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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    $\begingroup$ 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. $\endgroup$ Aug 13, 2011 at 19:23
  • $\begingroup$ oh you are right I forgot about -n $\endgroup$
    – sxd
    Aug 13, 2011 at 19:25
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    $\begingroup$ 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. $\endgroup$ Aug 13, 2011 at 19:27
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    $\begingroup$ @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? $\endgroup$ Aug 13, 2011 at 20:17
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    $\begingroup$ 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. $\endgroup$ Aug 14, 2011 at 3:39

2 Answers 2


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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    $\begingroup$ Sorry, -1: I think this a) Clearly not a suitable response for this asker, and b) Essentially hiding all of the essential mathematics behind a veil of language. (How do you prove that $\mathbb{Z}$ is an initial object?) $\endgroup$ Feb 1, 2012 at 16:03
  • $\begingroup$ 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 $R$). $\endgroup$
    – Loronegro
    Feb 1, 2012 at 18:20

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.


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