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 \(\displaystyle \phi (1) = 1_R\) and by the fact that \(\displaystyle \phi\) preserves addition ... ... "Can someone please explain to me (precisely, rigorously and formally) why the requirement that \(\displaystyle \phi (1) = 1_R\) and the fact that \(\displaystyle \phi\) 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 \(\displaystyle \phi\) is in fact a ring homomorphism, since \(\displaystyle \phi (1) = 1_R\), and

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

where the equality \(\displaystyle m(n1_R) = (m1_R) \cdot (n 1_R)\) holds by the distributivity axiom ... ... "Can someone explain how the equality \(\displaystyle m(n1_R) = (m1_R) \cdot (n 1_R)\) 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 \(\displaystyle \phi (1) = 1_R\) and by the fact that \(\displaystyle \phi\) preserves addition ... ... "Can someone please explain to me (precisely, rigorously and formally) why the requirement that \(\displaystyle \phi (1) = 1_R\) and the fact that \(\displaystyle \phi\) 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 \(\displaystyle \phi\) is in fact a ring homomorphism, since \(\displaystyle \phi (1) = 1_R\), and

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

where the equality \(\displaystyle m(n1_R) = (m1_R) \cdot (n 1_R)\) holds by the distributivity axiom ... ... "Can someone explain how the equality \(\displaystyle m(n1_R) = (m1_R) \cdot (n 1_R)\) 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 \(\displaystyle f(n\cdot 1) = nf(1)\)?

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]