Course MT4517 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 s1.
    The multiplicative identity of T is the function s goesto 1 for all s belongsS.
    Invertible functions are those for which f(s) noteq 0 for any s belongs S.
    Defining a function f from S = {a, b} to s1 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 s1 cross s1.
    Then the addition and multiplication on s1 cross s1 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 s1.
    (sqrt2 + 1)(sqrt2 - 1) = 2 - 1 = 1 and so (sqrt2 + 1)-1 = sqrt2 - 1. Similarly (2sqrt2 + 3)-1 = 2sqrt2 - 3.
    An element a + bsqrt2 has a multiplicative inverse if and only if a2 - 2b2 = plusminus1

  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 s1.
    The elements of R form an "equiangular lattice" as shown.
    Following the hint, the only invertible lements 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 goesto x. It is a standard result in set theory that the invertible functions are those which are one-one and onto (bijections).
    If x belongs s1 then (f + g)comph(x) = (f + g)(h(x)) = f(h(x)) + g(h(x)) = fcomph(x) + gcomph(x) = (fcomph + gcomph)(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 plusminus 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 = sqrtP and q = sqrtQ 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 = plusminus1 or equivalently (p2 - q2) - 2pq = plusminus1. We can write this as (p - q)2 - 2q2 = plusminus1 or, putting p - q = r as r2 - 2q2 = plusminus1. 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