Rings and Fields

Solution 1

  1. Since 0 + 0 = 0 we have (0 + 0).a = 0.a and so by the Distributive law we get 0.a + 0.a = 0.a and adding the additive inverse of 0.a to each side we get 0.a = 0.

    Then 0 = (-1).0 = (-1).(1 + (-1)) = -1 + (-1)2 and adding 1 to both sides gives 1 = (-1)2.

  2. The ring axioms for T follow from those for R.
    The multiplicative identity of T is the function s ↦ 1 for all s ∈S.
    Invertible functions are those for which f(s) ≠ 0 for any s ∈ S.
    Defining a function f from S = {a, b} to R means giving a pair of real numbers for x = f(a) and y = f(b) and so the set T can be regarded as the Cartesian product R × R.
    Then the addition and multiplication on R × R become:
    (x1 , y1) + (x2 , y2) = (x1 + x2 , y1 + y2) and (x1 , y1) . (x2 , y2) = (x1 . x2 , y1 . y2)

  3. The ring axioms follow because sums, differences and products of numbers of this form also have this form. The other ring axioms then follow from those in R.
    (√2 + 1)(√2 - 1) = 2 - 1 = 1 and so (√2 + 1)-1 = √2 - 1. Similarly (2√2 + 3)-1 = 2√2 - 3.
    An element a + b√2 has a multiplicative inverse if and only if a2 - 2b2 = ±1

  4. As in the last question sums, differences and products of numbers of this form also have this form and the other axioms follow from those in C.
    The elements of R form an "equiangular lattice" as shown.
    Following the hint, the only invertible elements are those for which |z| = 1 since there are no non-zero elements inside the unit circle. Thus the invertible elements are the six ones indicated.


  5. The multiplicative identity is the "identity function": x ↦ x. It is a standard result in set theory that the invertible functions are those which are one-one and onto (bijections).
    If x ∈ R then (f + g)h(x) = (f + g)(h(x)) = f(h(x)) + g(h(x)) = fh(x) + gh(x) = (fh + gh)(x).
    If (say) h(x) = x2 then it is easy to verify that for most functions f, g the other distributive law fails.
    Something like this is called a Near-ring.

  6. a)  a2 + b2 = (p2 - q2)2 + (2pq)2 = (p4 - 2p2q2 + q4)+ 4p2q2 = p4 + 2p2q2 + q4 = (p2 + q2)2 = c2.

    b)  Any square modulo 4 is either 0 or 1. Since a and b are coprime they are not both even and since a2 + b2 is a square this must be 1 mod 4. Hence a and b cannot both be odd.

    c)  We have both a and c odd and so a ± b are both even and we may take (c + a)/2 = P and (c - a)/2 = Q. We need to show that P and Q are exact squares.
    Then 4PQ = (c + a)(c - a) = c2 - a2 = b2. Since a, b are coprime, so are a, b and c and hence c + a and c - a have no common factors. So P and Q are coprime and since the product 4PQ is a square, both P and Q are squares. So take p = √P and q = √Q and we have b = 2pq as required.

    Taking small coprime values for p, q with p, q not both odd gives the eight primitive Pythagorean triples:
    (3, 4, 5), (15, 8, 17), (35, 12, 37), (5, 12, 13), (21, 20, 29), (45, 28, 53), (7, 24, 25), (9, 40, 41).

    d)  The above pair (a, b) = (21, 20) is pretty good. The next good ones are (119, 120), (697, 696), (4059, 4060), (23661, 23660), (137903, 137904), ... corresponding to p = 5, 12, 29, 70, 169, 408, ... and q = 2, 5, 12, 29, 70, 169, ... which you can find using Maple. Can you see how these sequences of p's and q's continue?
    e)  This suggests looking for a - b = ±1 or equivalently (p2 - q2) - 2pq = ±1. We can write this as (p - q)2 - 2q2 = ±1 or, putting p - q = r as r2 - 2q2 = ±1. This is a famous equation called Pell's equation and can be solved (with infinitely many solutions) by continued fractions methods.
    For example, take r = 577 and q = 408 giving p = 985 and (a, b) = (803761, 803760).
    You can find out more about this at:
    http://www-history.mcs.st-and.ac.uk/HistTopics/Pell.html