Question about all the homomorphisms from $\mathbb{Z}$ to $\mathbb{Z}$ An exercise in "A first course in Abstract Algebra" asked the following:
Describe all ring homomorphisms from the ring $\mathbb{Z},+,\cdot$ to itself.
I observed that for any such ring homomorphism the following has to hold: $$\varphi(1) = \varphi(1\cdot 1) = \varphi(1) \cdot \varphi(1)$$
In $\mathbb{Z}$ only two numbers exists so that their square equals itself: 0 and 1.
When $\varphi(1) = 0$ then $\varphi = 0$ hence $\forall n \in \mathbb{Z}$: $\varphi(n) = \varphi(n \cdot 1) = \varphi(n) \cdot \varphi(1) = \varphi(n) \cdot 0 = 0$.
Now, when $\varphi(1) = 1$ I showed that  $\varphi(n) = n$ using induction
Base case: $n = 1$, which is true by our assumption
Induction hypothesis: $\varphi(m) = m$ for $m < n$
Induction step: $\varphi(n) = \varphi((n-1) + 1) = \varphi(n-1) + \varphi(1) = n-1 + 1 = n$
Now I wonder whether you could show that $\varphi(n) = n$ when $\varphi(1) = 1$ without using induction, which seems overkill for this exercise.
EDIT:
Forgot about the negative n's.
Since $\varphi$ is also a group homomorphism under $\mathbb{Z},+$, we know that $\varphi(-n) = -\varphi(n)$. Thus,
 $$\varphi(-n) = -\varphi(n) = -n$$

I assume you are talking about Fraleigh's book. If so, he does not require that a ring homomorphism maps the multiplicative identity to itself. Follow his hint by concentrating on the possible values for $f(1)$. If $f$ is a (group) homomorphism for the group $(\mathbb{Z},+)$ and $f(1)=a$, then $f$ will reduce to multiplication by $a$. For what values of $a$ will you get a ring homomorphism? You will need to have $(mn)a=(ma)(na)$ for all pairs $(m,n)$ of integers. What can you conclude about the value of $a$? You still won't have a lot of homomorphisms.
