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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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    I would say that you need to do the induction at least once while learning. As you said, it is relatively straightforward, so you can skip it the next time it shows up (at least while self-learning). But you also need to check the negative numbers, and another trick is needed for that (induction takes care of the positive side only)! Also many (but not all) algebraists require that a ring homomorphism *must* map 1 to 1.2011-08-13
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    oh you are right I forgot about -n2011-08-13
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    Authors often require $1$ to map to $1$ under a ring homomorphism, so check your text (the zero ring has $0 = 1$, so there's no issue there). As for the induction, I think it's pretty rare for induction to be overkill. Most people would write $\varphi(n) = \varphi(\sum_{i = 1}^n 1) = \sum_{i = 1}^n \varphi(1)$, but at some point in the logical chain there is induction, even in these obvious equalities.2011-08-13
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    Induction is never "overkill". It's a basic instrument in math.2011-08-13
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    @Jyrki I added the part for the negative numbers now!:)2011-08-13
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    @Dimitri: The part about negative numbers looks good to me. You may also consider typing your completed argument as an answer. If you are happy with your understanding of this problem, you can accept that answer. The purpose of that exercise is twofold: 1) some of us may want to give you upvotes, and (more importantly) 2) the question won't be lingering in the ranks of unanswered ones (helping the system a bit). I don't know the proper protocol for this. May be it is polite to wait a little while in case somebody wants to add something that we may have overlooked before accepting?2011-08-13
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    @Jyrki The initial question was not whether my proof was correct but whether this could be done without using induction, so i'll wait until someone comes up with an answer about that.2011-08-13
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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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