# Hom

• Nov 9th 2008, 10:23 AM
roporte
Hom
Prove that:

a) $\displaystyle Hom_{\mathbb{Z}}(\mathbb{Z},\mathbb{Z}_n) \neq 0$
b) $\displaystyle Hom_{\mathbb{Z}}(\mathbb{Z}_5,\mathbb{Z}_7) = 0$
c) $\displaystyle Hom_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z}) = 0$

thanks!!!
• Nov 9th 2008, 10:34 AM
Jhevon
Quote:

Originally Posted by roporte
Prove that:

a) $\displaystyle Hom_{\mathbb{Z}}(\mathbb{Z},\mathbb{Z}_n) \neq 0$
b) $\displaystyle Hom_{\mathbb{Z}}(\mathbb{Z}_5,\mathbb{Z}_7) = 0$
c) $\displaystyle Hom_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z}) = 0$

thanks!!!

what does $\displaystyle \text{Hom}_{\mathbb{Z}} (A,B)$ mean?
• Nov 9th 2008, 09:36 PM
roporte
$\displaystyle Hom_{R}(A,B)$ are all R-module homomorphisms from A to B.

thanks!
• Nov 10th 2008, 06:59 PM
NonCommAlg
Quote:

Originally Posted by roporte
Prove that:

a) $\displaystyle Hom_{\mathbb{Z}}(\mathbb{Z},\mathbb{Z}_n) \neq 0$

the natural $\displaystyle \mathbb{Z}-$homomorphism $\displaystyle f: \mathbb{Z} \longrightarrow \mathbb{Z}_n$ defined by $\displaystyle f(k)=[k]_n$ is obviously nonzero.

Quote:

b) $\displaystyle Hom_{\mathbb{Z}}(\mathbb{Z}_5,\mathbb{Z}_7) = 0$
let $\displaystyle f \in \text{Hom}_{\mathbb{Z}}(\mathbb{Z}_5, \mathbb{Z}_7).$ then: $\displaystyle [0]_7=f([0]_5)=f([5]_5)=5f([1]_5).$ thus: $\displaystyle f([1]_5)=[0]_7.$ why? hence: $\displaystyle f=0.$

Quote:

c) $\displaystyle Hom_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z}) = 0$
let $\displaystyle f \in \text{Hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z}).$ then for any integer n > 0: $\displaystyle nf(1/n)=f(1).$ so all (positive) integers are divisors of the integer f(1), which is possible only if f(1) = 0. therefore: $\displaystyle f=0.$ why?