ALGEBRAIC NUMBERS Failure of Expected Laws.-The roots of an equation • • • = 0 having integral coefficients are called al gebraic numbers. If a= 1, the roots are integral algebraic num bers; examples are I5, ;15. Prior to a lengthy controversy in 1847, it was regarded as self-evident that the laws of divisibility which hold for whole numbers continue to hold also for integral algebraic numbers. Fundamental for mathematics was the recognition of the falsity of this intuition and the invention of "ideals" serving to restore those laws of divisibility.
A typical example is the set S of all quadratic integers a+b0, where 0 =1,1- 5, while a and b are whole numbers. By an in decomposable number n of this set S we mean one having no factor in S other than n, -n, 1, -1. If 2+ 0=(x+y0)(z+w0), where x, y, z, w are whole numbers, we multiply by the like equa tion having - 0 in place of 0, and get 9 = But 3 is evidently impossible in whole numbers. Hence we may take I, whence x+y0=± I. The same discus sion applies also if we replace 2+ 0 by 2 - 0 or 3. Hence 9 = 3.3 and 9 = (2+ 0)(2- 0) give two factorizations of 9 into indecom posable numbers of S. The law of unique factorization of whole numbers into primes therefore fails for numbers of S. Since the indecomposable number 3 divides the product 9 of 2+ 0 and 2 - e without dividing either of them, we have a second failure in S of laws of ordinary primes. A third more astonishing failure in S of the laws of arithmetic is the absence in S of a greatest common divisor of 9 and t= 3-60; their only common divisors are ± I, ± 3 and ± (2 - 0), no one of which is divisible by the remaining five.
of a number of an ideal by a number of S is a number of the ideal. Any set of numbers of S is called an ideal if it has these two prop erties and is not composed exclusively of the number zero.
When x and y range independently over all whole numbers, the totality x-1-yn is denoted by n]. It evidently has property (a). To show that (3= [3, I +0] is an ideal, it remains to verify that it has property (b). This will follow if we show that 3p and (1+0)p belong to for every p in S. But p=a+b0, where a and b are whole numbers, and 3p= 3(a-b)+3b(I-1- 0), (1- I- 0)p= -6b-Ka+b)(1+ 0) are numbers in Since we may replace 0 by - 0 in this proof, we conclude that also a = [3, I - 0] is an ideal.
If L ranges over the numbers of an ideal X and if M ranges over those of an ideal g, then the set composed of the products LM and all their sums and differences is seen to have properties (a) and (b) and hence is an ideal, called the product of X and /..t, and denoted by either X,u or /2X. In particular, a$= [9, 3+3 0, 3-30, I - = = [3, 30] = (3).
Similarly, (2- 0) and hence (2+6).
To remove the above difficulties concerning numbers 9, 3, • • • , 1=3-60 of S, we pass to the uniquely corresponding principal ideals (9), (3), • • • , (1). Now (9) = (3) (3) (2+ 0) • (2 - 0), and we have the single factorization (9) = instead of the former two factorizations of 9. Finally, y = [7, 3+0] is an ideal such that 07= (I - 20), so that (9) and (1) = (3)07 have the greatest common ideal divisor am.
If an ideal is different from the principal ideal (I) and if it is divisible by no ideal other than itself and (I), it is called a prime ideal. It is readily proved that every ideal which is not a prime can be expressed in one and but one way as a product of (a finite number of) prime ideals. Hence the laws of factorization in arithmetic hold for our ideals. Since a, 0, 7 are prime ideals, the above three difficulties have been completely removed.