I want to prove this theorem. I have stuck in proving this theorem, in homomorphism axiom.
Let $R$ be integral domain and $x\neq 0\in R$. Prove that if $x$ is unit then $R\cong Rx$ (as rings).
Proof.
Since $x$ is unit, there exist $y\in R$ such that $xy=1_R$ or $yx=1_R$. Let we construct the maps $$ \begin{array}{l@{$\;$}c@{$\;$}c@{$\;$}c} f\colon & R &\to& Rx\\ & r &\mapsto &rx. \end{array} $$
Take $a,b\in R$. We want to show $f$ is a ring homomorphism. Consider that \begin{align*} f(a+b)&=(a+b)x\\ &=ax+bx\\ &= f(a)+f(b) \end{align*} and \begin{align*} f(ab)&=(ab)x\\ \end{align*} (I'm stuck here, because I can't associate $x$ as unit with the axiom of homomorphism). How to prove it?
Now we want to prove $f$ is onto and 1-1 maps. For all $rx\in Rx$, there exist $r\in R$ such that $f(r)=rx$. So, $f$ is onto. Let $r_1,r_2\in R$. If $f(r_1)=f(r_2)$ then $r_1 x= r_2 x$. Since $R$ is integral domain, imply $r_1=r_2$. Thus, $f$ 1-1 maps.
Does my proof for $f$ is onto and 1-1 maps is right?
So, we (hope) can conclude $R\cong Rx$ (as rings).
The problem is that the map $r \mapsto xr$ is not a homomorphism. This is because $x$ is a unit, but not $1_R$.
Still, $R \cong Rx$, but that's because actually $R = Rx$.