# Show that the mod p map is a ring homomorphism

Let p be a prime and let (mod $p$)$: Z[x] \mapsto Z_p[x]$ be the mod-p map which sends any polynomial...

$f(x) = a_0 + a_1x + a_2x^2 + · · · + a_nx^n \in Z[x]$

... to the polynomial...

$f(x)$(mod $p$) $= a_0 + a_1x + a_2x^2 + · · · + a_nx^n$ where $a_i := a_i$ (mod $p$).

I must show whether or not the map is a ring homomorphism, and whether or not it's onto.

My first instinct is to create another function $g(x)= b_0 + b_1x +b_2x^2 + ... + b_nx^n$ to check if the map is closed under addition and multiplication. Is this a correct approach?

• Do you know about quotient rings, i.e. how to mod out by ideals? Feb 27, 2014 at 15:47
• I don't believe so
– Alex
Feb 27, 2014 at 16:00

Hint Define $\pi_p: \mathbb{Z}[x] \rightarrow \mathbb{Z}_p[x]$ to be the mod map you describe. It suffices to show that $\pi_p(f+g) = \pi_p(f)+\pi_p(g)$ and $\pi_p(fg) = \pi_p(f)\pi_p(g)$ to prove that $\pi_p$ is a ring homomorphism. This can be done more easily if you know that the mod map taking $\mathbb{Z} \rightarrow \mathbb{Z}_p$ is itself a ring homomorphism.