this field possess another automorphism, consisting of the non-trivial mapping To prove that this is an automorphism, we note that it is one-to-one, and we have the relations and Obviously the composition of J with itself yields the identity I, so the group of automorphisms for this field {I,J}, with the group operation table For a slightly more complicated example, consider the field , which consists of all numbers of the form where q1, q2, q3, q4 are rational numbers. (For a demonstration that this is indeed a field, see Platonions.) Clearly the additive condition for an automorphism is satisfied by a mapping that negates any one or more of the terms. The product two numbers in this field is given by If we negate the coefficients with indices 2 and 4 in the arguments, the coefficients of the product with indices 2 and 4 (i.e., the coefficients of the square roots of 2 and 6) are negated whereas the others are unchanged. Likewise if we negate the coefficients with indices 2 and 3, the coefficients of the product with indices 2 and 3 are negated. Also, negating the terms in the arguments with indices 3 and 4 negates the corresponding terms in the product. If we denote these three mappings by N, J, and U, along with the identity mapping I, the group of automorphisms of the field is {I,N,J,U}, with the group operation table shown below Notice that of the four mappings the mappings I and U leave elements of the sub-field unchanged. In other words, I(a) = a and J(a) = a for all a Î . Thus the group of automorphisms of the field that leave the sub-field fixed consists of just {I, U}, with the group operation table shown below (Naturally this is isomorphic to the previous group {I,J}, since there is essentially just a single group of order two.) Compare this with the case of the field , whose only automorphisms are I and J, and obviously the only one of these that leaves elements of unchanged is the identity mapping I. Hence the group of automorphisms of that leave fixed is simply the identity group I of order 1. Now, for any given polynomial f(x) with coefficients in a field F (called the coefficient field), there exists an extension field E that contains all the roots of f. To construct this field we merely need to adjoin to the elements of F the roots of f(x). For example, consider the general quadratic polynomial f(x) = x2 + ax + b, the roots of which are The discriminant is D = a2 - 4b, so we need only adjoin the number to the set Q of rationals to give the extension field E = . This is the minimal extension field containing the roots of f(x), and it is called the splitting field of f (because f "splits" into linear factors in this field). Of course, if the discriminant D happens to be the square of a rational number, then E = F, but if D is not a square then E is a proper extension of F. Definition: The Galois group of a polynomial f with respect to the coefficient field F is defined as the group of automorphisms of the splitting field of f that leaves F fixed. From this definition we can see that the Galois group of a quadratic polynomial relative to the coefficient field Q depends on whether the discriminant is the square of a rational number. If it is, then the splitting field E of the polynomial is simply Q itself, and we've already seen that the group of automorphisms of Q that leave Q fixed is nothing but the identity mapping. Thus, if the discriminant of the quadratic f(x) is not a squared rational number, the Galois group of f(x) is the trivial group of order 1. On the other hand, if the discriminant is not the square of a rational number, the splitting field E is more complicated, i.e., it is of the form for some non-square D, and hence has a group of automorphisms of order four, as discussed above. Also, the sub-group of mappings that leave Q fixed is the group of order two. For any polynomial with coefficients cj in some field F, we refer to F as the coefficient field, and the splitting field E is some extension of F (or simply F itself, if the polynomial happens to be completely factorable in F[x]). The Galois group of the polynomial is then the group G of automorphisms of E that leave F fixed. Notice that if m is one of these automorphisms (mappings), and if q is a root of f, then m(q) is also a root of f. This follows immediately from the fact that automorphisms (by definition) preserve sums and products, and the fact that these particular automorphisms leave elements of F unchanged, so we have Hence m(f(x)) = f(m(x)), and since f(q) = 0 and the zero element is fixed under m, we have f(m(q)) = m(0) = 0, and thus m(q) is a root of f. For example, the group of automorphisms of the splitting field of an irreducible quadratic that leaves the coefficients fixed consists of the identity mapping m(a + b) = a + band the conjugation mapping m(a + b) = a - b, and of course with either of these mappings, if q is a root then so is m(q). For polynomials of higher degree, the automorphisms that generate the Galois group are just generalizations of the simple identity and conjugation mappings of quadratics. The preceding definition of a Galois group is actually a modernized version introduced by Emil Artin in the late 1920's (based on the earlier ideas of Emmy Noether and Richard Dedekind). When Evariste Galois himself first described the Galois group, he did so in terms of permutations of the roots of a given polynomial. The definitions are equivalent, but it is sometimes helpful to return to the original formulation of the subject. According to Galois, for any given polynomial f(x) of degree d with coefficients in a field F, there exists a unique group G of permutations of d entities such that every F-valued rational function of the roots of f is invariant under each of the permutations in G, and conversely, such that every rational function of the roots that is invariant under each of the permutations in G is F-valued. The group G is called the Galois group of the polynomial f relative to the coefficient field F. To illustrate, consider again the general quadratic polynomial f(x) = x2 + ax + b where the coefficients a,b are elements of a field F. Letting r1 and r2 denote the roots of f, we immediately have the F-valued rational functions -(r1 + r2) = a and r1 r2 = b, both of which are fully symmetrical in the roots. It can be shown that every symmetrical rational function of the roots can be expressed as a rational function of these two elementary symmetric functions. If these are the only F-valued rational functions of the roots, then the group of permutations is the fully symmetric group S2 of order two. However, if f(x) factors over the field F, this implies that r1 and r2 are individually rational numbers, although in general they are distinct, so the function r1 is F-valued but it is not invariant under permutations of the roots. Therefore, in this case, the group of the polynomial is not S2, it is a subgroup of S2, namely, the identity group of order 1. More generally, we see that the coefficients of a polynomial (of any degree) are really nothing but the elementary symmetric functions of the roots. For example, the cubic with roots r1, r2, r3 can be expressed as In general, for a polynomial of degree d, the coefficient of xd-j is the sum of all products of j distinct roots. It can be shown that every symmetric rational function of the roots can be expressed as a rational function of the coefficients of f(x), so if these coefficients are in the field F, then so are all the fully symmetrical functions of the roots. In this context, the term "rational function" signifies a function involving only the basic arithmetical operations of addition, subtraction multiplication, division, and constants that are rational numbers. Also, the term "fully symmetrical" signifies that the function is invariant with respect to every permutation of the d roots. Thus if a rational function of the roots is invariant under every permutation of those roots, we know that the invariant value is in the field F of the coefficients. However, there may be other rational functions of the roots of a given polynomial that are elements of F, even though these functions are not fully symmetrical in the roots. To illustrate, consider the quintic f(x) = x5 - 4x4 - 15x3 - 94x2 - 61x - 22 with coefficients in the set of rational numbers. We immediately know five fully symmetrical functions of the roots of this polynomial that yield rational values, namely the coefficient functions. For example, we have r1 + r2 + r3 + r4 + r5 = 4 This is obviously invariant under any permutation of the five roots, as are all the other coefficient functions. But the polynomial f(x) happens to factor into two polynomials with rational coefficients, i.e., we have f(x) = (x2 + 3x + 11) (x3 - 7x2 - 5x - 2) It follows that, letting r1 and r2 denote the roots of the quadratic factor, we have and for the roots of the cubic factor we have These functions are F-valued, but they are clearly not fully symmetrical in the roots, because they will be made false if we transpose r1 with r3 (for example). They do, however, possess symmetry for the roots {r1, r2} and for the roots {r3, r4, r5}. Consequently the Galois group for this polynomial over the rationals is the product of the symmetric groups S2 and S3, of order 2 and 6 respectively, and so the overall group is of order 12. In this case of the above quintic we found rational-valued functions of just certain subsets of the roots, so we could immediately infer that the original polynomial was factorable over the rationals. However, even if a polynomial is irreducible over the rationals, it is still possible to have non-symmetrical rational functions of the roots that are elements of F. As an example, consider the quartic polynomial f(x) = x4 - 2x3 + 4x + 2 This is irreducible over the rationals, but there are non-symmetrical rational functions of the roots that yield rational values. The four roots are where u and v are the complex numbers and the overbars indicate complex conjugation. It is easily verified that there are several non-symmetrical rational functions of these roots that yield rational values, such as We find that these functions all have lesser symmetries, specifically, that q1 and q2 can be transposed, and q3 and q4 can be transposed. In addition, the pair q1,q2 can be transposed with the pair q3,q4 . Thus, of the 24 possible permutations of these four roots, the Galois group consists only of the subgroup consisting of those permutations that conform to these extra conditions. This limits us to the group of eight permutations shown below. {abcd}, {bacd}, {abdc}, {badc}, {cdab}, {dcab}, {cdba}, {dcba} This, therefore, is the Galois group of the preceding polynomial relative to the coefficient field Q. One of the most important applications of Galois theory (indeed, the reason it was invented) is to provide the criterion for deciding when a polynomial is solvable by means of rational operations and root extractions. This is done by exploiting the correspondence between fields and their respective automorphism groups. In general there is a sequence of sub-fields between the splitting field and the coefficient field, and these correspond to a sequence of subgroups. It is possible to construct, by rational operations and root extractions, an extension from one field to the next only if the group of automorphisms of the larger group that leave the elements of the sub-field fixed is abelian, so solvability by radicals requires that we can decompose the sequence of groups from the splitting field down to the coefficient field into a sequence of Abelian steps. From group theory it can be shown that this is possible for the fully symmetric permutation groups of two, three, or four entities (of orders 2, 6, and 24 respectively), but not for the fully symmetric group of five (or more) entities. It can also be shown that for each degree d there exist polynomials whose Galois group is the fully symmetric group Sd. Consequently, there can be no general algebraic formula (involving just rational operations and root extractions) for the roots of polynomials of degree greater than four. Return to MathPages Main Menu