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

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In summary, Peter is trying to solve an equation that has the form $f(n) = nf(1)$, but is having trouble.
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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

Last edited:
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 $n\cdot1 = n$.

Opalg said:
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

## 1. What is a ring homomorphism?

A ring homomorphism is a function that preserves the algebraic structure of a ring. In other words, it is a function between two rings that respects the operations of addition and multiplication.

## 2. What are the two rings involved in a ring homomorphism from Z to Z?

The two rings involved are the ring of integers (Z) and another ring, which can also be Z in this specific case. The ring homomorphism maps elements from the first ring (Z) to elements in the second ring (Z).

## 3. How is a ring homomorphism from Z to Z defined?

A ring homomorphism from Z to Z is defined as a function f: Z -> Z, where f(a + b) = f(a) + f(b) and f(ab) = f(a)f(b) for all elements a and b in Z.

## 4. Can you provide an example of a ring homomorphism from Z to Z?

One example of a ring homomorphism from Z to Z is the identity function, which maps every element in Z to itself. This function preserves the operations of addition and multiplication, making it a homomorphism.

## 5. What is the significance of ring homomorphisms in mathematics?

Ring homomorphisms are important in abstract algebra and other areas of mathematics because they allow us to study the properties of rings by looking at their structure and how they relate to other rings. They also help us understand the behavior of algebraic operations and their connections to other mathematical concepts.

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