Questions about the function $f:\Bbb Z_{8}\rightarrow \Bbb Z_4$

Question

I have the function $f:\Bbb Z_{8}\rightarrow \Bbb Z_4$ without any particular expression associated.

• How many surjective functions $f:\Bbb Z_{8}\rightarrow \Bbb Z_4$?
• How many of them are a homomorphism of rings? What are they?

For the first question, I reasoned as follows. The elements of $\Bbb Z_8 $ are 0,1,2,3,4,5,6,7 and the elements of $\Bbb Z_4$ are 0,1,2,3,4. So if I calculate the preimage of each element in $\Bbb Z_4$ I obtain two elements of codomain for each. For example for $[0]_4$ I have $[0]_8, [4]_8$. Is it a good approach?

For the second question, I'm a bit confused. I need to provide and count how many possible expressions are there such that the function is a ring homomorphism?

Answer 1

There exist two Ring Homomorphisms which are: f(a) = 0 (the trivial map) and f(a) = a (Identity map)

I hope this is helpful, let me know if you need proper explaination to the approach.

Edit : We use 2 properties of homomorphism: P1)Homomorphic image of idempotent is idempotent. P2)Order of 'f(a)' divides order of 'a'. Basically, we try to find images of the idempotents in Z8. Aim: Find all possible images of '1' (an idempotent in Z8). Collect all the idempotents in Z4 i.e. 0 & 1. So, these can be the only "possible" images of 1 in Z4(by P1). Now we find their orders in Z4 which are 1 & 4 respectively. By P2 order of f(1) must divide order of 1(in Z8) which is 8. Choose all those idempotents in Z4 whose order divides 8.Here, 0 & 1both satisfy the required condition.Therefore 1->0,1 & so a->0,a (ofc under operation mod4)

Answer 2

The first question is purely a combinatorial one. But the preimage of each element in $\mathbb{Z}_4$ does not necessarily have exactly two elements. Think of the first question in the following way: You want to assign each element in $\mathbb{Z}_8$ an element in $\mathbb{Z}_4$, so that all elements in $\mathbb{Z}_4$ are taken care of. You have four boxes numbered 0 through 3, and you have 8 balls numbered 0 through 7. The number of different ways to put these balls into the boxes such that all boxes are nonempty is actually the number of surjective functions from $\mathbb{Z}_8$ to $\mathbb{Z}_4$.

The formulation of this model is not a trivial one. You may look it up here, where it is referred to as theorem 5.

Regarding the second question, remember that a ring homomorphism is at first place a group homomorphism. $\mathbb{Z}_8$ has one generator as an abelian group, i.e., 1. The group homomorphism is decided by where 1 is sent. Now note that being a ring homomorphism requires 1 to be sent to 1. Hence we have at most one ring homomorphism which sends 0 to 0 and 1 to 1. Now check if this function respects the ring structure. We see it does. Thus there is one ring homomorphism.

Hope this addresses your question.