MHB Initial Objects in the Category Ring

[Math Amateur]

I am reading Paolo Aluffi's book, Algebra: Chapter 0.

I have a question related to Aluffi's description of initial objects in the category ring ... ...

In Chapter III, Section 2. on the category ring, we read the following:View attachment 4480
View attachment 4481QUESTION 1

In the above text, we read the following:

" ... ... This ring homomorphism is unique, since it is determined by the requirement that $$\phi (1) = 1_R$$ and by the fact that $$\phi$$ preserves addition ... ... "Can someone please explain to me (precisely, rigorously and formally) why the requirement that $$\phi (1) = 1_R$$ and the fact that $$\phi$$ preserves addition imply that ring homomorphism is unique?

(Intuitively the above seems true ... but how do you show this exactly and precisely ... wonder if I am overthinking this matter ... )QUESTION 2

In the above text, we read the following:

" ... ... But $$\phi$$ is in fact a ring homomorphism, since $$\phi (1) = 1_R$$, and

$$\phi (mn) = (mn) 1_R = m(n1_R) = (m1_R) \cdot (n 1_R) = \phi (m) \cdot \phi (n)
$$

where the equality $$m(n1_R) = (m1_R) \cdot (n 1_R)$$ holds by the distributivity axiom ... ... "Can someone explain how the equality $$m(n1_R) = (m1_R) \cdot (n 1_R)$$ follows from the distributivity axiom

Help will be much appreciated ... ...

Peter

 

[Euge]

QUESTION 1

In the above text, we read the following:

" ... ... This ring homomorphism is unique, since it is determined by the requirement that $$\phi (1) = 1_R$$ and by the fact that $$\phi$$ preserves addition ... ... "Can someone please explain to me (precisely, rigorously and formally) why the requirement that $$\phi (1) = 1_R$$ and the fact that $$\phi$$ preserves addition imply that ring homomorphism is unique?

(Intuitively the above seems true ... but how do you show this exactly and precisely ... wonder if I am overthinking this matter ... )

Let $f : \Bbb Z \to R$ be a ring homomoprhism. Define $a := f(1)$. Since $f$ is a $\Bbb Z$-homomorphism, $f(n) = f(n\cdot 1) = nf(1) = na$ for all $n\in \Bbb Z$. Choosing $a = 1_R$ forces $f(n) = \phi(n)$ for all $n \in \Bbb Z$; due to this choice, $f = \phi$.

QUESTION 2

In the above text, we read the following:

" ... ... But $$\phi$$ is in fact a ring homomorphism, since $$\phi (1) = 1_R$$, and

$$\phi (mn) = (mn) 1_R = m(n1_R) = (m1_R) \cdot (n 1_R) = \phi (m) \cdot \phi (n)
$$

where the equality $$m(n1_R) = (m1_R) \cdot (n 1_R)$$ holds by the distributivity axiom ... ... "Can someone explain how the equality $$m(n1_R) = (m1_R) \cdot (n 1_R)$$ follows from the distributivity axiom

They key here is to use induction on $m$ or $n$. ;)

 

[Euge]

Let $f : \Bbb Z \to R$ be a ring homomoprhism. Define $a := f(1)$. Since $f$ is a $\Bbb Z$-homomorphism, $f(n) = f(n\cdot 1) = nf(1) = na$ for all $n\in \Bbb Z$. Choosing $a = 1_R$ forces $f(n) = \phi(n)$ for all $n \in \Bbb Z$; due to this choice, $f = \phi$.

Let me just note that if you're not allowed to assume that a ring homomorphism is a $\Bbb Z$-homomorphism, then use induction to show that $f(n) = nf(1)$ for all $n\in \Bbb Z$.

 

[Math Amateur]

Let $f : \Bbb Z \to R$ be a ring homomoprhism. Define $a := f(1)$. Since $f$ is a $\Bbb Z$-homomorphism, $f(n) = f(n\cdot 1) = nf(1) = na$ for all $n\in \Bbb Z$. Choosing $a = 1_R$ forces $f(n) = \phi(n)$ for all $n \in \Bbb Z$; due to this choice, $f = \phi$.They key here is to use induction on $m$ or $n$. ;)

Thanks for the help Euge ...

Two clarifications ...

1. What is a $\Bbb Z$-homomorphism?

2. how do we justify the statement $$f(n\cdot 1) = nf(1)$$?

Hope you can help further ...

Thanks again ...

Peter

 

[Euge]

A $\Bbb Z$-homomorphism is a $\Bbb Z$-module homomorphism. I already gave the answer to 2. -- take a look back at my last post. [emoji2]