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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, $\varphi(-n) = -\varphi(n) = -n$

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    Dear Dimitri, I'd say that the answer is no because $\mathbb Z$ itself is characterized by a property which involves induction. So, you may be able to hide induction, but not to remove it.2011-08-14

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If you are working in the category of unitary associative rings, the morphisms $\varphi : R\to S$ must satisfy $\varphi (1)=1$. In this category the ring $\mathbb{Z}$ is an initial object, that is, for any ring $R$ there is exactly one morphism (i.e. ring homomorphism) $\chi : \mathbb{Z}\to R$ (which defines the characteristic of the ring $R$). In particular there is exactly one ring homomorphism $\mathbb{Z}\to\mathbb{Z}$, which is the identity map.

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    OK, I forgot to comment that @Dimitri 's proof by induction works well to prove that $\mathbb{Z}$ is an initial object. The statement to prove is: for each $n \in\mathbb{Z}$, $\varphi (n)=n1_R$ ($n$ times the identity of $R$).2012-02-01
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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.