Solutions 4

1. It is easy to verify the conditions that I J is an ideal.

   An element of IJ is of the form i₁j₁ + i₂j₂ + ... + i_kj_k and each of these terms is in both I and J and hence in I J.

   If I = < 4 > in and J = < 6 > then all products ij IJ are divisible by 24. Thus IJ = < 24 >.
   If an ideal contains both I and J all its elements must be divisible by 12. Thus I J = < 12 >.
   The integer 2 I + J and it is easy to see that all elements of I + J are divisble by 2. Thus I + J = < 2 >.
   More generally, If I = < m > in and J = < n > then IJ = < mn > , I J = < lcm(m, n) > and I + J = < gcd(m, n > .

2. The proof is similar to that showing that is a principal ideal domain. It is equivalent to the group theoretic result that any subgroup of a cyclic group is cyclic.
   To find all the ideals of _n (or subgroups of this cyclic group) take the ideals generated by all the divisors of n.
   For example, ₁₂ has subgroups generated by 1, 2, 3, 4, 6 and 12 = 0 of orders (respectively) 12, 6, 4, 3, 2 and 1.

3. As groups 2 and 3 are both infinite cyclic groups generated by 2 and 3 respectively.
   Any ring homomorphism would have to take an additive generator to an additive generator and thus map 2 to 3. However 2² = 4 would then have to map to 6 (3)².

   A suitable group isomorphism is 1 (1, 1). Note that since 2, 3 are coprime the (additive) order of (1, 1) = 6.
   Then an element a = (1 + 1 + ... + 1) maps to (a, a) and b maps to (b, b). The product ab maps to (ab, ab) which is where it has to go to be a ring isomorphism.

   More generally, the same proof works for any pair of integers m, n which are coprime.

4. You need to verify that the map is well-defined. If m = n mod 12 then 12 divides m - n and so 4 divides m - n and so the image of m and n in ₄ is the same. The other properties follow easily. Its kernel is the ideal generated by 3.

   This "well-definition" fails working modulo 14. For example 2 = 16 mod 14 but not modulo 7. This means you get problems with the homomorphism property. For example 8 0 but 8 + 8 = 2 2 0 + 0.

5. Invariance of multiplication fails for the first and invariance of addition fails for the third. Note that since is a field the kernel of any homomorphism is either {0} or and so these clearly cannot be homomorphisms.
   The second is an isomorphism from to as you can verify.
   The fourth is not a homomorphism since, for example, 1 i but 1² 1 i² = -1.

6. The product of units is a unit and so the set U_n forms a multiplicative group. We'll write them as products of cyclic groups C_i
   U₂ C₁ , U₃ C₂ , U₄ C₂ , U₅ C₄ , U₆ C₂ , U₇ C₆ , U₈ C₂ C₂ , U₉ = C₂ C₃ C₆ , U₁₀ C₄ , ...
   The pattern you should notice is that if n is prime the group U_n C_n-1 . Then from Question 3 above if m, n are coprime U_mn U_m U_n and that should enable you to take the calculation quite far.

7. A ring homomorphism from ₁₂ onto ₄ will have the ideal < 3 > as fernel and is the map of Qu 4 above.
   Look at the (additive) orders of elements to conclude that the only homomorphism is x 0 for all x.
   One can only get a homomorphism from _m onto _n if n divides m and then it is the map x x mod n.

8. There is only one possible addition table for a ring {0, 1} of order 2 and then either 1.1 = 0 or 1.1 = 1 give two rings.

   If the additive group is cyclic every element is of the form na for some integer n and the (ma).(na) = (mn)a² and so if you know a² the multipication is determined.

   The additive group of a ring of order 3 must be cyclic so R = {0, a, 2a}. Then we can either take a² = 0 or a² = a. If we take a² = 2a then we have (2a)² = 4a²= 2a and so if we swap the generators a and 2a we get isomorphic rings. Thus there are two possible rings of order 3

   For rings of order 4 with a cyclic additive group {0, a, 2a, 3a} we have a choice of 0, a or 2a for a². Choosing a² = 3a gives a ring isomorphic to the a² = a case.
   There are in fact eight rings whose additive group is the Klein 4-group.