Solutions 3

1. The subring must contain 1, 1+1, ..., -1, -1 + (-1), ... and hence all the integers. Hence the subring is .
   Since the sub-field must also contain all integers and their inverses and the products of these, it must contain all rational numbers and hence is .
   In any commutative ring with an identity, the ideal < 1 > is the whole ring.
   The subring of generated by 2 contains all integer multiples of 2 as well as the integer 2. It is {a + b2 | a 2, b }.
   The subfield of R generated by 2 is {r + s2 | r, s 2 }.

2. The subring generated by A is {0, A} since this is closed under + and .
   The ideal generated by A contains A B for any subset B of S. Hence the ideal is the Boolean ring of all subsets of A.

3. To show that a 0 has an inverse, look at the ideal generated by a. Since this is the whole ring, 1 < a > and so 1 = ab for some b R as required.

   The non-invertible elements do not form an ideal. For example, in the elements 2, 3 are not invertible but 3 - 2 = 1 is invertible.

4. The gcd = 1 as the following calculation shows.
   
   It follows that 1 = -(x + 1)/8 (x³ + 5x - 2) + (x² + x + 2)/8 (x² + 3).

   Maple does it very painlessly:

5. In we have (2 + i)/(1 + i) = (3 - i)/2 which is "in the middle" of one of the integer squares. So any of the corners: 1, 2, 1 - i, 2 - i will do as quotients. The corresponding remainders are: 1, -i, i, -1.

   Use MacTutor or trial and error to find a "three quotient" example. e.g. u = 1 + i, v = 4.
   For a "two quotient" example take for example: u = 1, v = 2 giving 0 and 1 as possible quotients.

   The only time the quotient is unique is when the remainder is 0

6. Write down the prime factorisations: a = p₁^a₁p₂^a₂ ... and b = p₁^b₁p₂^b₂ ... Then gcd(a, b) = p₁^{min(a₁ , b₁)}p₂^{min(a₂ , b₂)} ... and lcm(a, b) = p₁^{max(a₁ , b₁)}p₂^{max(a₂ , b₂)} ... and the result follows.
   A lcm of two polynomials is a polynomial of least degree divisible by both and the lcm of two Gaussian integers is a Gaussian integer of lowest norm divisible by both.
   The theorems are then just as stated for the integers. The proofs can wait until you have seen prime decompositions for the rings [x] and [i].

7. We have 0 = (x₁ - x₂)a + (y₁ - y₂)b and so if we have a = ds and b = dt for integers s, t we must have t dividing x₁ - x₂ and s dividing y₁ - y₂ . In particular, if a, b are coprime, the x's differ by a multiple of b and the y's differ by the same multiple of a.

8. It is easy to verify the subring criterion.
   The determinant of the given matrix is a² - 2b² and since 2 is irrational this is never 0 if either a, b - {0}.
   Taking x = a/b in a field _p , all the non-zero matrices of the given form will have inverses if x² 2 in _p . It is easy to verify this for p = 3, 5, 11, 13, 19, 29, 37, ... but not for p = 7, 17, 23, 31, 41, ...

   If you choose k to be an integer which does not have a square root modulo p (about half the integers don't) then one can get a field out of this process.