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I am trying to find all homomorphisms from $\mathbb{Z}$ to $\mathbb{Z}$. I am looking at a solution, but i do not understand it. Here it is:

Let $\Phi\colon \mathbb{Z}\to\mathbb{Z}$ be a ring homomorphism. Because $1^2 = 1$, we see that $\Phi(1)$ must be an integer whose square is itself, namely either $0$ or $1$. If $\Phi(1) = 1$ then $\Phi(n)= \Phi(n · 1)= n$, so $\Phi$ is the identity map of $\mathbb{Z}$ onto itself which is a homomorphism. If $\Phi(1)= 0$, then $\Phi(n)=\Phi(n · 1)0$, so maps everthing onto $0$, which also yields a homomorphism.

I do not understand here what "Because $1^2= 1$, we see that $\Phi(1)$ must be an integer whose square is itself, namely either $0$ or $1$.". All proof is built upon this and I do not understand what it means. Can someone help me with this?

Thank you

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Since $1=1\cdot 1$, and then by definition of a homomorphism, $\Phi(1)=\Phi(1\cdot 1) = \Phi(1)\cdot \Phi(1)$.

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  • $\begingroup$ Thanks, but why do we start the proof from here? Is it because 1 is the multiplicative identity? $\endgroup$ – Yasin Razlık May 12 '13 at 20:41
  • $\begingroup$ @bigO Yes, the point is that both the multiplicative and additive identies must map to themselves and this forces, as the proof notes, all other $n$ to go to themselves too. Checking what the images of the identities are is a standard thing to do first. $\endgroup$ – Quinn Culver May 12 '13 at 20:43
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    $\begingroup$ Since $1$ generates $\mathbb Z$, $\Phi(1)$ generates the image of $\mathbb Z$. It is often convenient to examine the effect of a homomorphism on a generating set - if you know one. It is one of the most convenient ways of converting an apparently infinite problem into a finite one - and why finitely generated things are often relatively easy to study. In a similar way, for example, a homomorphism of a vector space is defined for every element once its effect is known on a basis. This property of $\mathbb Z$ will undoubtedly come in handy again. $\endgroup$ – Mark Bennet May 12 '13 at 20:52
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Remember the basic properties of a ring homomorphism: $\Phi$ must be such that

$$\Phi(m+n)=\Phi(m)+\Phi(n)$$

and

$$\Phi(mn)=\Phi(m)\Phi(n)$$

for all $m,n\in\Bbb Z$. Since $1\cdot1=1$, we must have

$$\Phi(1)=\Phi(1\cdot1)=\Phi(1)\Phi(1)\;,$$

i.e., $\Phi(1)^2=\Phi(1)$. The only two integers with the property that $n^2=n$ are $0$ and $1$, so $\Phi(1)$ must be one of them.

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By additivity you must have $\phi(n)\phi(1)= \phi(n)=n \phi(1)$ thus if $\phi(1) \neq0$ you get $\phi(n)=n$ and so $\phi$ is the identity on $\mathbb{Z}$, otherwise $\phi=0$.

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  • $\begingroup$ Thank you, but i am confused how do we know that phi(n)=nphi(1)? $\endgroup$ – Yasin Razlık May 12 '13 at 20:57
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    $\begingroup$ $\phi(n)=\phi(1+...+1)$; written in this way should be clear, but it's mathematically freakish.. A correct argument is to proceed by induction on $n$ : $\phi(1 \cdot 1)= \phi(1)= 1\cdot \phi(1)$ and $ \phi(n)= \phi(n-1 +1)=\phi(n-1) + \phi(1)= (n-1)\phi(1) + \phi(1)=n \cdot \phi(1)$ $\endgroup$ – Edoardo Lanari May 12 '13 at 21:11

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