- It is easy to verify the conditions that
*I*∩*J*is an ideal.An element of

*IJ*is of the form*i*_{1}*j*_{1}+*i*_{2}*j*_{2}+ ... +*i*_{k}*j*_{k}and each of these terms is in both*I*and*J*and hence in*I*∩*J*.If

*I*= < 4 > in**Z**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**Z**and*J*= <*n*> then*IJ*= <*mn*> ,*I*∩*J*= <*lcm*(*m*,*n*) > and*I*+*J*= <*gcd*(*m*,*n*> .

- The proof is similar to that showing that
**Z**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**Z**_{n}(or subgroups of this cyclic group) take the ideals generated by all the divisors of*n*.

For example,**Z**_{12}has subgroups generated by 1, 2, 3, 4, 6 and 12 = 0 of orders (respectively) 12, 6, 4, 3, 2 and 1.

- As groups 2
**Z**and 3**Z**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 2^{2}= 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.

- 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**Z**_{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.

- Invariance of multiplication fails for the first and invariance of addition fails for the third. Note that since
**C**is a field the kernel of any homomorphism is either {0} or**C**and so these clearly cannot be homomorphisms.

The second is an isomorphism from**C**to**C**as you can verify.

The fourth is not a homomorphism since, for example, 1 ↦*i*but 1^{2}↦ 1 ≠*i*^{2}= -1.

- 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}

*U*_{2}*C*_{1},*U*_{3}*C*_{2},*U*_{4}*C*_{2},*U*_{5}*C*_{4},*U*_{6}*C*_{2},*U*_{7}*C*_{6},*U*_{8}*C*_{2}×*C*_{2},*U*_{9}=*C*_{2}×*C*_{3}*C*_{6},*U*_{10}*C*_{4}, ...

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.

- A ring homomorphism from
**Z**_{12}onto**Z**_{4}will have the ideal < 3 > as kernel 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**Z**_{m}onto**Z**_{n}if*n*divides*m*and then it is the map*x*↦*x**mod**n*.

- 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*)*a*^{2}and so if you know*a*^{2}the multipication is determined.The additive group of a ring of order 3 must be cyclic so

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

*a*, 2*a*, 3*a*} we have a choice of 0,*a*or 2*a*for*a*^{2}. Choosing*a*^{2}= 3*a*gives a ring isomorphic to the*a*^{2}=*a*case.

There are in fact eight rings whose additive group is the Klein 4-group.