Rings and Fields

Solution 2

  1. The associativity and distributivity axioms follow from the diagrams


    The zero element is the empty set and each element is its own additive inverse.
    The set S itself is the multiplicative identity and is the only element with a multiplicative inverse.

    With A + B = AB we do not get a ring. The only thing that goes wrong is additive inverses.

  2. Let u, v be the vectors (x1 , y1 , z1) and (x2 , y2 , z2).
    The product of the corresponding "pure quaternions"
    x1 i + y1 j + z1 k and x2 i + y2 j + z2 k is
    (x1x2 + y1y2 + z1z2 + (y1z2 - z1y2)i + (z1x2 - x1z2)j + (x1y2 - y1x2)k
    and this corresponds to u.v + u × v.

    The only thing that stops (R3, +, ×) being a ring is that × is not associative.
    To see this observe that (u × v) × w is perpendicular to u × v and so in the plane spanned by u and v, while u × (v × w) is perpendicular to v × w and so in the plane spanned by v and w. It is then easy to find a counterexample.

  3. This is an ideal (and hence a subring) since it is generated by x2.

    Check it directly [using the fact that if p(1) = q(1) = 0 then (p - q)(1) = 0 and p(1)r(1) = 0 for any polynomial r] or (cleverer!) observe that this is the ideal < x - 1 > generated by the polynomial x - 1.

  4. It is easy to check the axioms. The multiplicative inverse of a + b√2 is a/(a2 - 2b2) - b/(a2 - 2b2)√2.

  5. This is a ring of order 4 but (x + 1)2 = 0 and so it has a zero divisor and is not a field.

    As usual, this is a commutative ring with identity and so we need only find multiplicative inverses.

    1-1 = 1 and 2-1 = 2 since -1 = 2 the "rule" gives x(x + 2) = 1 and we have inverses for two more elements. Squaring this gives (x + 1)(2x + 2) = 1 and so we have two more. Also -x(-x - 2) = 1 and this gives 2x(2x + 1) = 1 and this is the last pair.

    A cleverer way to work is to observe that powers of x generate the eight non-zero elements of the ring and then (xm)-1 = x8-m.

  6. If the additive order of an element a is m = pq then (writing 1 + 1 = 2, etc) we get ma = 0 = p.qa and so p and qa are zero divisors. The only way of avoiding this is for m to be a prime number.

    The order of any non-zero element is then the same as the order of 1.

  7. There is a choice of p elements for each coefficient and hence there are pk elements. Since the multiplication rule tells us p(x) = 0 in this ring, if we can split p(x) as a product of two lower degree polynomials, these elements of the ring will be zero-divisors and so we will not have a field.

    In fact, if the polynomial p(x) cannot be factorised ("is irreducible") then using the "rule" p(x) = 0 will give a field.

    If the polynomial x2 - x - 1 ∈ Z3[x] could be factored it would have to have linear factors and hence would have a root. But it is easy to check that substituting x = 0, 1, 2 will not give 0.