Integral Domains: Strange Integers

Z is the set of all integers ..., -2, -1, 0, 1, 2, ... Form the set Z[3] = {a + b3: a, b ∈ Z}. For example, 99999 + 2222223Z[3].

As a subset of the set of real numbers R, Z[3] is closed under operations of addition, subtraction, and multiplication. In terms of the two components associated with every number in Z[3], the arithmetic operations are expressed as following.

  • Addition:
      (a1 + b13) + (a2 + b23) = (a1 + a2) + (b1 + b2)3.
  • Subtraction:
      (a1 + b13) - (a2 - b23) = (a1 - a2) + (b1 + b2)3.
  • Multiplication:
      (a1 + b13)·(a2 + b23) = (a1·a2 + 3b1·b2) + (a1b2 + a2b1)3.

Both addition and multiplication are commutative. 0 and 1 are elements of Z[3] (0 = 0 + 03 and 1 = 1 + 03) Since the operations are actually the usual arithmetic operations on real numbers, we also have the distributive law.This makes Z[3] a commutative ring just like Z. In Z, from ab = 0 we can conclude that either a = 0 or b = 0. In Algebra, nonzero elements for which ab = 0 are known as divisors of zero. So no element of Z is a divisor of zero. Rings with this property are called integral domains. Z is one example of integral domain. Z[3] is another. To show this, it is convenient to introduce a couple of notations.

For any A = a + b3 define its conjugate A' = a - b3. The conjugate has several important properties. For example,

  1. (A + B)' = A' + B', and
  2. (A · B)' = A' · B'.
Note that AA' = a² - 3b². Define the function N(A) = a² - 3b². N(A) = N(A'). Since 3 is irrational, N(A) ≠ 0, except for A = 0.

Lemma

Z[3] is an integral domain.

Proof

Note that N(AB) = N(A)N(B). Indeed, N(AB) = (AB)·(AB)' = ABA'B' = AA'BB' = N(A)N(B).

Assume AB = 0. Then also N(AB) = 0. From the preceding sentence, N(A)N(B) = 0, i.e., either N(A) = 0 or N(B) = 0. This implies that either A = 0 or B = 0;

(Write down the identity N(AB) = N(A)N(B) explicitly. You may be surprised to discover what actually has been proven so easily in an abstract form.)

Z and Z[3] have other features in common. Methods that work in Z may be useful for solving problems in Z[3]. Here is a problem that has at least two solutions, one borrowed from Z, another that makes use of the peculiar structure of Z[3].

Problem

Is 99999 + 2222223 a square of a number from Z[3]?

Solution 1

Is integer 3726125 a square of an integer? This is the kind of questions to answer which K.F.Gauss (1777-1855) in 1801 invented Modular Arithmetic. Consider operations modulo 9. Look at the main diagonal of the multiplication table. Unless modulo 9 a number is one of 0, 1, 4, 7, it can't be a square of an integer. 3726125 = 8 (mod 9). (Casting out 9s is easy; for 3 + 6 = 9 and 7 + 2 = 9.) Therefore, the answer to the question is negative: 3726125 is not a square of an integer.

Let's apply this approach to solving our problem.

(a + b3)² = (a² + 3b²) + 2ab3

If 99999 = a² + 3b² for a couple of integers a and b, then a² is divisible by 3. Then by Euclid's Proposition VII.30 a is divisible by 3. Thus a² = 0 (mod 9), and necessarily 3b² = 0 (mod 9). Which implies that b = 0 (mod 3) also. We finally get 2ab = 0 (mod 9). However, 222222 = 3 (mod 9). The answer to our problem is negative, 99999 + 2222223 can't be a square of a number from Z[3].

Solution 2

Assume that 99999 + 2222223 = (a + b3)² for some integer a and b. Then it must be that also 99999 - 2222223 = (a - b3)². But the latter is impossible since 99999 - 2222223 < 0. And no square of a real number (a - b3, in our case) can be less than 0.

In Z[3], the two methods lead to different generalizations of the problem.

  1. From Solution 1, if integers a and b are such that a = 0 (mod 9) while b is not then a + b3 is not a square of a number from Z[3].
  2. From Solution 2, if A' < 0, then a + b3 is not a square of a number from Z[3]. (This is of course true also if A < 0.)

Both admit further generalizations. For example, we can define Z[m] where m is not a square of an integer. #1 extends easily to the case of an odd m. For #2, consider Q[m] = {a + bm: a, b∈Q}, Q being the set of all rational numbers. All the operations and the conjugate operator are defined analogously to Z[3]. Q[m] is a field. It is the smallest field that contains all rational numbers and m. It is known as the extension (or Galois) field of Q by m. (We already encountered a similar construct in the discussion on Orthogonal Latin Squares.) Obviously, #2 remains true for Q[m], where m is not a complete square of an integer.

Remark

To establish that Q[m] is a field, one has to verify that it is closed under division. I.e., the result of division of two numbers from Q[m] is itself an element of Q[m]. (This has to do with removing irrationality from the denominator.)

Problem

Show that the set Z[3] is dense on the real line.

Solution

We have to show that for any real number r and any positive ε there exists a number a + b3, with integer a and b such that

|r - (a + b3)| < ε.

In other words, we have to show that, by a proper choice of integers a and b, |r - (a + b3)| can be made arbitrarily small.

First of all, note that we can easily achieve this goal for r = 0. Indeed, since m = (-1 + 3) < 1, the integer powers mn of m become, as n grows, as small as we wish. Of course, mnZ! Thus, we can find integer a and b such that

(*) |k| < ε, k = a + b3.

Assume k > 0. (If it's not, take -k instead.)

For a given real number r

[r]≤ r < [r] + 1,

where [r] is the whole part of r. Consider the sequence

[r], [r] + k, [r] + 2k, [r] + 3k, ...

Since k is not 0, r will fall between a pair of successive terms of the sequence. It's distance to either of the terms would not exceed k and, from (*), ε.

And we are done.

Constructible Numbers, Geometric Construction, Gauss' and Galois' Theories

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