Ring Homomorphisms from Z to Z .... Lovett, Ex. 1, Section 5.4 .... ....

[Math Amateur]

I am reading Stephen Lovett's book, "Abstract Algebra: Structures and Applications" and am currently focused on Section 5.4 Ring Homomorphisms ...

I need some help with Exercise 1 of Section 5.4 ... ... ...

Exercise 1 reads as follows:View attachment 6452Relevant Definitions

A ring homomorphism is defined by Lovett as follows:https://www.physicsforums.com/attachments/6453Thoughts so far ... ...

One ring homomorphism, \(\displaystyle f_1 \ : \ \mathbb{Z} \rightarrow \mathbb{Z}\) would be the Zero Homomorphism defined by \(\displaystyle f_1(r) = 0 \ \forall r \in \mathbb{Z}\) ...

(\(\displaystyle f_1\) is clearly a homomorphism ... )

Another ring homomorphism \(\displaystyle f_2 \ : \ \mathbb{Z} \rightarrow \mathbb{Z}\) would be the Identity Homomorphism defined by \(\displaystyle f_2(r) = r \ \forall r \in \mathbb{Z}\) ...

(\(\displaystyle f_2\) is clearly a homomorphism ... )
Now presumably ... ... ? ... ... \(\displaystyle f_1\) and \(\displaystyle f_2\) are the only ring homomorphisms from \(\displaystyle \mathbb{Z} \rightarrow \mathbb{Z}\) ... ... but how do we formally and rigorously show that there are no further homomorphisms ... ...

Hope that someone can help ...

Peter

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Have no tried the following idea ... but no luck ...Suppose \(\displaystyle \exists \ f_3 \ : \ \mathbb{Z} \rightarrow \mathbb{Z}\) ... ... try to show no such \(\displaystyle f_3\) exists ... at least no \(\displaystyle f_3\) that is different from \(\displaystyle f_1 , f_2\) exists ... ...Let \(\displaystyle f_3(2) = x \) where \(\displaystyle x \in \mathbb{Z}\) ...Then \(\displaystyle f_3(4) = f_3( 2 \cdot 2 ) = f_3( 2) f_3( 2) = x^2 \)
and
\(\displaystyle f_3(4) = f_3( 2 + 2) = f_3( 2) + f_3( 2) = x + x = 2x\)
Then we must have \(\displaystyle x^2 = 2x\) ... ... ... (1)
I was hoping that there would be no integer solution to (1) ... but \(\displaystyle x = 0\) satisfies ... so ... problems ..

Maybe a similar approach with different numbers will work ... ...

Can anyone comment on this type of approach ...

Peter

 

[Opalg]

Here's an outline of how I would do it.

First, $f(n+1) = f(n) + f(1)$. Use this to prove by induction that $f(n) = nf(1)$ for all positive $n$.

Next, $-n + n = 0$, so $f(-n) + f(n) = f(0) = 0$. Therefore $f(-n) = -f(n) = -nf(1)$. Thus $f(n) = nf(1)$ for all $n$, positive or negative.

Finally, $1^2 = 1$, so $\bigl(f(1)\bigr)^2 = f(1)$. It follows that $f(1)$ must be either $0$ or $1$. So $f(n)$ must be either $n\cdot0 = 0$ or $n\cdot1 = n$.

 

[Math Amateur]

Here's an outline of how I would do it.

First, $f(n+1) = f(n) + f(1)$. Use this to prove by induction that $f(n) = nf(1)$ for all positive $n$.

Next, $-n + n = 0$, so $f(-n) + f(n) = f(0) = 0$. Therefore $f(-n) = -f(n) = -nf(1)$. Thus $f(n) = nf(1)$ for all $n$, positive or negative.

Finally, $1^2 = 1$, so $\bigl(f(1)\bigr)^2 = f(1)$. It follows that $f(1)$ must be either $0$ or $1$. So $f(n)$ must be either $n\cdot0 = 0$ or $n\cdot1 = n$.

Thanks Opalg ...

Have the idea now ... very grateful for your help!

Peter