Free Trial

Safari Books Online is a digital library providing on-demand subscription access to thousands of learning resources.


Share this Page URL
Help

Part V Theorems and Problems > V.22 Liouville's Theorem and Roth's Theorem - Pg. 710

710 G = Gal(f ) has a fixed field H , which is defined to be the set of all numbers a E such that (a) = a for all H. Galois proved that the association between H and H gives a one-to-one correspondence between subgroups of G and fields which lie between Q and E (the so-called intermediate subfields of E). The con- dition that f (x) has a radical formula for its roots leads to certain special kinds of intermediate subfields, and hence to certain special subgroups of G, and even- tually to Galois's most famous theorem: the polyno- mial f (x) has a radical formula for its roots if and only if its Galois group Gal(f ) is a soluble group. (This means that G = Gal(f ) has a sequence of subgroups 1 = G 0 < G 1 < · · · < G r = G such that for each i, G i is a normal subgroup [I.3 §3.3] of G i+1 and the factor group G i+1 /G i is Abelian.) It follows from Galois's theorem that to demonstrate the insolubility of the quintic, it is enough to produce a quintic f (x) such that Gal(f ) is not a soluble group. An example of such a quintic is f (x) = 2x 5 - 5x 4 + 5: one can show first that Gal(f ) is isomorphic to the symmet- ric group S 5 ; and second that S 5 is not a soluble group. Here is a brief sketch of how the argument goes. First one establishes that f (x) is an irreducible polynomial (i.e., is not the product of two rational polynomials of smaller degree) with five distinct complex roots. Thus, as observed above, Gal(f ) can be regarded as a sub- group of S 5 that permutes the five roots. By sketch- ing the graph of f (x) one can easily see that three of its roots are real and that the other two, call them 1 and 2 , are complex conjugates of each other. Since ¯ the complex conjugation map z z always gives an automorphism in Gal(f ), it follows that Gal(f ) is a subgroup of S 5 that contains a 2-cycle, namely ( 1 2 ). Another basic general fact is that the Galois group of an irreducible polynomial permutes the roots transitively, meaning that for any two roots i , j there exists an automorphism in Gal(f ) that sends i to j . Thus, our group Gal(f ) is a subgroup of S 5 that permutes the five roots transitively and contains a 2-cycle. At this point some fairly elementary group theory shows that Gal(f ) must actually be the whole of S 5 . Finally, the fact that S 5 is not a soluble group follows easily from the fact that the alternating group A 5 is a non-Abelian simple group (i.e., it has no normal subgroups apart from the identity subgroup and A 5 itself). These ideas can be extended to produce polynomials 5 that have Galois group S n , and of any degree n that are therefore not soluble by radicals. The reason V. Theorems and Problems this cannot be done for quartics, cubics, and quadratics is that S 4 and all its subgroups are soluble groups. V.22 Liouville's Theorem and Roth's Theorem One of the most famous theorems in mathematics is the statement that 2 is irrational. This means that there is no pair of integers p and q such that 2 = p/q, or equivalently that the equation p 2 = 2q 2 has no integer solutions apart from the trivial solution p = q = 0. The argument that proves this can be considerably general- ized, and, in fact, if P (x) is any polynomial with integer coefficients and leading coefficient 1, then all its roots are either integers or irrational numbers. For example, since x 3 + x - 1 is negative when x = 0 and positive when x = 1 it must have a root strictly between 0 and 1. This root is not an integer, so it must be irrational. Once one has proved that a number is irrational, it may seem as though not much more can be said. How- ever, this is very far from true: given an irrational num- ber, one can ask how close it is to being rational, and fascinating and extremely difficult questions arise as soon as one does so. It is not immediately obvious what this question means, since every irrational number can be approx- imated as closely as you like by rational numbers. For example, the decimal expansion of 2 begins 1.414213 . . . , which tells us that 2 is within 1/100 000 of the rational number 141 421/100 000. More gener- ally, for any positive integer q we can let p be the largest integer such that p/q < 2, and then p/q will be within 1/q of 2. In other words, if we want an approximation to 2 with accuracy 1/q, we can obtain it if we use a denominator of q. However, we can now ask the following question: are there denominators q for which one can one obtain an accuracy much better than 1/q? The answer turns out to be yes. To see this, let N be a positive integer and con- sider the numbers 0, 2, 2 2, . . . , N 2. Each of these can be written in the form m + , where m is an integer and , the fractional part, lies between 0 and 1. Since there are N + 1 numbers, at least two of their fractional parts must be within 1/N of each other. That is, we can find integers r < s between 0 and N such that if we write r 2 = n+ and s 2 = m+, then |-| 1/N. Thus, if we set = -, we have (s -r ) 2 = n-m+ and || 1/N. If we now let q = s - r and p = n - m, 1/qN. Since then 2 = p/q + /q, so | 2 - p/q| q, 1/qN 1/q 2 , so for at least some positive N