TITLE: Is $\varphi$ a homomorphism?
QUESTION [5 upvotes]: Define $\varphi:\mathbb{Z}\times \mathbb{Z}\rightarrow \mathbb{Z}_6$ by $\varphi(a,b)=[a+b]_6$. 

To show $\varphi$ preserves addition, 
$\hspace{100pt} \varphi((a,b)+(c,d))=\varphi((a,b))+\varphi((c,d))$.
$\varphi((a,b)+(c,d))=\varphi((a+c),(b+d))=[a+c]_6+[b+d]_6$
$\hspace{73pt}$$=[a+b]_6+[c+d]_6=\varphi((a,b))+\varphi((c,d))$
$\\$
To show $\varphi$ preserves multiplication, 
$\hspace{100pt} \varphi((a,b)\cdot(c,d))=\varphi((a,b))\cdot\varphi((c,d))$.
$\varphi((a,b)\cdot(c,d))=\varphi(ac,bd)=[ac]_6+[bd]_6$
$\varphi((a,b))\cdot \varphi((c,d))=[a+b]_6\cdot[c+d]_6=[(a+b)\cdot(c+d)]_6=[ac+ad+bc+bd]_6$
$\\$

From what I can tell, multiplication is not preserved. However, my book says that $\varphi$ is a ring homomorphism, so either multiplication must be preserved or the book is wrong. Am I missing something?

REPLY [3 votes]: You are right, this $\phi$ is not a ring homomorphism, only a homomorphism between the underlying ableian groups. Consider for example $(1,0)$ and $(0,1)$, then $(1,0)\cdot(0,1)=(0,0)$ so
$$\phi((1,0)\cdot(0,1))=[0]_6\ \ne \  [1]_6=\phi(1,0)\cdot\phi(0,1)\,.$$