# Thread: Z[x] -> Z_m[x] defined by ... is a homomorphism

1. ## Z[x] -> Z_m[x] defined by ... is a homomorphism

The question is:

Let m be a fixed positive int. For any integer a, let
$\displaystyle \bar{a}$ denote (a mod m) Show that the mapping of $\displaystyle \Phi: Z[x] \to Z_{m}[x]$ given by [Math]\Phi(a_nx^n + a_{n-1}x^{n-1}+...+a_0) = \bar{a_n}x^n + \bar{a_{n-1}}x^{n-1}+...+\bar{a_0}[/tex]
is a ring homomorphism.

All i want to know is that is it enough to show that $\displaystyle \phi : Z \to Z_m$ defined by $\displaystyle \phi(x) =$ x mod m is a homomorphism ?

2. Originally Posted by jakncoke
The question is:

Let m be a fixed positive int. For any integer a, let
$\displaystyle \bar{a}$ denote (a mod m) Show that the mapping of $\displaystyle \Phi: Z[x] \to Z_{m}[x]$ given by [Math]\Phi(a_nx^n + a_{n-1}x^{n-1}+...+a_0) = \bar{a_n}x^n + \bar{a_{n-1}}x^{n-1}+...+\bar{a_0}[/tex]
is a ring homomorphism.

All i want to know is that is it enough to show that $\displaystyle \phi : Z \to Z_m$ defined by $\displaystyle \phi(x) =$ x mod m is a homomorphism ?

Well, yes, but you've to clarify how this is kept by the polynomial ring's operations.

Tonio