Note: In this class, a ring homomorphism must map multiplicative and additive identities to multiplicative and additive identities. This is different from our textbook's requirement, and often means there are fewer situations to consider.
I always have a pretty hard time answering these types of questions:
Let $\phi: \mathbb{Z~ \times ~Z} \rightarrow \mathbb{Z~ \times ~Z}$ be a ring homomorphism. We know, then, by definition of a ring homomorphism, that $\phi(1,1) = (1,1)$ (because $(1,1)$ is the multiplicative identity of $\mathbb{Z~ \times ~Z}$). Any ring homomorphism must then have the form $\phi(a,b) = (a,b)$ or $\phi(a,b) = (b,a)$. Any addition/multiplication to elements would cease to send $(1,1)$ to $(1,1)$.
Is… this correct? It seems too simple, but I'm pretty sure it covers the possibilities.
Best Answer
You know that $(1,0) \cdot (0,1)=(0,0)$ then $f(1,0) \cdot f(0,1)=f(0,0)=0$. This implies that $f(1,0)=(a,0)$ or $(0,a)$, for some $a\in\mathbb{Z}$ and similarly for $f(0,1)=(b,0)$ or $(0,b)$. Since $f(1,1)=(1,1)$ you get that $a=b=1$ and that only two of the four options are valid. In this way you get exactly your two functions.