Proof of the First Isomorphism Theorem for Rings The statement is the First Isomorphism Theorem for Rings from Abstract Algebra by Dummit and Foote. I'd like to check if all is ok. In particular I'm a bit worried about the (*) line. It looks a bit awkward, but is it too bad? Any suggestions?
Theorem:
1) If $\varphi:R \rightarrow S$ is a homomorphism of rings, then the kernel of $\varphi$ is an ideal of $R$, the image of $\varphi$ is a subring of $S$ and $R/ker \varphi$ is isomorphic as a ring to $\varphi(R)$. 
2) If $I$ is any ideal of $R$, then the map $R \rightarrow R/I$ defined by $r \rightarrow r+I$ is a surjective ring homomorphism with kernel $I$. Thus every ideal is the kernel of a ring homomorphism and vice versa.
Proof: Let $\varphi: R \rightarrow S$ be a ring homomorfism. If $r \in R$ and $r' \in ker \varphi$, then we have $rr',r'r \in ker \varphi$ (so that it is closed under multiplication by elements of $R$) since $$\varphi (rr')=\varphi (r) \varphi (r')=\varphi (r)0 =0=0 \varphi (r)=\varphi (r') \varphi (r)=\varphi (r'r);$$ since $ker \varphi$ is also a subring of $R$, it is an ideal of $R$. It's clear that $\varphi (R)$ is a subring of $S$. Now, let $I$ be an ideal of $R$, so that $R/I$ is also a ring, and define $\pi:R \rightarrow R/I$ by $\pi (r)=r+I$. We know $\pi$ is a group homomorphism with kernel $I$, and for $r,s \in R$, we have $$\pi (rs)=(rs)+I=(r+I)(s+I)=\pi (r) \pi (s),$$ so that $\pi$ is in fact a ring homomorphism. Define then $\phi: R/ker \varphi \rightarrow \varphi(R)$ by $$\phi (r+(ker \phi))= \varphi (r),$$ for each $(r+(ker \phi)) \in R/ker \varphi$, for some $r \in R$. This is well defined because  if $r' \in (r+(ker \varphi)),$ then $$\phi (r'+(ker \varphi))=\varphi(r')=\varphi(r)=\phi (r+(ker \varphi)).$$ Also, this is a ring isomorphism because for each $\varphi (s) \in \varphi (R)$ for some $s \in R$, we have $$(*) \ \phi^{-1}\{\varphi (s)\}= \phi^{-1} \varphi[r+(ker \varphi)]= \phi^{-1} \varphi[\pi^{-1} \{r+(ker \varphi)\}]= \{r+(ker \varphi)\},$$ a set with a single element of $R/ker \varphi$, so that it is a bijection, and for every $r+(ker \varphi), r'+(ker \varphi) \in R/ker \varphi$, for some $r,r' \in R$, we have $$\phi [(r+(ker \varphi))+(r'+(ker \varphi))]=\phi [(r+r')+(ker \varphi)]=\varphi(r+r')=\varphi(r)+ \varphi(r')=\phi [r+(ker \varphi)]+ \phi [r'+(ker \varphi)],$$ and $$\phi [(r+(ker \varphi)) (r'+(ker \varphi))]=\phi [(rr')+(ker \varphi)]=\varphi(rr')=\varphi(r) \varphi(r')=\phi [r+(ker \varphi)] \phi [r'+(ker \varphi)],$$ so that it is a ring homomorphism.
 A: Basically, I think you know what you are proving. But it's not really "straight to the point", and not very clear.


*

*For example, when you are proving the well-defineness of $\phi$, you haven't stated explicitly why $ \dots = \phi(r') = \phi(r) = \dots$. So, I'll fix that for you. To prove that $\phi$ is well-defined, you must prove that $\phi$ maps 1 element of $R/\text{ker}\varphi$ to 1 and only 1 element of $\varphi(R)$. So, let $r' + \text{ker}\varphi = r + \text{ker}\varphi$, we'll prove that $\phi(r' + \text{ker}\varphi) = \phi(r + \text{ker}\varphi)$. Since we have $r' + \text{ker}\varphi = r + \text{ker}\varphi$, that means $r' - r \in \text{ker}\varphi$, so $\varphi(r' - r) = 0$, hence $\varphi(r') = \varphi(r)$, and finally, this yields $\phi(r' + \text{ker}\varphi) = \phi(r + \text{ker}\varphi)$

*It's obviously that $\phi$ is a ring homomorphism.

*$\phi$ is injective, say, we have: $r + \text{ker}\varphi \in \ker \phi$, i.e $0 = \phi(r + \text{ker}\varphi) = \varphi(r)$, which means $r \in \text{ker}\varphi$, which then implies $r + \text{ker}\varphi = 0 + \text{ker}\varphi$. So $\ker \phi = \{0 + \ker \varphi\}$.

*$\phi$ is surjective, since for every $y \in \varphi(R)$, there exists $r \in R$, such that $y = \varphi(r) = \phi(r + \ker \varphi)$.
