MHB Ring Homomorphisms - Rotman - Theorem 3.33

[Math Amateur]

I am reading Joseph J.Rotman's book, A First Course in Abstract Algebra.

I am currently focused on Section 3.4 Homomorphisms (of Rings)

I need help with the proof of Theorem 3.33 ...

Theorem 3.33 and the start of its proof reads as follows:
https://www.physicsforums.com/attachments/4529

In the above text, we read the following:" ... ... $$ \tilde{ \phi } \ : \ r_0 + r_1 x + \ ... \ ... \ + r_n x^n \longmapsto \phi (r_0) + \phi (r_1)s + \ ... \ ... \ + \phi (r_n) s^n$$This formula shows that $$ \tilde{ \phi } (x) = s$$ and $$\tilde{ \phi } (r) = \phi (r)$$ ... ... "
My question is as follows:

How do we show (formally and rigorously) that $$\tilde{ \phi } (x) = s$$ and $$\tilde{ \phi } (r) = \phi (r)$$ from the above definition of $$\tilde{ \phi }$$ ... ...
*** NOTE ***

We have that

$$x = 0 + 1.x + 0.x^2 + 0.x^3 + \ ... \ ... \ + 0.x^n
$$

So it seems that, from the definition of \tilde{ \phi } that

$$\tilde{ \phi } (x) = \phi (0) + \phi (1)s + \phi (0) s^2 + \phi (0) s^3 + \ ... \ ... \ + \phi (0) s^n $$ ... BUT ... for this to give us $$\tilde{ \phi } (x) = s$$ we would need to show $$\phi (0) = 0$$ ... but why is this true?Hope someone can help ...

PeterPossible Solution to my question ...

Since $$\phi$$ is a homomorphism we have ... ...

$$\phi (0) = \phi (0 + 0) = \phi (0) + \phi (0)$$

Thus $$\phi (0) = 0$$ ...

and then the required results follow ...

Can someone please confirm the above analysis is correct?

Peter

 

[Euge]

Hi Peter,

You are correct; $\varphi(0) = 0$, and so $\tilde{\varphi}(x) = \varphi(1)s = s$. I'm assuming you knew $\varphi(1) = 1$.

 

[Math Amateur]

Hi Peter,

You are correct; $\varphi(0) = 0$, and so $\tilde{\varphi}(x) = \varphi(1)s = s$. I'm assuming you knew $\varphi(1) = 1$.

Thanks Euge ... that gives me some confidence ...

Yes, knew that $\varphi(1) = 1$ ...

Thanks again ...

Peter