(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Exhibit two examples of a ring homomorphism [itex]\phi[/itex] from Z4 to Z8, one that is one-to-one and another that is not. For each case, find ker([itex]\phi[/itex]) and describe Z4/ker([itex]\phi[/itex])

2. The attempt at a solution

Let [itex]\phi : \mathbb{Z}_4 \longrightarrow \mathbb{Z}_8[/itex] be the identity mapping such that a = a mod 4 for [itex]a \in \mathbb{Z}_4[/itex] Clearly, this is a ring homomorphism since for any a,b in Z4,

[itex]\phi (a+b)=(a+b) mod 4=a mod 4+b mod 4=\phi(a)+\phi(b)[/itex]

and

[itex]\phi(ab)=ab mod 4=a mod 4 \cdot b mod 4 = \phi(a) \cdot \phi(b)[/itex]

This homomorphism is injective (or one-to-one) as every element in Z4 is carried to exactly one element in Z8, but not surjective as the range Z4 a proper subset of the codomain Z8.

The kernel of [itex]\phi[/itex] is defined as the elements of Z4 that are mapped by [itex]\phi[/itex] to the zero element in Z8, so ker [itex]\phi[/itex] is the trivial ring {0} and Z4/ker [itex]\phi[/itex] is the quotient ring formed by modding out the trivial ring.

Let us define another homomorphism [itex]\varphi : \mathbb{Z}_4 \longrightarrow \mathbb{Z}_8[/itex] by [itex]\varphi(a)=(8a)[/itex] mod 8 for [itex]a \in \mathbb{Z}_4[/itex]. Now,

[itex]\varphi(a+b)=(8a+8b) mod 8=8a mod 8+8b mod 8=\phi(a)+\phi(b)[/itex]

Also,

[itex]\varphi(ab)=(64ab) mod 8=0=0 \cdot 0=(8a mod 8)(8b mod 8)[/itex]

Observe that all of the elements in Z4 map to the zero element in Z8 so the kernel of [itex]\varphi[/itex] is simply the whole of Z4 and Z4/ker [itex]\varphi[/itex] is simply the trivial ring {0}.

My first homomorphism seems to be wrong... a mod 4 * b mod 4 can be, say 4, 6 or 9 while ab mod 4 cannot when a,b belong to the subset {2,3}. What should I do to rectify this? Also, I cannot think of a homomorphism between these two that isn't injective. Any suggestions? Thanks! :D

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# Homework Help: Ring homomorphism from Z4 to Z8

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