Question

How can we determine all ring homomorphisms from Z12 to Z30?

Answer 1

Let ϕ:Z12→Z30 be an homomorphism. We know that ϕ is fully determined by the value of ϕ(1) , because

ϕ(n)=n⋅ϕ(1)∀n∈Z12. 

Taking n=30 , we get that

ϕ(30)=30⋅ϕ(1)=0. 

But, on the other hand, we have that

ϕ(30)=ϕ(6)=6⋅ϕ(1). 

From 6⋅ϕ(1)=0 we conclude that ϕ(1) must be a multiple of 5 . So the candidate functions are of the form x↦5k⋅x , where k∈{0,1,2,3,4,5} . Any such function is well defined, because

ϕ(n+12t)=5k⋅(n+12t)=5kn+60kt=5kn .

It is also easy to show that these functions preserve the addition operation. We need to decide which of them also preserve multiplication. Note that

5k=ϕ(1)=ϕ(1⋅1)=ϕ(1)ϕ(1)=25k2. 

So 25k2−5k is a multiple of 30 , and 5k2−k is a multiple of 6 . The values of k for which this is true are k∈{0,2,3,5} . For any of these values, we have that

ϕ(a)ϕ(b)−ϕ(ab)=25k2ab−5kab=5ab(5k2−k) 

is a multiple of 30 , as we wanted.

So we have 4 homomorphisms.

It is also easy to show that these functions preserve the addition operation. We need to decide which of them also preserve multiplication. Note that

5k=ϕ(1)=ϕ(1⋅1)=ϕ(1)ϕ(1)=25k2. 

So 25k2−5k is a multiple of 30 , and 5k2−k is a multiple of 6 . The values of k for which this is true are k∈{0,2,3,5} . For any of these values, we have that

 

ϕ(a)ϕ(b)−ϕ(ab)=25k2ab−5kab=5ab(5k2−k) 

 

is a multiple of 30 , as we wanted.

 we have 4 homomorphisms:

x↦0,x↦10x,x↦15x,x↦25x.